OP Malhotra Class 10 Maths Solutions Chapter 7 Factor Theorem Factorization Exercise 7

S Chand Class 10 ICSE Maths Solutions Chapter 7 Factor Theorem - Factorization Ex 7

Question 1. Find the remainder when the expression
(i) \( 3x^3 + 8x^2 - 6x + 1 \) is divided by \( x + 3 \).
(ii) \( 5x^3 - 8x^2 + 3x - 4 \) is divided by \( x - 1 \).
(iii) \( x^3 + 3x^2 - 1 \) is divided by \( 3x + 2 \).
(iv) \( 4x^3 - 12x^2 + 14x - 3 \) is divided by \( 2x - 1 \).
Answer:
(i) Let \( f(x) = 3x^3 + 8x^2 - 6x + 1 \). When \( x + 3 = 0 \), we get \( x = -3 \).
So, the remainder is \( f(-3) \). We substitute \( x = -3 \) into the expression:
\( f(-3) = 3(-3)^3 + 8(-3)^2 - 6(-3) + 1 \)
\( = 3(-27) + 8(9) + 18 + 1 \)
\( = -81 + 72 + 18 + 1 \)
\( = -81 + 91 \)
\( = 10 \)
(ii) Let \( f(x) = 5x^3 - 8x^2 + 3x - 4 \). When \( x - 1 = 0 \), we get \( x = 1 \).
So, the remainder is \( f(1) \). We substitute \( x = 1 \) into the expression:
\( f(1) = 5(1)^3 - 8(1)^2 + 3(1) - 4 \)
\( = 5 - 8 + 3 - 4 \)
\( = 8 - 12 \)
\( = -4 \)
(iii) Let \( f(x) = x^3 + 3x^2 - 1 \). When \( 3x + 2 = 0 \), we get \( 3x = -2 \), which means \( x = \frac{-2}{3} \).
So, the remainder is \( f(\frac{-2}{3}) \). We substitute \( x = \frac{-2}{3} \) into the expression:
\( f\left(\frac{-2}{3}\right) = \left(\frac{-2}{3}\right)^3 + 3\left(\frac{-2}{3}\right)^2 - 1 \)
\( = \frac{-8}{27} + 3\left(\frac{4}{9}\right) - 1 \)
\( = \frac{-8}{27} + \frac{12}{9} - 1 \)
To combine these, we find a common denominator, which is 27.
\( = \frac{-8}{27} + \frac{4 \times 9}{3 \times 9} - \frac{1 \times 27}{1 \times 27} \)
\( = \frac{-8 + 36 - 27}{27} \)
\( = \frac{36 - 35}{27} \)
\( = \frac{1}{27} \)
(iv) Let \( f(x) = 4x^3 - 12x^2 + 14x - 3 \). When \( 2x - 1 = 0 \), we get \( 2x = 1 \), which means \( x = \frac{1}{2} \).
So, the remainder is \( f(\frac{1}{2}) \). We substitute \( x = \frac{1}{2} \) into the expression:
\( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3 \)
\( = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 14\left(\frac{1}{2}\right) - 3 \)
\( = \frac{4}{8} - \frac{12}{4} + \frac{14}{2} - 3 \)
\( = \frac{1}{2} - 3 + 7 - 3 \)
\( = \frac{1}{2} + 1 \)
\( = 1\frac{1}{2} = \frac{3}{2} \)
In simple words: To find the remainder when a polynomial is divided by a linear expression, we use the Remainder Theorem. This means we set the divisor to zero to find the value of x, then substitute this x-value into the polynomial. The result is the remainder. This method is very efficient for these types of division problems.

🎯 Exam Tip: When using the Remainder Theorem, make sure to correctly solve for x in the divisor (e.g., \( x+3=0 \implies x=-3 \)) and substitute it carefully, especially with negative numbers and fractions to avoid calculation errors.

Question 2. When \( x^3 + 3x^2 - kx + 4 \) is divided by \( x - 2 \), the remainder is k. Find the value of the constant k.
Answer:
Let the polynomial be \( f(x) = x^3 + 3x^2 - kx + 4 \).
We are told that when \( f(x) \) is divided by \( x - 2 \), the remainder is \( k \).
According to the Remainder Theorem, if \( x - 2 = 0 \), then \( x = 2 \). The remainder is \( f(2) \).
So, \( f(2) = k \).
Substitute \( x = 2 \) into the polynomial:
\( f(2) = (2)^3 + 3(2)^2 - k(2) + 4 \)
\( = 8 + 3(4) - 2k + 4 \)
\( = 8 + 12 - 2k + 4 \)
\( = 24 - 2k \)
Since the remainder is \( k \), we set this expression equal to \( k \):
\( 24 - 2k = k \)
Now, we solve for \( k \):
\( 24 = k + 2k \)
\( 24 = 3k \)
\( k = \frac{24}{3} \)
\( k = 8 \)
In simple words: We are given a polynomial and told the remainder when it's divided by \( x-2 \) is \( k \). We find the remainder by putting \( x=2 \) into the polynomial. Then, we set this result equal to \( k \) and solve the equation to find the exact number for \( k \).

🎯 Exam Tip: When the remainder itself is given as a variable (like k), remember to set the expression for the remainder equal to that variable and solve the resulting equation carefully.

Question 3. Find the value of a if the division of \( ax^3 + 9x^2 + 4x - 10 \) by \( x + 3 \) leaves a remainder 5.
Answer:
Let the polynomial be \( f(x) = ax^3 + 9x^2 + 4x - 10 \).
We are given that when \( f(x) \) is divided by \( x + 3 \), the remainder is 5.
Using the Remainder Theorem, if \( x + 3 = 0 \), then \( x = -3 \). The remainder is \( f(-3) \).
So, \( f(-3) = 5 \).
Substitute \( x = -3 \) into the polynomial:
\( f(-3) = a(-3)^3 + 9(-3)^2 + 4(-3) - 10 \)
\( = a(-27) + 9(9) - 12 - 10 \)
\( = -27a + 81 - 12 - 10 \)
\( = -27a + 81 - 22 \)
\( = -27a + 59 \)
Since the remainder is 5, we set this expression equal to 5:
\( -27a + 59 = 5 \)
Now, we solve for \( a \):
\( -27a = 5 - 59 \)
\( -27a = -54 \)
\( a = \frac{-54}{-27} \)
\( a = 2 \)
In simple words: We put the value of \( x \) (found by setting the divisor to zero) into the given polynomial. The answer we get is the remainder. Since we already know the remainder is 5, we set our polynomial result equal to 5 and solve for the unknown letter \( a \).

🎯 Exam Tip: Pay close attention to signs when substituting negative values and performing calculations involving exponents, as this is a common source of errors.

Question 4. If the polynomials \( ax^3 + 4x^2 + 3x - 4 \) and \( x^3 - 4x - a \) leave the same remainder when divided by \( x - 2 \), find the value of a.
Answer:
Let the first polynomial be \( f(x) = ax^3 + 4x^2 + 3x - 4 \).
Let the second polynomial be \( q(x) = x^3 - 4x - a \).
We are told that both polynomials leave the same remainder when divided by \( x - 2 \).
If \( x - 2 = 0 \), then \( x = 2 \). So, \( f(2) = q(2) \).
First, find \( f(2) \):
\( f(2) = a(2)^3 + 4(2)^2 + 3(2) - 4 \)
\( = 8a + 4(4) + 6 - 4 \)
\( = 8a + 16 + 6 - 4 \)
\( = 8a + 18 \)
Next, find \( q(2) \):
\( q(2) = (2)^3 - 4(2) - a \)
\( = 8 - 8 - a \)
\( = -a \)
Since the remainders are equal, we set \( f(2) = q(2) \):
\( 8a + 18 = -a \)
Now, we solve for \( a \):
\( 8a + a = -18 \)
\( 9a = -18 \)
\( a = \frac{-18}{9} \)
\( a = -2 \)
In simple words: We find the remainder for each polynomial by putting \( x=2 \) (because of the \( x-2 \) divisor). Since these two remainders are stated to be equal, we set the two expressions for the remainders equal to each other. This creates an equation that we can solve to find the value of \( a \).

🎯 Exam Tip: When comparing remainders from different polynomials, ensure you set up two separate remainder expressions first and then equate them to form an equation to solve for the unknown variable.

Question 5. Use factor theorem in each of the following to find whether g (x) is a factor f(x) or not :
(i) \( f(x) = x^3 - 6x^2 + 11x - 6 \); \( g (x) = x - 3 \)
(ii) \( f(x) = 2x^3 - 9x^2 + x + 12 \); \( g (x) = x + 1 \)
(iii) \( f(x) = 7x^2 - 2\sqrt{8}x - 6 \); \( g (x) = x - \sqrt{2} \)
(iv) \( f(x) = 3x^3 + x^2 - 20x + 12 \); \( g (x) = 3x - 2 \).
Answer:
The Factor Theorem states that \( g(x) \) is a factor of \( f(x) \) if \( f(a) = 0 \) when \( x-a \) is \( g(x) \).
(i) Let \( f(x) = x^3 - 6x^2 + 11x - 6 \) and \( g(x) = x - 3 \).
Set \( x - 3 = 0 \), so \( x = 3 \).
Now, calculate \( f(3) \):
\( f(3) = (3)^3 - 6(3)^2 + 11(3) - 6 \)
\( = 27 - 6(9) + 33 - 6 \)
\( = 27 - 54 + 33 - 6 \)
\( = 60 - 60 \)
\( = 0 \)
Since the remainder is 0, \( x - 3 \) is a factor of \( f(x) \).
(ii) Let \( f(x) = 2x^3 - 9x^2 + x + 12 \) and \( g(x) = x + 1 \).
Set \( x + 1 = 0 \), so \( x = -1 \).
Now, calculate \( f(-1) \):
\( f(-1) = 2(-1)^3 - 9(-1)^2 + (-1) + 12 \)
\( = 2(-1) - 9(1) - 1 + 12 \)
\( = -2 - 9 - 1 + 12 \)
\( = -12 + 12 \)
\( = 0 \)
Since the remainder is 0, \( x + 1 \) is a factor of \( f(x) \).
(iii) Let \( f(x) = 7x^2 - 2\sqrt{8}x - 6 \) and \( g(x) = x - \sqrt{2} \).
Set \( x - \sqrt{2} = 0 \), so \( x = \sqrt{2} \).
Now, calculate \( f(\sqrt{2}) \):
\( f(\sqrt{2}) = 7(\sqrt{2})^2 - 2\sqrt{8}(\sqrt{2}) - 6 \)
\( = 7(2) - 2\sqrt{16} - 6 \)
\( = 14 - 2(4) - 6 \)
\( = 14 - 8 - 6 \)
\( = 14 - 14 \)
\( = 0 \)
Since the remainder is 0, \( x - \sqrt{2} \) is a factor of \( f(x) \).
(iv) Let \( f(x) = 3x^3 + x^2 - 20x + 12 \) and \( g(x) = 3x - 2 \).
Set \( 3x - 2 = 0 \), so \( 3x = 2 \), which means \( x = \frac{2}{3} \).
Now, calculate \( f(\frac{2}{3}) \):
\( f\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^3 + \left(\frac{2}{3}\right)^2 - 20\left(\frac{2}{3}\right) + 12 \)
\( = 3\left(\frac{8}{27}\right) + \frac{4}{9} - \frac{40}{3} + 12 \)
\( = \frac{8}{9} + \frac{4}{9} - \frac{40}{3} + 12 \)
To combine these, find a common denominator, which is 9.
\( = \frac{8}{9} + \frac{4}{9} - \frac{40 \times 3}{3 \times 3} + \frac{12 \times 9}{1 \times 9} \)
\( = \frac{8 + 4 - 120 + 108}{9} \)
\( = \frac{120 - 120}{9} \)
\( = \frac{0}{9} \)
\( = 0 \)
Since the remainder is 0, \( 3x - 2 \) is a factor of \( f(x) \).
In simple words: The Factor Theorem helps us check if one expression can divide another with no remainder. We take the divisor, set it to zero to find the value of \( x \). If putting that \( x \) value into the original expression makes the whole thing equal to zero, then it's a factor. If the result is not zero, it's not a factor.

🎯 Exam Tip: To prove a linear expression is a factor, the key is to show that the remainder is exactly zero when substituting the root of the linear expression into the polynomial. Double-check your arithmetic, especially with fractions and square roots.

Question 6. Find the value of a if \( x^3 + ax + 2a - 2 \) is exactly divisible by \( x + 1 \).
Answer:
Let the polynomial be \( f(x) = x^3 + ax + 2a - 2 \).
We are told that \( f(x) \) is exactly divisible by \( x + 1 \). This means the remainder is 0.
Using the Remainder Theorem, if \( x + 1 = 0 \), then \( x = -1 \). The remainder is \( f(-1) \).
Since it's exactly divisible, \( f(-1) = 0 \).
Substitute \( x = -1 \) into the polynomial:
\( f(-1) = (-1)^3 + a(-1) + 2a - 2 \)
\( = -1 - a + 2a - 2 \)
\( = a - 3 \)
Now, set this expression equal to 0:
\( a - 3 = 0 \)
\( a = 3 \)
In simple words: When one expression exactly divides another, it means there's no remainder left. We use the Factor Theorem by finding the value of \( x \) that makes the divisor zero. Then we put this \( x \) value into the main expression. Since the remainder must be zero, we set the result to zero and solve to find the value of \( a \).

🎯 Exam Tip: Remember that "exactly divisible" implies a remainder of zero, which is the core condition for applying the Factor Theorem. This simplifies the problem to solving a linear equation for the unknown constant.

Question 7. Find the values of a and b so that the expression \( x^3 + 10x^2 + ax + b \) is exactly divisible by \( x - 1 \) as well as \( x - 2 \).
Answer:
Let the polynomial be \( f(x) = x^3 + 10x^2 + ax + b \).
We are given that \( f(x) \) is exactly divisible by \( x - 1 \). This means \( f(1) = 0 \).
Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = (1)^3 + 10(1)^2 + a(1) + b \)
\( = 1 + 10 + a + b \)
\( = 11 + a + b \)
Since \( f(1) = 0 \):
\( 11 + a + b = 0 \)
\( a + b = -11 \) ... (i)
We are also given that \( f(x) \) is exactly divisible by \( x - 2 \). This means \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^3 + 10(2)^2 + a(2) + b \)
\( = 8 + 10(4) + 2a + b \)
\( = 8 + 40 + 2a + b \)
\( = 48 + 2a + b \)
Since \( f(2) = 0 \):
\( 48 + 2a + b = 0 \)
\( 2a + b = -48 \) ... (ii)
Now we have a system of two linear equations:
1) \( a + b = -11 \)
2) \( 2a + b = -48 \)
Subtract equation (i) from equation (ii):
\( (2a + b) - (a + b) = -48 - (-11) \)
\( 2a + b - a - b = -48 + 11 \)
\( a = -37 \)
Substitute the value of \( a \) into equation (i):
\( -37 + b = -11 \)
\( b = -11 + 37 \)
\( b = 26 \)
Thus, \( a = -37 \) and \( b = 26 \).
In simple words: Because the polynomial can be divided by two different expressions with no remainder, we get two separate equations. We substitute the \( x \) values from both divisors into the polynomial and set each result to zero. Then, we solve these two equations together (like a pair of simultaneous equations) to find the specific values for \( a \) and \( b \).

🎯 Exam Tip: Problems involving two divisors that lead to a zero remainder will typically require setting up and solving a system of two linear equations. Carefully execute the substitution and algebraic steps for both equations.

Question 8. If \( (x - 2) \) is a factor of \( x^2 + ax - 6 = 0 \) and \( x^2 - 9x + b = 0 \), find the values of a and b.
Answer:
Let the first polynomial be \( f(x) = x^2 + ax - 6 \).
Let the second polynomial be \( q(x) = x^2 - 9x + b \).
We are given that \( (x - 2) \) is a factor of both polynomials.
If \( x - 2 = 0 \), then \( x = 2 \).
Since \( (x - 2) \) is a factor of \( f(x) \), then \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^2 + a(2) - 6 \)
\( = 4 + 2a - 6 \)
\( = 2a - 2 \)
Set this expression to 0:
\( 2a - 2 = 0 \)
\( 2a = 2 \)
\( a = \frac{2}{2} \)
\( a = 1 \)
Since \( (x - 2) \) is also a factor of \( q(x) \), then \( q(2) = 0 \).
Substitute \( x = 2 \) into \( q(x) \):
\( q(2) = (2)^2 - 9(2) + b \)
\( = 4 - 18 + b \)
\( = -14 + b \)
Set this expression to 0:
\( -14 + b = 0 \)
\( b = 14 \)
Therefore, the values are \( a = 1 \) and \( b = 14 \).
In simple words: We know that \( x-2 \) divides both expressions completely, which means putting \( x=2 \) into each expression will give a remainder of zero. We use this fact for the first expression to find \( a \), and for the second expression to find \( b \). This gives us two simple equations to solve.

🎯 Exam Tip: Remember that if a term like \( (x-k) \) is a factor, then substituting \( x=k \) into the polynomial must yield a remainder of zero. This is a direct application of the Factor Theorem for solving such problems.

Question 9. If both \( x - 2 \) and \( x - \frac{1}{2} \) are factors of \( px^2 + 5x + r \), show that \( p = r \).
Answer:
Let the polynomial be \( f(x) = px^2 + 5x + r \).
Given that \( (x - 2) \) is a factor of \( f(x) \).
If \( x - 2 = 0 \), then \( x = 2 \). By the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = p(2)^2 + 5(2) + r \)
\( = 4p + 10 + r \)
Since \( f(2) = 0 \):
\( 4p + 10 + r = 0 \)
\( 4p + r = -10 \) ... (i)
Given that \( (x - \frac{1}{2}) \) is also a factor of \( f(x) \).
If \( x - \frac{1}{2} = 0 \), then \( x = \frac{1}{2} \). By the Factor Theorem, \( f(\frac{1}{2}) = 0 \).
Substitute \( x = \frac{1}{2} \) into \( f(x) \):
\( f\left(\frac{1}{2}\right) = p\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) + r \)
\( = p\left(\frac{1}{4}\right) + \frac{5}{2} + r \)
\( = \frac{p}{4} + \frac{5}{2} + r \)
Since \( f(\frac{1}{2}) = 0 \):
\( \frac{p}{4} + \frac{5}{2} + r = 0 \)
To eliminate fractions, multiply the entire equation by 4:
\( 4\left(\frac{p}{4}\right) + 4\left(\frac{5}{2}\right) + 4(r) = 4(0) \)
\( p + 10 + 4r = 0 \)
\( p + 4r = -10 \) ... (ii)
Now we have two equations:
1) \( 4p + r = -10 \)
2) \( p + 4r = -10 \)
Subtract equation (ii) from equation (i):
\( (4p + r) - (p + 4r) = -10 - (-10) \)
\( 4p + r - p - 4r = 0 \)
\( 3p - 3r = 0 \)
Divide by 3:
\( p - r = 0 \)
\( p = r \)
Hence, it is shown that \( p = r \).
In simple words: If two different terms are factors of a polynomial, then putting the \( x \) value from each factor into the polynomial must give a remainder of zero. We do this for both factors to get two equations. Then, we solve these two equations to show that \( p \) and \( r \) must be the same number.

🎯 Exam Tip: When showing \( p=r \), ensure your algebraic manipulation of the simultaneous equations is precise. Remember to clear any fractions by multiplying by the least common multiple of the denominators.

Question 10. If \( x^3 + ax^2 + bx + 6 \) has \( (x - 2) \) as a factor and leaves a remainder 3 when divided by \( (x - 3) \), find the values of a and b.
Answer:
Let the polynomial be \( f(x) = x^3 + ax^2 + bx + 6 \).
Given that \( (x - 2) \) is a factor of \( f(x) \).
If \( x - 2 = 0 \), then \( x = 2 \). By the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^3 + a(2)^2 + b(2) + 6 \)
\( = 8 + 4a + 2b + 6 \)
\( = 4a + 2b + 14 \)
Set this expression to 0:
\( 4a + 2b + 14 = 0 \)
Divide the entire equation by 2 to simplify:
\( 2a + b + 7 = 0 \)
\( 2a + b = -7 \) ... (i)
Given that \( f(x) \) leaves a remainder 3 when divided by \( (x - 3) \).
If \( x - 3 = 0 \), then \( x = 3 \). By the Remainder Theorem, \( f(3) = 3 \).
Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = (3)^3 + a(3)^2 + b(3) + 6 \)
\( = 27 + 9a + 3b + 6 \)
\( = 9a + 3b + 33 \)
Set this expression to 3:
\( 9a + 3b + 33 = 3 \)
\( 9a + 3b = 3 - 33 \)
\( 9a + 3b = -30 \)
Divide the entire equation by 3 to simplify:
\( 3a + b = -10 \) ... (ii)
Now we have a system of two linear equations:
1) \( 2a + b = -7 \)
2) \( 3a + b = -10 \)
Subtract equation (i) from equation (ii):
\( (3a + b) - (2a + b) = -10 - (-7) \)
\( 3a + b - 2a - b = -10 + 7 \)
\( a = -3 \)
Substitute the value of \( a \) into equation (i):
\( 2(-3) + b = -7 \)
\( -6 + b = -7 \)
\( b = -7 + 6 \)
\( b = -1 \)
Thus, the values are \( a = -3 \) and \( b = -1 \).
In simple words: We use two pieces of information: first, that \( x-2 \) is a factor (meaning the remainder is 0 when \( x=2 \)), and second, that dividing by \( x-3 \) gives a remainder of 3. Each piece of information helps us create an equation. We then solve these two equations together to find the values of \( a \) and \( b \).

🎯 Exam Tip: Distinguish carefully between cases where an expression is a "factor" (remainder is 0) and when it "leaves a remainder" (remainder is a specific number). Both situations use the Remainder Theorem to form equations.

Question 11. Factorise :
(i) \( x^3 + 13x^2 + 32x + 20 \), if it is given that \( x + 2 \) is its factor.
(ii) \( 4x^3 + 20x^2 + 33x + 18 \), if it is given that \( 2x + 3 \) is its factor.
Answer:
(i) Let \( f(x) = x^3 + 13x^2 + 32x + 20 \).
Given that \( x + 2 \) is a factor. This means \( f(-2) = 0 \).
Check: \( f(-2) = (-2)^3 + 13(-2)^2 + 32(-2) + 20 \)
\( = -8 + 13(4) - 64 + 20 \)
\( = -8 + 52 - 64 + 20 \)
\( = 72 - 72 = 0 \).
Since \( f(-2) = 0 \), \( (x+2) \) is indeed a factor.
Now, we divide \( f(x) \) by \( (x+2) \) using polynomial long division:
\[ \require{enclose} \begin{array}{r} x^2+11x+10 \\ x+2 \enclose{longdiv}{x^3+13x^2+32x+20} \\ \underline{-x^3-2x^2\phantom{+32x+20}} \\ 11x^2+32x\phantom{+20} \\ \underline{-11x^2-22x\phantom{+20}} \\ 10x+20 \\ \underline{-10x-20} \\ 0 \end{array} \]
So, \( f(x) = (x+2)(x^2 + 11x + 10) \).
Now, we factorise the quadratic part \( x^2 + 11x + 10 \).
We look for two numbers that multiply to 10 and add up to 11 (which are 1 and 10).
\( x^2 + 11x + 10 = x^2 + x + 10x + 10 \)
\( = x(x+1) + 10(x+1) \)
\( = (x+1)(x+10) \)
Therefore, the complete factorisation of \( f(x) \) is \( (x+2)(x+1)(x+10) \).
(ii) Let \( f(x) = 4x^3 + 20x^2 + 33x + 18 \).
Given that \( 2x + 3 \) is a factor. This means \( f(-\frac{3}{2}) = 0 \).
Set \( 2x + 3 = 0 \), so \( x = -\frac{3}{2} \).
Check: \( f\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^3 + 20\left(-\frac{3}{2}\right)^2 + 33\left(-\frac{3}{2}\right) + 18 \)
\( = 4\left(-\frac{27}{8}\right) + 20\left(\frac{9}{4}\right) - \frac{99}{2} + 18 \)
\( = -\frac{27}{2} + 45 - \frac{99}{2} + 18 \)
\( = \frac{-27 + 90 - 99 + 36}{2} \)
\( = \frac{126 - 126}{2} = 0 \).
Since \( f(-\frac{3}{2}) = 0 \), \( (2x+3) \) is indeed a factor.
Now, we divide \( f(x) \) by \( (2x+3) \) using polynomial long division:
\[ \require{enclose} \begin{array}{r} 2x^2+7x+6 \\ 2x+3 \enclose{longdiv}{4x^3+20x^2+33x+18} \\ \underline{-4x^3-6x^2\phantom{+33x+18}} \\ 14x^2+33x\phantom{+18} \\ \underline{-14x^2-21x\phantom{+18}} \\ 12x+18 \\ \underline{-12x-18} \\ 0 \end{array} \]
So, \( f(x) = (2x+3)(2x^2 + 7x + 6) \).
Now, we factorise the quadratic part \( 2x^2 + 7x + 6 \).
We look for two numbers that multiply to \( 2 \times 6 = 12 \) and add up to 7 (which are 3 and 4).
\( 2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6 \)
\( = 2x(x+2) + 3(x+2) \)
\( = (x+2)(2x+3) \)
Therefore, the complete factorisation of \( f(x) \) is \( (2x+3)(x+2)(2x+3) \), which can be written as \( (2x+3)^2(x+2) \).
In simple words: To factorise a polynomial when one factor is given, first confirm it's a true factor by checking if the remainder is zero. Then, use long division to divide the polynomial by that factor. This will give you a simpler polynomial (a quadratic). Finally, factorise this simpler polynomial into its linear parts.

🎯 Exam Tip: Always show the remainder check using the Factor Theorem, as it confirms the given factor and can earn you marks. Be careful with long division steps, especially with signs and carrying down terms.

Question 12. Show that:
(i) \( (x - 10) \) is a factor of \( x^3 - 23x^2 + 142x - 120 \) and hence factorise it completely.
(ii) \( (3z + 10) \) is factor of \( 9z^3 - 27z^2 - 100z + 300 \) and factorise it completely.
Answer:
(i) Let \( f(x) = x^3 - 23x^2 + 142x - 120 \).
To show \( (x - 10) \) is a factor, we check if \( f(10) = 0 \).
If \( x - 10 = 0 \), then \( x = 10 \).
\( f(10) = (10)^3 - 23(10)^2 + 142(10) - 120 \)
\( = 1000 - 23(100) + 1420 - 120 \)
\( = 1000 - 2300 + 1420 - 120 \)
\( = 2420 - 2420 = 0 \).
Since \( f(10) = 0 \), \( (x - 10) \) is a factor of \( f(x) \).
Now, divide \( f(x) \) by \( (x - 10) \) using polynomial long division:
\[ \require{enclose} \begin{array}{r} x^2-13x+12 \\ x-10 \enclose{longdiv}{x^3-23x^2+142x-120} \\ \underline{-x^3+10x^2\phantom{+142x-120}} \\ -13x^2+142x\phantom{-120} \\ \underline{-(-13x^2+130x)\phantom{-120}} \\ 12x-120 \\ \underline{-(12x-120)} \\ 0 \end{array} \]
So, \( f(x) = (x-10)(x^2 - 13x + 12) \).
Now, factorise the quadratic part \( x^2 - 13x + 12 \).
We look for two numbers that multiply to 12 and add up to -13 (which are -1 and -12).
\( x^2 - 13x + 12 = x^2 - x - 12x + 12 \)
\( = x(x-1) - 12(x-1) \)
\( = (x-1)(x-12) \)
Therefore, the complete factorisation of \( f(x) \) is \( (x-10)(x-1)(x-12) \).
(ii) Let \( f(z) = 9z^3 - 27z^2 - 100z + 300 \).
To show \( (3z + 10) \) is a factor, we check if \( f(-\frac{10}{3}) = 0 \).
If \( 3z + 10 = 0 \), then \( z = -\frac{10}{3} \).
\( f\left(-\frac{10}{3}\right) = 9\left(-\frac{10}{3}\right)^3 - 27\left(-\frac{10}{3}\right)^2 - 100\left(-\frac{10}{3}\right) + 300 \)
\( = 9\left(-\frac{1000}{27}\right) - 27\left(\frac{100}{9}\right) + \frac{1000}{3} + 300 \)
\( = -\frac{1000}{3} - 3(100) + \frac{1000}{3} + 300 \)
\( = -\frac{1000}{3} - 300 + \frac{1000}{3} + 300 \)
\( = 0 \).
Since \( f(-\frac{10}{3}) = 0 \), \( (3z + 10) \) is a factor of \( f(z) \).
Now, divide \( f(z) \) by \( (3z + 10) \) using polynomial long division:
\[ \require{enclose} \begin{array}{r} 3z^2-19z+30 \\ 3z+10 \enclose{longdiv}{9z^3-27z^2-100z+300} \\ \underline{-9z^3-30z^2\phantom{-100z+300}} \\ -57z^2-100z\phantom{+300} \\ \underline{-(-57z^2-190z)\phantom{+300}} \\ 90z+300 \\ \underline{-(90z+300)} \\ 0 \end{array} \]
So, \( f(z) = (3z+10)(3z^2 - 19z + 30) \).
Now, factorise the quadratic part \( 3z^2 - 19z + 30 \).
We look for two numbers that multiply to \( 3 \times 30 = 90 \) and add up to -19 (which are -9 and -10).
\( 3z^2 - 19z + 30 = 3z^2 - 9z - 10z + 30 \)
\( = 3z(z-3) - 10(z-3) \)
\( = (z-3)(3z-10) \)
Therefore, the complete factorisation of \( f(z) \) is \( (3z+10)(z-3)(3z-10) \).
In simple words: To show that an expression is a factor, we simply substitute the root of that factor into the polynomial and verify if the result is zero. Once confirmed, we use long division to break down the polynomial into a simpler one. Then, we factorise this simpler polynomial further, if possible, to get the complete factorisation.

🎯 Exam Tip: Always perform a remainder check first to confirm the given factor, especially in "show that" questions. Be meticulous with polynomial long division, as errors here will affect the final factorization.

Question 13. Given that \( (x - 2) \) and \( (x + 1) \) are factors of \( x^3 + 3x^2 + ax + b \), calculate the values of a and b, and hence find the remaining factor.
Answer:
Let the polynomial be \( f(x) = x^3 + 3x^2 + ax + b \).
Given that \( (x - 2) \) is a factor of \( f(x) \).
If \( x - 2 = 0 \), then \( x = 2 \). By the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^3 + 3(2)^2 + a(2) + b \)
\( = 8 + 3(4) + 2a + b \)
\( = 8 + 12 + 2a + b \)
\( = 20 + 2a + b \)
Since \( f(2) = 0 \):
\( 20 + 2a + b = 0 \)
\( 2a + b = -20 \) ... (i)
Given that \( (x + 1) \) is also a factor of \( f(x) \).
If \( x + 1 = 0 \), then \( x = -1 \). By the Factor Theorem, \( f(-1) = 0 \).
Substitute \( x = -1 \) into \( f(x) \):
\( f(-1) = (-1)^3 + 3(-1)^2 + a(-1) + b \)
\( = -1 + 3(1) - a + b \)
\( = -1 + 3 - a + b \)
\( = 2 - a + b \)
Since \( f(-1) = 0 \):
\( 2 - a + b = 0 \)
\( -a + b = -2 \)
\( a - b = 2 \) ... (ii)
Now we have a system of two linear equations:
1) \( 2a + b = -20 \)
2) \( a - b = 2 \)
Add equation (i) and equation (ii):
\( (2a + b) + (a - b) = -20 + 2 \)
\( 3a = -18 \)
\( a = \frac{-18}{3} \)
\( a = -6 \)
Substitute the value of \( a \) into equation (ii):
\( -6 - b = 2 \)
\( -b = 2 + 6 \)
\( -b = 8 \)
\( b = -8 \)
So, \( a = -6 \) and \( b = -8 \).
The polynomial is now \( f(x) = x^3 + 3x^2 - 6x - 8 \).
Since \( (x - 2) \) is a factor, we can divide \( f(x) \) by \( (x - 2) \) to find the remaining quadratic factor.
\[ \require{enclose} \begin{array}{r} x^2+5x+4 \\ x-2 \enclose{longdiv}{x^3+3x^2-6x-8} \\ \underline{-x^3+2x^2\phantom{-6x-8}} \\ 5x^2-6x\phantom{-8} \\ \underline{-5x^2+10x\phantom{-8}} \\ 4x-8 \\ \underline{-4x+8} \\ 0 \end{array} \]
So, \( f(x) = (x-2)(x^2 + 5x + 4) \).
Now, factorise the quadratic part \( x^2 + 5x + 4 \).
We look for two numbers that multiply to 4 and add up to 5 (which are 1 and 4).
\( x^2 + 5x + 4 = x^2 + x + 4x + 4 \)
\( = x(x+1) + 4(x+1) \)
\( = (x+1)(x+4) \)
Thus, \( f(x) = (x-2)(x+1)(x+4) \).
The remaining factor is \( (x+4) \).
In simple words: First, we use the fact that \( x-2 \) and \( x+1 \) are factors to set up two equations for \( a \) and \( b \). We solve these equations to find the values of \( a \) and \( b \). Once we have the full polynomial, we divide it by one of the known factors (like \( x-2 \)) to get a quadratic expression. Then, we factorise this quadratic expression to find the last factor.

🎯 Exam Tip: When given multiple factors, use each to form an equation using the Factor Theorem. After finding the unknown coefficients, use polynomial long division with one of the factors to reduce the polynomial to a simpler form for complete factorization.

Question 13. Given that (x – 2) and (x + 1) are factors of \( x^3 + 3x^2 + ax + b \), calculate the values of a and b, and hence find the remaining factor.
Answer: Let the polynomial be \( f(x) = x^3 + 3x^2 + ax + b \).
Since \( (x-2) \) is a factor, \( f(2) = 0 \).
So, \( (2)^3 + 3(2)^2 + a(2) + b = 0 \)
\( 8 + 3(4) + 2a + b = 0 \)
\( 8 + 12 + 2a + b = 0 \)
\( 20 + 2a + b = 0 \)
\( 2a + b = -20 \) ... (i)

Since \( (x+1) \) is a factor, \( f(-1) = 0 \).
So, \( (-1)^3 + 3(-1)^2 + a(-1) + b = 0 \)
\( -1 + 3(1) - a + b = 0 \)
\( -1 + 3 - a + b = 0 \)
\( 2 - a + b = 0 \)
\( -a + b = -2 \)
\( a - b = 2 \) ... (ii)

Now, add equations (i) and (ii):
\( (2a + b) + (a - b) = -20 + 2 \)
\( 3a = -18 \)
\( a = \frac{-18}{3} \)
\( a = -6 \)

Substitute the value of \( a \) into equation (ii):
\( -6 - b = 2 \)
\( -b = 2 + 6 \)
\( -b = 8 \)
\( b = -8 \)

So, \( a = -6 \) and \( b = -8 \).
The polynomial is \( f(x) = x^3 + 3x^2 - 6x - 8 \).
We know \( (x-2) \) is a factor. Now we divide \( f(x) \) by \( (x-2) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {+5x} & {+4} \\ \cline{2-6} x-2 & x^3 & {+3x^2} & {-6x} & {-8} \\ \multicolumn{2}{r}{x^3} & {-2x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {5x^2} & {-6x} \\ \multicolumn{2}{r}{~} & {5x^2} & {-10x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {4x} & {-8} \\ \multicolumn{2}{r}{~} & {~} & {4x} & {-8} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x-2)(x^2+5x+4) \).
Now, factorise the quadratic part \( x^2+5x+4 \):
\( x^2+5x+4 = x^2+x+4x+4 \)
\( = x(x+1)+4(x+1) \)
\( = (x+1)(x+4) \)
Thus, \( f(x) = (x-2)(x+1)(x+4) \).
The remaining factor is \( (x+4) \).
In simple words: First, we use the fact that if \( (x-k) \) is a factor, then \( f(k)=0 \). This helps us set up two equations and find the values of 'a' and 'b'. Then, we divide the polynomial by one of the known factors to find the remaining quadratic expression, which we factorize further to get all factors.

🎯 Exam Tip: Remember to set up a system of simultaneous equations when finding multiple unknown constants (like 'a' and 'b'). A correct long division is key to finding the remaining factors for higher degree polynomials.

Question 14. Given that (x + 2) and (x – 3) are factors of \( x^3 + ax + b \), calculate the values of a and b, and find the remaining factor.
Answer: Let the polynomial be \( f(x) = x^3 + ax + b \).
Since \( (x+2) \) is a factor, \( f(-2) = 0 \).
So, \( (-2)^3 + a(-2) + b = 0 \)
\( -8 - 2a + b = 0 \)
\( -2a + b = 8 \) ... (i)

Since \( (x-3) \) is a factor, \( f(3) = 0 \).
So, \( (3)^3 + a(3) + b = 0 \)
\( 27 + 3a + b = 0 \)
\( 3a + b = -27 \) ... (ii)

Subtract equation (i) from equation (ii):
\( (3a + b) - (-2a + b) = -27 - 8 \)
\( 3a + b + 2a - b = -35 \)
\( 5a = -35 \)
\( a = \frac{-35}{5} \)
\( a = -7 \)

Substitute the value of \( a \) into equation (i):
\( -2(-7) + b = 8 \)
\( 14 + b = 8 \)
\( b = 8 - 14 \)
\( b = -6 \)

So, \( a = -7 \) and \( b = -6 \).
The polynomial is \( f(x) = x^3 - 7x - 6 \).
We know \( (x+2) \) is a factor. Now we divide \( f(x) \) by \( (x+2) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {-2x} & {-3} \\ \cline{2-6} x+2 & x^3 & {+0x^2} & {-7x} & {-6} \\ \multicolumn{2}{r}{x^3} & {+2x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {-2x^2} & {-7x} \\ \multicolumn{2}{r}{~} & {-2x^2} & {-4x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-3x} & {-6} \\ \multicolumn{2}{r}{~} & {~} & {-3x} & {-6} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x+2)(x^2-2x-3) \).
Now, factorise the quadratic part \( x^2-2x-3 \):
\( x^2-2x-3 = x^2-3x+x-3 \)
\( = x(x-3)+1(x-3) \)
\( = (x-3)(x+1) \)
Thus, \( f(x) = (x+2)(x-3)(x+1) \).
The remaining factor is \( (x+1) \).
In simple words: We first find the values of 'a' and 'b' by using the factor theorem twice, creating two equations. After finding 'a' and 'b', we divide the polynomial by one of the given factors to get a simpler expression, which we then factorize completely to find the last factor.

🎯 Exam Tip: When a polynomial has missing terms (like \( x^2 \) in \( x^3+ax+b \)), remember to include them with a zero coefficient (e.g., \( 0x^2 \)) during long division to keep terms aligned.

Question 15. Factorise, using remainder theorem :
(i) \( x^3 - 19x - 30 \)
(ii) \( x^3 + 7x^2 - 21x - 27 \)
(iii) \( x^3 - 3x^2 - 9x - 5 \)
(iv) \( 2x^3 + 9x^2 + 7x - 6 \)
Answer:
(i) Let \( f(x) = x^3 - 19x - 30 \).
The factors of the constant term 30 are \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 \).
By trial and error:
If \( x = -2 \): \( f(-2) = (-2)^3 - 19(-2) - 30 = -8 + 38 - 30 = 0 \).
So, \( (x+2) \) is a factor.
If \( x = -3 \): \( f(-3) = (-3)^3 - 19(-3) - 30 = -27 + 57 - 30 = 0 \).
So, \( (x+3) \) is another factor.
If \( x = 5 \): \( f(5) = (5)^3 - 19(5) - 30 = 125 - 95 - 30 = 0 \).
So, \( (x-5) \) is the third factor.
Thus, \( x^3 - 19x - 30 = (x+2)(x+3)(x-5) \).

(ii) Let \( f(x) = x^3 + 7x^2 - 21x - 27 \).
The factors of the constant term 27 are \( \pm 1, \pm 3, \pm 9, \pm 27 \).
By trial and error:
If \( x = -1 \): \( f(-1) = (-1)^3 + 7(-1)^2 - 21(-1) - 27 = -1 + 7 + 21 - 27 = 0 \).
So, \( (x+1) \) is a factor.
Divide \( f(x) \) by \( (x+1) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {+6x} & {-27} \\ \cline{2-6} x+1 & x^3 & {+7x^2} & {-21x} & {-27} \\ \multicolumn{2}{r}{x^3} & {+x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {6x^2} & {-21x} \\ \multicolumn{2}{r}{~} & {6x^2} & {+6x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-27x} & {-27} \\ \multicolumn{2}{r}{~} & {~} & {-27x} & {-27} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x+1)(x^2+6x-27) \).
Factorise the quadratic part \( x^2+6x-27 \):
\( x^2+6x-27 = x^2+9x-3x-27 \)
\( = x(x+9)-3(x+9) \)
\( = (x+9)(x-3) \)
Thus, \( x^3 + 7x^2 - 21x - 27 = (x+1)(x+9)(x-3) \).

(iii) Let \( f(x) = x^3 - 3x^2 - 9x - 5 \).
The factors of the constant term 5 are \( \pm 1, \pm 5 \).
By trial and error:
If \( x = -1 \): \( f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3(1) + 9 - 5 = -1 - 3 + 9 - 5 = 0 \).
So, \( (x+1) \) is a factor.
Divide \( f(x) \) by \( (x+1) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {-4x} & {-5} \\ \cline{2-6} x+1 & x^3 & {-3x^2} & {-9x} & {-5} \\ \multicolumn{2}{r}{x^3} & {+x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {-4x^2} & {-9x} \\ \multicolumn{2}{r}{~} & {-4x^2} & {-4x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-5x} & {-5} \\ \multicolumn{2}{r}{~} & {~} & {-5x} & {-5} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x+1)(x^2-4x-5) \).
Factorise the quadratic part \( x^2-4x-5 \):
\( x^2-4x-5 = x^2-5x+x-5 \)
\( = x(x-5)+1(x-5) \)
\( = (x-5)(x+1) \)
Thus, \( x^3 - 3x^2 - 9x - 5 = (x+1)(x-5)(x+1) = (x+1)^2(x-5) \).

(iv) Let \( f(x) = 2x^3 + 9x^2 + 7x - 6 \).
The factors of the constant term 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \). We also consider factors of \( p/q \) where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient (2).
By trial and error:
If \( x = -2 \): \( f(-2) = 2(-2)^3 + 9(-2)^2 + 7(-2) - 6 = 2(-8) + 9(4) - 14 - 6 = -16 + 36 - 14 - 6 = 0 \).
So, \( (x+2) \) is a factor.
Divide \( f(x) \) by \( (x+2) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{2x^2} & {+5x} & {-3} \\ \cline{2-6} x+2 & 2x^3 & {+9x^2} & {+7x} & {-6} \\ \multicolumn{2}{r}{2x^3} & {+4x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {5x^2} & {+7x} \\ \multicolumn{2}{r}{~} & {5x^2} & {+10x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-3x} & {-6} \\ \multicolumn{2}{r}{~} & {~} & {-3x} & {-6} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x+2)(2x^2+5x-3) \).
Factorise the quadratic part \( 2x^2+5x-3 \):
We need two numbers that multiply to \( 2 \times -3 = -6 \) and add to 5. These are 6 and -1.
\( 2x^2+5x-3 = 2x^2+6x-x-3 \)
\( = 2x(x+3)-1(x+3) \)
\( = (x+3)(2x-1) \)
Thus, \( 2x^3 + 9x^2 + 7x - 6 = (x+2)(x+3)(2x-1) \).
In simple words: To factorize these polynomials, we first look for possible factors by testing simple numbers like \( \pm 1, \pm 2, \pm 3 \), etc., using the remainder theorem. If plugging in a number makes the polynomial zero, then \( (x - \text{number}) \) is a factor. Once we find one factor, we use long division to get a simpler polynomial, usually a quadratic one. Then, we factorize that quadratic expression to find the remaining factors.

🎯 Exam Tip: When using the Remainder Theorem for factorization, always try integers that are factors of the constant term first. For polynomials with a leading coefficient other than 1, also consider fractions \( \frac{p}{q} \).

Question 16. (x – 3) is the H.C.F. of \( x^3 – 2x^2 + px + 6 \) and \( x^2 – 5x + q \). Find \( 6p + 5q \).
Answer: Let \( P(x) = x^3 - 2x^2 + px + 6 \) and \( Q(x) = x^2 - 5x + q \).
Given that \( (x-3) \) is the H.C.F. (Highest Common Factor) of \( P(x) \) and \( Q(x) \).
This means \( (x-3) \) is a factor of both \( P(x) \) and \( Q(x) \).
So, by the Factor Theorem, \( P(3) = 0 \) and \( Q(3) = 0 \).

For \( P(x) \):
\( P(3) = (3)^3 - 2(3)^2 + p(3) + 6 = 0 \)
\( 27 - 2(9) + 3p + 6 = 0 \)
\( 27 - 18 + 3p + 6 = 0 \)
\( 9 + 3p + 6 = 0 \)
\( 15 + 3p = 0 \)
\( 3p = -15 \)
\( p = \frac{-15}{3} \)
\( p = -5 \)

For \( Q(x) \):
\( Q(3) = (3)^2 - 5(3) + q = 0 \)
\( 9 - 15 + q = 0 \)
\( -6 + q = 0 \)
\( q = 6 \)

Now, we need to find the value of \( 6p + 5q \):
\( 6p + 5q = 6(-5) + 5(6) \)
\( = -30 + 30 \)
\( = 0 \)
In simple words: Since \( (x-3) \) is the Highest Common Factor, it means \( x=3 \) will make both polynomials equal to zero. We plug \( x=3 \) into each polynomial, which helps us find the values of 'p' and 'q'. Once we have 'p' and 'q', we just put them into the expression \( 6p + 5q \) to get the final answer.

🎯 Exam Tip: The H.C.F. of two polynomials is also a factor of each polynomial. This crucial property allows you to apply the Factor Theorem to solve for unknown coefficients.

Question 17.
(i) What number must be subtracted from \( x^3 – 6x^2 – 15x + 80 \), so that the result is exactly divisible by \( x + 4 \).
(ii) What number must be added to \( x^3 – 3x^2 – 12x + 19 \), so that the result is exactly divisible by \( x – 2 \).
Answer:
(i) Let \( f(x) = x^3 - 6x^2 - 15x + 80 \).
Let \( k \) be the number to be subtracted. The new polynomial is \( P(x) = x^3 - 6x^2 - 15x + 80 - k \).
For \( P(x) \) to be exactly divisible by \( (x+4) \), the remainder must be 0 when divided by \( (x+4) \).
So, \( P(-4) = 0 \).
\( (-4)^3 - 6(-4)^2 - 15(-4) + 80 - k = 0 \)
\( -64 - 6(16) + 60 + 80 - k = 0 \)
\( -64 - 96 + 60 + 80 - k = 0 \)
\( -160 + 140 - k = 0 \)
\( -20 - k = 0 \)
\( k = -20 \)
So, the number to be subtracted is -20. This means we should add 20 to the polynomial. If the question asks to subtract 'k', then 'k' itself is -20.

(ii) Let \( f(x) = x^3 - 3x^2 - 12x + 19 \).
Let \( k \) be the number to be added. The new polynomial is \( Q(x) = x^3 - 3x^2 - 12x + 19 + k \).
For \( Q(x) \) to be exactly divisible by \( (x-2) \), the remainder must be 0 when divided by \( (x-2) \).
So, \( Q(2) = 0 \).
\( (2)^3 - 3(2)^2 - 12(2) + 19 + k = 0 \)
\( 8 - 3(4) - 24 + 19 + k = 0 \)
\( 8 - 12 - 24 + 19 + k = 0 \)
\( -4 - 24 + 19 + k = 0 \)
\( -28 + 19 + k = 0 \)
\( -9 + k = 0 \)
\( k = 9 \)
So, the number to be added is 9.
In simple words: For part (i), we find the remainder when the polynomial is divided by \( (x+4) \). To make it exactly divisible, this remainder must be zero. The number we need to subtract from the polynomial is equal to this remainder. For part (ii), we find the remainder when the polynomial is divided by \( (x-2) \). The number we need to add to the polynomial is the negative of this remainder to make it zero.

🎯 Exam Tip: If a polynomial \( f(x) \) gives a remainder \( R \) when divided by \( (x-a) \), then to make it exactly divisible, you must subtract \( R \) from \( f(x) \) or add \( -R \) to \( f(x) \). This is a direct application of the Remainder Theorem.

Self Evaluation And Revision (Latest ICSE Questions)

Question 1. Find the remainder when \( 2x^3 – 3x^2 + 7x – 8 \) is divided by \( x – 1 \).
Answer: Let \( f(x) = 2x^3 - 3x^2 + 7x - 8 \).
We are dividing by \( (x-1) \), so we need to find \( f(1) \).
According to the Remainder Theorem, the remainder is \( f(1) \).
\( f(1) = 2(1)^3 - 3(1)^2 + 7(1) - 8 \)
\( = 2(1) - 3(1) + 7 - 8 \)
\( = 2 - 3 + 7 - 8 \)
\( = 9 - 11 \)
\( = -2 \)
The remainder is -2.
In simple words: To find the leftover amount when we divide the polynomial by \( (x-1) \), we simply replace every 'x' in the polynomial with '1' and calculate the result. That number is our remainder.

🎯 Exam Tip: The Remainder Theorem states that if a polynomial \( f(x) \) is divided by \( (x-a) \), the remainder is \( f(a) \). This is a quick way to find remainders without performing long division.

Question 2. Find the value of the constants a and b if \( (x – 2) \) and \( (x + 3) \) are both factors of the expression \( x^3 + ax^2 + bx – 12 \).
Answer: Let \( f(x) = x^3 + ax^2 + bx - 12 \).
Since \( (x-2) \) is a factor, \( f(2) = 0 \).
\( (2)^3 + a(2)^2 + b(2) - 12 = 0 \)
\( 8 + 4a + 2b - 12 = 0 \)
\( 4a + 2b - 4 = 0 \)
Divide by 2:
\( 2a + b - 2 = 0 \)
\( 2a + b = 2 \) ... (i)

Since \( (x+3) \) is a factor, \( f(-3) = 0 \).
\( (-3)^3 + a(-3)^2 + b(-3) - 12 = 0 \)
\( -27 + 9a - 3b - 12 = 0 \)
\( 9a - 3b - 39 = 0 \)
Divide by 3:
\( 3a - b - 13 = 0 \)
\( 3a - b = 13 \) ... (ii)

Add equations (i) and (ii):
\( (2a + b) + (3a - b) = 2 + 13 \)
\( 5a = 15 \)
\( a = \frac{15}{5} \)
\( a = 3 \)

Substitute \( a = 3 \) into equation (i):
\( 2(3) + b = 2 \)
\( 6 + b = 2 \)
\( b = 2 - 6 \)
\( b = -4 \)
Thus, \( a = 3 \) and \( b = -4 \).
In simple words: We are given that two expressions divide the polynomial exactly, meaning the remainder is zero for both. We use the Remainder Theorem twice by setting \( x=2 \) and \( x=-3 \), which gives us two simple equations. Solving these two equations together helps us find the exact values for 'a' and 'b'.

🎯 Exam Tip: When a polynomial has multiple factors, each factor can be used with the Factor Theorem to generate an equation. Solving the resulting system of linear equations will determine the unknown coefficients.

Question 3. Using factor theorem, show that \( (x – 3) \) is a factor of \( x^3 – 7x^2 + 15x – 9 \). Hence factorise the given expression completely.
Answer: Let \( f(x) = x^3 - 7x^2 + 15x - 9 \).
To show that \( (x-3) \) is a factor, we must show that \( f(3) = 0 \).
\( f(3) = (3)^3 - 7(3)^2 + 15(3) - 9 \)
\( = 27 - 7(9) + 45 - 9 \)
\( = 27 - 63 + 45 - 9 \)
\( = 72 - 72 \)
\( = 0 \)
Since \( f(3) = 0 \), \( (x-3) \) is a factor of \( f(x) \).

Now, we divide \( f(x) \) by \( (x-3) \) to find the other factor:
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {-4x} & {+3} \\ \cline{2-6} x-3 & x^3 & {-7x^2} & {+15x} & {-9} \\ \multicolumn{2}{r}{x^3} & {-3x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {-4x^2} & {+15x} \\ \multicolumn{2}{r}{~} & {-4x^2} & {+12x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {3x} & {-9} \\ \multicolumn{2}{r}{~} & {~} & {3x} & {-9} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x-3)(x^2-4x+3) \).
Now, factorise the quadratic part \( x^2-4x+3 \):
\( x^2-4x+3 = x^2-x-3x+3 \)
\( = x(x-1)-3(x-1) \)
\( = (x-1)(x-3) \)
Therefore, \( f(x) = (x-3)(x-1)(x-3) = (x-3)^2(x-1) \).
In simple words: First, we prove that \( (x-3) \) is a factor by putting \( x=3 \) into the polynomial and showing the result is zero. Then, we divide the polynomial by \( (x-3) \) to get a simpler expression. Finally, we break down this simpler expression into its own factors to get the polynomial fully factorized.

🎯 Exam Tip: After confirming a factor with the Remainder Theorem, use polynomial long division or synthetic division to find the quotient. The complete factorization involves factoring this quotient further if it's not linear.

Question 4. Find the value of a, if \( (x-a) \) is a factor of \( x^3 – a^2x + x + 2 \).
Answer: Let \( f(x) = x^3 - a^2x + x + 2 \).
If \( (x-a) \) is a factor of \( f(x) \), then by the Factor Theorem, \( f(a) = 0 \).
Substitute \( x = a \) into the polynomial:
\( f(a) = (a)^3 - a^2(a) + (a) + 2 = 0 \)
\( a^3 - a^3 + a + 2 = 0 \)
\( a + 2 = 0 \)
\( a = -2 \)
In simple words: Since \( (x-a) \) is a factor, we know that if we put 'a' in place of 'x' in the polynomial, the whole thing should become zero. By doing this, we create a simple equation that helps us solve for 'a'.

🎯 Exam Tip: The Factor Theorem is direct: if \( (x-a) \) is a factor of \( f(x) \), then \( f(a) = 0 \). Substitute 'a' into the polynomial and solve the resulting equation for the unknown constant.

Question 5. Use the factor theorem to factorise completely : \( x^3 + x^2 - 4x – 4 \).
Answer: Let \( f(x) = x^3 + x^2 - 4x - 4 \).
We look for factors of the constant term -4, which are \( \pm 1, \pm 2, \pm 4 \).
By trial and error:
If \( x = -1 \): \( f(-1) = (-1)^3 + (-1)^2 - 4(-1) - 4 \)
\( = -1 + 1 + 4 - 4 \)
\( = 0 \)
Since \( f(-1) = 0 \), \( (x+1) \) is a factor of \( f(x) \).

Now, divide \( f(x) \) by \( (x+1) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {+0x} & {-4} \\ \cline{2-6} x+1 & x^3 & {+x^2} & {-4x} & {-4} \\ \multicolumn{2}{r}{x^3} & {+x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {0x^2} & {-4x} \\ \multicolumn{2}{r}{~} & {~} & {0x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-4x} & {-4} \\ \multicolumn{2}{r}{~} & {~} & {-4x} & {-4} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x+1)(x^2-4) \).
The term \( x^2-4 \) is a difference of squares, \( a^2-b^2 = (a-b)(a+b) \), where \( a=x \) and \( b=2 \).
So, \( x^2-4 = (x-2)(x+2) \).
Therefore, \( x^3 + x^2 - 4x - 4 = (x+1)(x-2)(x+2) \).
In simple words: We first find one factor by testing simple numbers that divide the constant term, making the polynomial equal to zero. Once we have one factor, we divide the original polynomial by it to get a simpler quadratic expression. Then, we factorize this quadratic part completely, in this case using the difference of squares rule, to find all factors.

🎯 Exam Tip: After finding one factor using the Remainder Theorem, perform long division carefully. Always look for standard algebraic identities (like difference of squares or perfect squares) to factorize the resulting quadratic quotient quickly.

Question 6. The expression \( x^3 + ax^2 + bx + 6 \). When this expression is divided by \( (x – 3) \), it leaves a remainder 3. Find the values of a and b.
Answer: Let \( f(x) = x^3 + ax^2 + bx + 6 \).
Given that \( (x-2) \) is a factor of \( f(x) \). This means \( f(2) = 0 \).
\( (2)^3 + a(2)^2 + b(2) + 6 = 0 \)
\( 8 + 4a + 2b + 6 = 0 \)
\( 4a + 2b + 14 = 0 \)
Divide by 2:
\( 2a + b + 7 = 0 \)
\( 2a + b = -7 \) ... (i)

Given that when \( f(x) \) is divided by \( (x-3) \), the remainder is 3. This means \( f(3) = 3 \).
\( (3)^3 + a(3)^2 + b(3) + 6 = 3 \)
\( 27 + 9a + 3b + 6 = 3 \)
\( 9a + 3b + 33 = 3 \)
\( 9a + 3b = 3 - 33 \)
\( 9a + 3b = -30 \)
Divide by 3:
\( 3a + b = -10 \) ... (ii)

Subtract equation (i) from equation (ii):
\( (3a + b) - (2a + b) = -10 - (-7) \)
\( 3a + b - 2a - b = -10 + 7 \)
\( a = -3 \)

Substitute \( a = -3 \) into equation (i):
\( 2(-3) + b = -7 \)
\( -6 + b = -7 \)
\( b = -7 + 6 \)
\( b = -1 \)
Therefore, \( a = -3 \) and \( b = -1 \).
In simple words: We use the Remainder Theorem twice. First, because \( (x-2) \) is a factor, we know that putting \( x=2 \) into the polynomial makes it zero. Second, because dividing by \( (x-3) \) leaves a remainder of 3, putting \( x=3 \) into the polynomial gives us 3. These two facts give us two equations with 'a' and 'b'. We then solve these equations together to find the values of 'a' and 'b'.

🎯 Exam Tip: Carefully distinguish between "is a factor" (remainder is 0) and "leaves a remainder R". Each condition generates an equation, and solving these simultaneous equations is the standard approach for finding multiple unknowns.

Question 7. Show that \( 2x + 7 \) is a factor of \( 2x^3 + 5x^2 – 11x – 14 \). Hence factorise the given expression completely, using factor theorem.
Answer: Let \( f(x) = 2x^3 + 5x^2 - 11x - 14 \).
To show that \( (2x+7) \) is a factor, we must show that \( f(-\frac{7}{2}) = 0 \).
From \( 2x+7=0 \), we get \( x = -\frac{7}{2} \).
\( f(-\frac{7}{2}) = 2(-\frac{7}{2})^3 + 5(-\frac{7}{2})^2 - 11(-\frac{7}{2}) - 14 \)
\( = 2(-\frac{343}{8}) + 5(\frac{49}{4}) + \frac{77}{2} - 14 \)
\( = -\frac{343}{4} + \frac{245}{4} + \frac{77}{2} - 14 \)
To combine, find a common denominator, which is 4:
\( = -\frac{343}{4} + \frac{245}{4} + \frac{77 \times 2}{2 \times 2} - \frac{14 \times 4}{1 \times 4} \)
\( = -\frac{343}{4} + \frac{245}{4} + \frac{154}{4} - \frac{56}{4} \)
\( = \frac{-343 + 245 + 154 - 56}{4} \)
\( = \frac{399 - 399}{4} \)
\( = \frac{0}{4} \)
\( = 0 \)
Since \( f(-\frac{7}{2}) = 0 \), \( (2x+7) \) is a factor of \( f(x) \).

Now, we divide \( f(x) \) by \( (2x+7) \) to find the other factor:
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {-x} & {-2} \\ \cline{2-6} 2x+7 & 2x^3 & {+5x^2} & {-11x} & {-14} \\ \multicolumn{2}{r}{2x^3} & {+7x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {-2x^2} & {-11x} \\ \multicolumn{2}{r}{~} & {-2x^2} & {-7x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {-4x} & {-14} \\ \multicolumn{2}{r}{~} & {~} & {-4x} & {-14} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (2x+7)(x^2-x-2) \).
Now, factorise the quadratic part \( x^2-x-2 \):
\( x^2-x-2 = x^2-2x+x-2 \)
\( = x(x-2)+1(x-2) \)
\( = (x-2)(x+1) \)
Therefore, \( 2x^3 + 5x^2 - 11x - 14 = (2x+7)(x-2)(x+1) \).
In simple words: To show \( (2x+7) \) is a factor, we set \( 2x+7=0 \) and find the value of \( x \), then plug that value into the polynomial. If the result is zero, it's a factor. Next, we divide the original polynomial by \( (2x+7) \) using long division. This gives us a quadratic expression, which we then factorize completely to find all the individual factors.

🎯 Exam Tip: When the factor is of the form \( (ax+b) \), ensure you correctly find the value of \( x \) (i.e., \( x = -b/a \)) for substitution in the Remainder Theorem. Long division must be performed with this \( (ax+b) \) factor, not just a simple \( (x-k) \).

Question 8. Show that \( (x – 1) \) is a factor of \( x^3 – 7x^2 + 14x – 8 \). Hence completely factorise the above expression.
Answer: Let \( f(x) = x^3 - 7x^2 + 14x - 8 \).
To show that \( (x-1) \) is a factor, we must show that \( f(1) = 0 \).
\( f(1) = (1)^3 - 7(1)^2 + 14(1) - 8 \)
\( = 1 - 7 + 14 - 8 \)
\( = 15 - 15 \)
\( = 0 \)
Since \( f(1) = 0 \), \( (x-1) \) is a factor of \( f(x) \).

Now, we divide \( f(x) \) by \( (x-1) \) to find the other factor:
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{x^2} & {-6x} & {+8} \\ \cline{2-6} x-1 & x^3 & {-7x^2} & {+14x} & {-8} \\ \multicolumn{2}{r}{x^3} & {-x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {-6x^2} & {+14x} \\ \multicolumn{2}{r}{~} & {-6x^2} & {+6x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {8x} & {-8} \\ \multicolumn{2}{r}{~} & {~} & {8x} & {-8} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x-1)(x^2-6x+8) \).
Now, factorise the quadratic part \( x^2-6x+8 \):
\( x^2-6x+8 = x^2-2x-4x+8 \)
\( = x(x-2)-4(x-2) \)
\( = (x-2)(x-4) \)
Therefore, \( x^3 - 7x^2 + 14x - 8 = (x-1)(x-2)(x-4) \).
In simple words: First, we use the factor theorem to prove that \( (x-1) \) is indeed a factor by showing that the polynomial becomes zero when \( x=1 \). After that, we perform long division to divide the polynomial by this factor, which leaves us with a quadratic expression. Finally, we factorize this quadratic expression to get the full list of factors for the original polynomial.

🎯 Exam Tip: The prompt explicitly asks to "completely factorise." This means you must continue factoring until all factors are linear (i.e., of the form \( (ax+b) \)). Do not stop after finding the quadratic quotient.

Question 9. If \( (x – 2) \) is a factor of \( 2x^3 – x^2 – px – 2 \).
(i) Find the value of p.
(ii) With the value of p, factorise the above expression completely.
Answer: Let \( f(x) = 2x^3 - x^2 - px - 2 \).
(i) Since \( (x-2) \) is a factor of \( f(x) \), by the Factor Theorem, \( f(2) = 0 \).
\( 2(2)^3 - (2)^2 - p(2) - 2 = 0 \)
\( 2(8) - 4 - 2p - 2 = 0 \)
\( 16 - 4 - 2p - 2 = 0 \)
\( 10 - 2p = 0 \)
\( 2p = 10 \)
\( p = \frac{10}{2} \)
\( p = 5 \)

(ii) Now, substitute \( p=5 \) back into the polynomial:
\( f(x) = 2x^3 - x^2 - 5x - 2 \)
We know \( (x-2) \) is a factor. Now we divide \( f(x) \) by \( (x-2) \):
\[ \begin{array}{c|cc cc cc} \multicolumn{2}{r}{2x^2} & {+3x} & {+1} \\ \cline{2-6} x-2 & 2x^3 & {-x^2} & {-5x} & {-2} \\ \multicolumn{2}{r}{2x^3} & {-4x^2} \\ \cline{2-3} \multicolumn{2}{r}{~} & {3x^2} & {-5x} \\ \multicolumn{2}{r}{~} & {3x^2} & {-6x} \\ \cline{3-4} \multicolumn{2}{r}{~} & {~} & {x} & {-2} \\ \multicolumn{2}{r}{~} & {~} & {x} & {-2} \\ \cline{4-5} \multicolumn{2}{r}{~} & {~} & {~} & {0} \\ \end{array} \]
So, \( f(x) = (x-2)(2x^2+3x+1) \).
Now, factorise the quadratic part \( 2x^2+3x+1 \):
We need two numbers that multiply to \( 2 \times 1 = 2 \) and add to 3. These are 2 and 1.
\( 2x^2+3x+1 = 2x^2+2x+x+1 \)
\( = 2x(x+1)+1(x+1) \)
\( = (x+1)(2x+1) \)
Therefore, \( f(x) = (x-2)(x+1)(2x+1) \).
In simple words: For part (i), we use the factor theorem: if \( (x-2) \) is a factor, then putting \( x=2 \) into the polynomial should make it zero. This allows us to find the value of 'p'. For part (ii), once 'p' is known, we write down the full polynomial and divide it by the known factor \( (x-2) \) to get a quadratic expression. We then break this quadratic into two simpler factors to complete the factorization.

🎯 Exam Tip: When solving for 'p' and then factoring, ensure the value of 'p' is correctly substituted into the polynomial before proceeding with long division. Any error in 'p' will lead to incorrect factors.

Question 10. Given that \( (x + 2) \) and \( (x + 3) \) are factors of \( 2x^3 + ax^2 + 7x – b \). Determine the values of a and b.
Answer: Let \( f(x) = 2x^3 + ax^2 + 7x - b \).
Since \( (x+2) \) is a factor, \( f(-2) = 0 \).
\( 2(-2)^3 + a(-2)^2 + 7(-2) - b = 0 \)
\( 2(-8) + a(4) - 14 - b = 0 \)
\( -16 + 4a - 14 - b = 0 \)
\( 4a - b - 30 = 0 \)
\( 4a - b = 30 \) ... (i)

Since \( (x+3) \) is a factor, \( f(-3) = 0 \).
\( 2(-3)^3 + a(-3)^2 + 7(-3) - b = 0 \)
\( 2(-27) + a(9) - 21 - b = 0 \)
\( -54 + 9a - 21 - b = 0 \)
\( 9a - b - 75 = 0 \)
\( 9a - b = 75 \) ... (ii)

Subtract equation (i) from equation (ii):
\( (9a - b) - (4a - b) = 75 - 30 \)
\( 9a - b - 4a + b = 45 \)
\( 5a = 45 \)
\( a = \frac{45}{5} \)
\( a = 9 \)

Substitute \( a = 9 \) into equation (i):
\( 4(9) - b = 30 \)
\( 36 - b = 30 \)
\( -b = 30 - 36 \)
\( -b = -6 \)
\( b = 6 \)
Therefore, \( a = 9 \) and \( b = 6 \).
In simple words: Given that two expressions are factors of the polynomial, we apply the Factor Theorem twice. For each factor, we find the value of 'x' that makes the factor zero, and then substitute this value into the polynomial, setting the result to zero. This gives us two linear equations with 'a' and 'b'. We then solve these two equations simultaneously to find the values of 'a' and 'b'.

🎯 Exam Tip: This type of problem often leads to a system of two linear equations in two variables. Be careful with signs when substituting values and solving the system. Double-check your arithmetic to avoid errors.

Question 11. When divided by \( x - 3 \), the polynomials \( x^3 - px^2 + x + 6 \) and \( 2x^3 - x^2 - (p + 3)x - 6 \) leave the same remainder. Find the value of 'p'.
Answer: Let the first polynomial be \( f(x) = x^3 - px^2 + x + 6 \).
When \( f(x) \) is divided by \( x - 3 \), the remainder is \( f(3) \).
\( f(3) = (3)^3 - p(3)^2 + (3) + 6 \)
\( = 27 - 9p + 3 + 6 \)
\( = 36 - 9p \)
Let the second polynomial be \( g(x) = 2x^3 - x^2 - (p + 3)x - 6 \).
When \( g(x) \) is divided by \( x - 3 \), the remainder is \( g(3) \).
\( g(3) = 2(3)^3 - (3)^2 - (p + 3)(3) - 6 \)
\( = 2(27) - 9 - 3p - 9 - 6 \)
\( = 54 - 9 - 3p - 9 - 6 \)
\( = 30 - 3p \)
Since both remainders are equal:
\( 36 - 9p = 30 - 3p \)
\( \implies \) \( 36 - 30 = 9p - 3p \)
\( \implies \) \( 6 = 6p \)
\( \implies \) \( p = 1 \)
Therefore, the value of 'p' is 1. This means the common root works for both polynomial expressions.
In simple words: We find the remainder for both number patterns when divided by \( x-3 \). Since the problem says these remainders are the same, we set them equal to each other. Solving this simple equation gives us the value of 'p'.

🎯 Exam Tip: Remember to substitute the value of 'x' (obtained from setting the divisor to zero) into all terms of the polynomial, including those with the unknown constant, to find the correct remainder.

Question 12. Use factor theorem to factorise completely the following expression: \( 2x^3 + x^2 - 13x + 6 \).
Answer: Let \( f(x) = 2x^3 + x^2 - 13x + 6 \).
We need to find a value for \( x \) that makes \( f(x) = 0 \). We can test integer factors of the constant term, which is 6: \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Let's try \( x = 2 \):
\( f(2) = 2(2)^3 + (2)^2 - 13(2) + 6 \)
\( = 2(8) + 4 - 26 + 6 \)
\( = 16 + 4 - 26 + 6 \)
\( = 26 - 26 = 0 \)
Since \( f(2) = 0 \), by the Factor Theorem, \( (x - 2) \) is a factor of \( f(x) \).
Now, we divide \( f(x) \) by \( (x - 2) \) using long division to find the other factor:
\[ \begin{array}{r} x^2 + 5x - 3 \\ x-2\overline{|2x^3 + x^2 - 13x + 6} \\ - \underline{(2x^3 - 4x^2)} \\ 5x^2 - 13x \\ - \underline{(5x^2 - 10x)} \\ -3x + 6 \\ - \underline{(-3x + 6)} \\ 0 \\ \end{array} \] So, \( f(x) = (x - 2)(2x^2 + 5x - 3) \).
Next, we factorise the quadratic expression \( 2x^2 + 5x - 3 \). We look for two numbers that multiply to \( 2 \times (-3) = -6 \) and add up to 5. These numbers are 6 and -1.
\( 2x^2 + 5x - 3 = 2x^2 + 6x - x - 3 \)
\( = 2x(x + 3) - 1(x + 3) \)
\( = (2x - 1)(x + 3) \)
Thus, the complete factorisation of \( 2x^3 + x^2 - 13x + 6 \) is \( (x - 2)(2x - 1)(x + 3) \). This method helps break down complex polynomials into simpler forms.
In simple words: First, we guess a number that makes the whole expression zero. If it works, then \( x \) minus that number is a factor. Then, we divide the original expression by this factor. This gives us a simpler quadratic expression. Finally, we break down this quadratic into two more factors.

🎯 Exam Tip: When using the Factor Theorem, systematically try simple integer factors of the constant term (like \( \pm 1, \pm 2, \ldots \)). Once a factor is found, long division or synthetic division is crucial to find the remaining polynomial to factorise further.

Question 13. Find the value of 'k' if \( (x - 2) \) is a factor of \( x^3 + 2x^2 - kx + 10 \). Hence determine whether \( (x + 5) \) is also a factor.
Answer: Let the polynomial be \( f(x) = x^3 + 2x^2 - kx + 10 \).
Given that \( (x - 2) \) is a factor of \( f(x) \). By the Factor Theorem, if \( (x - 2) \) is a factor, then \( f(2) = 0 \).
Substitute \( x = 2 \) into the polynomial:
\( f(2) = (2)^3 + 2(2)^2 - k(2) + 10 = 0 \)
\( 8 + 2(4) - 2k + 10 = 0 \)
\( 8 + 8 - 2k + 10 = 0 \)
\( 26 - 2k = 0 \)
\( \implies \) \( 2k = 26 \)
\( \implies \) \( k = 13 \)
So, the value of \( k \) is 13.
Now, the polynomial is \( f(x) = x^3 + 2x^2 - 13x + 10 \).
We need to determine if \( (x + 5) \) is also a factor. If \( (x + 5) \) is a factor, then \( f(-5) \) should be 0.
Substitute \( x = -5 \) into the polynomial:
\( f(-5) = (-5)^3 + 2(-5)^2 - 13(-5) + 10 \)
\( = -125 + 2(25) + 65 + 10 \)
\( = -125 + 50 + 65 + 10 \)
\( = -125 + 125 \)
\( = 0 \)
Since \( f(-5) = 0 \), \( (x + 5) \) is also a factor of the polynomial. This indicates that \( k=13 \) is correctly found and the polynomial can be further factored.
In simple words: First, we use the given factor \( (x-2) \) to find \( k \) by setting the polynomial to zero when \( x=2 \). Once we have \( k \), we check if \( (x+5) \) is another factor by setting \( x=-5 \). If the polynomial also becomes zero with \( x=-5 \), then \( (x+5) \) is indeed a factor.

🎯 Exam Tip: When you find the value of a constant like 'k', it's good practice to re-write the full polynomial before testing for additional factors. This reduces errors in calculation.

Question 14. Using a Remainder Theorem factorise completely the following polynomial: \( 3x^3 + 2x^2 - 19x + 6 \).
Answer: Let the polynomial be \( P(x) = 3x^3 + 2x^2 - 19x + 6 \).
We need to find a value of \( x \) that makes \( P(x) = 0 \). We can test integer factors of the constant term, which is 6: \( \pm 1, \pm 2, \pm 3, \pm 6 \). We also consider factors of \( \frac{\text{constant term}}{\text{leading coefficient}} \), i.e., \( \frac{6}{3} = 2 \).
Let's try \( x = 2 \):
\( P(2) = 3(2)^3 + 2(2)^2 - 19(2) + 6 \)
\( = 3(8) + 2(4) - 38 + 6 \)
\( = 24 + 8 - 38 + 6 \)
\( = 38 - 38 = 0 \)
Since \( P(2) = 0 \), by the Factor Theorem, \( (x - 2) \) is a factor of \( P(x) \).
Now, we divide \( P(x) \) by \( (x - 2) \). We can do this by grouping terms:
\( 3x^3 + 2x^2 - 19x + 6 \)
We want to introduce \( (x-2) \) as a common factor.
\( = 3x^3 - 6x^2 + 8x^2 - 16x - 3x + 6 \) (We split \( 2x^2 \) into \( -6x^2+8x^2 \) and \( -19x \) into \( -16x-3x \))
\( = 3x^2(x - 2) + 8x(x - 2) - 3(x - 2) \)
\( = (x - 2)(3x^2 + 8x - 3) \)
Next, we factorise the quadratic expression \( 3x^2 + 8x - 3 \). We look for two numbers that multiply to \( 3 \times (-3) = -9 \) and add up to 8. These numbers are 9 and -1.
\( 3x^2 + 8x - 3 = 3x^2 + 9x - x - 3 \)
\( = 3x(x + 3) - 1(x + 3) \)
\( = (3x - 1)(x + 3) \)
Thus, the complete factorisation of \( 3x^3 + 2x^2 - 19x + 6 \) is \( (x - 2)(3x - 1)(x + 3) \). Breaking a polynomial down this way simplifies solving related equations.
In simple words: We find one number that makes the expression zero, which gives us the first factor. Then, we use special grouping to pull out this factor from the rest of the expression, leaving a simpler quadratic part. Finally, we break down this quadratic part into two more factors.

🎯 Exam Tip: When factorising by grouping after finding the first factor, adjust the middle terms (like \( 2x^2 \) and \( -19x \)) so that the factor \( (x-2) \) can be easily pulled out from each group. This requires a bit of foresight.

Question 15. If \( (x - 2) \) is a factor of the expression \( 2x^3 + ax^2 + bx - 14 \) and when the expression is divided by \( (x - 3) \), it leaves a remainder 52, find the values of a and b.
Answer: Let the polynomial be \( f(x) = 2x^3 + ax^2 + bx - 14 \).
Since \( (x - 2) \) is a factor of \( f(x) \), by the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( 2(2)^3 + a(2)^2 + b(2) - 14 = 0 \)
\( 2(8) + 4a + 2b - 14 = 0 \)
\( 16 + 4a + 2b - 14 = 0 \)
\( 4a + 2b + 2 = 0 \)
Dividing by 2 gives: \( 2a + b + 1 = 0 \)
\( \implies \) \( 2a + b = -1 \) ... (i)
When \( f(x) \) is divided by \( (x - 3) \), the remainder is 52. By the Remainder Theorem, \( f(3) = 52 \).
Substitute \( x = 3 \) into \( f(x) \):
\( 2(3)^3 + a(3)^2 + b(3) - 14 = 52 \)
\( 2(27) + 9a + 3b - 14 = 52 \)
\( 54 + 9a + 3b - 14 = 52 \)
\( 9a + 3b + 40 = 52 \)
\( \implies \) \( 9a + 3b = 52 - 40 \)
\( \implies \) \( 9a + 3b = 12 \)
Dividing by 3 gives: \( 3a + b = 4 \) ... (ii)
Now we have a system of two linear equations:
1. \( 2a + b = -1 \)
2. \( 3a + b = 4 \)
Subtract equation (i) from equation (ii):
\( (3a + b) - (2a + b) = 4 - (-1) \)
\( 3a - 2a + b - b = 4 + 1 \)
\( a = 5 \)
Substitute the value of \( a = 5 \) into equation (i):
\( 2(5) + b = -1 \)
\( 10 + b = -1 \)
\( \implies \) \( b = -1 - 10 \)
\( \implies \) \( b = -11 \)
Therefore, the values are \( a = 5 \) and \( b = -11 \). This process involves solving a system of linear equations derived from polynomial properties.
In simple words: We use the given conditions (one factor and one remainder) to make two equations with 'a' and 'b'. Then, we solve these two equations together to find the exact values for 'a' and 'b'.

🎯 Exam Tip: Carefully set up the system of linear equations from the given conditions. A small error in signs or arithmetic in the initial substitution can lead to incorrect values for 'a' and 'b'.

Question 16. Using the Remainder and Factor theorem, factorise completely the following polynomial: \( x^3 + 10x^2 - 37x + 26 \).
Answer: Let the polynomial be \( f(x) = x^3 + 10x^2 - 37x + 26 \).
We need to find a value of \( x \) that makes \( f(x) = 0 \). We test integer factors of the constant term, 26: \( \pm 1, \pm 2, \pm 13, \pm 26 \).
Let's try \( x = 1 \):
\( f(1) = (1)^3 + 10(1)^2 - 37(1) + 26 \)
\( = 1 + 10 - 37 + 26 \)
\( = 37 - 37 = 0 \)
Since \( f(1) = 0 \), by the Factor Theorem, \( (x - 1) \) is a factor of \( f(x) \).
Now, we perform long division of \( f(x) \) by \( (x - 1) \) to find the other factor:
\[ \begin{array}{r} x^2 + 11x - 26 \\ x-1\overline{|x^3 + 10x^2 - 37x + 26} \\ - \underline{(x^3 - x^2)} \\ 11x^2 - 37x \\ - \underline{(11x^2 - 11x)} \\ -26x + 26 \\ - \underline{(-26x + 26)} \\ 0 \\ \end{array} \] So, \( f(x) = (x - 1)(x^2 + 11x - 26) \).
Next, we factorise the quadratic expression \( x^2 + 11x - 26 \). We look for two numbers that multiply to -26 and add up to 11. These numbers are 13 and -2.
\( x^2 + 11x - 26 = x^2 + 13x - 2x - 26 \)
\( = x(x + 13) - 2(x + 13) \)
\( = (x - 2)(x + 13) \)
Thus, the complete factorisation of \( x^3 + 10x^2 - 37x + 26 \) is \( (x - 1)(x - 2)(x + 13) \). Each factor represents a root of the polynomial equation.
In simple words: First, we find a number that makes the expression zero, which gives us one factor. Then, we divide the polynomial by this factor to get a simpler quadratic expression. Finally, we break down this quadratic into two more factors to get the full list of factors.

🎯 Exam Tip: After finding one factor, remember to perform long division carefully. Ensure all terms are aligned and signs are correctly handled during subtraction to arrive at the correct quadratic quotient.

Question 17. Find 'a' if the two polynomials \( ax^3 + 3x^2 - 9 \) and \( 2x^3 + 4x + a \), leave the same remainder when divided by \( x + 3 \).
Answer: Let the first polynomial be \( p(x) = ax^3 + 3x^2 - 9 \).
Let the second polynomial be \( q(x) = 2x^3 + 4x + a \).
When divided by \( x + 3 \), the remainder for \( p(x) \) is \( p(-3) \) and for \( q(x) \) is \( q(-3) \).
First, calculate \( p(-3) \):
\( p(-3) = a(-3)^3 + 3(-3)^2 - 9 \)
\( = a(-27) + 3(9) - 9 \)
\( = -27a + 27 - 9 \)
\( = -27a + 18 \)
Next, calculate \( q(-3) \):
\( q(-3) = 2(-3)^3 + 4(-3) + a \)
\( = 2(-27) - 12 + a \)
\( = -54 - 12 + a \)
\( = -66 + a \)
Given that both polynomials leave the same remainder, we can set \( p(-3) = q(-3) \):
\( -27a + 18 = -66 + a \)
\( \implies \) \( 18 + 66 = a + 27a \)
\( \implies \) \( 84 = 28a \)
\( \implies \) \( a = \frac{84}{28} \)
\( \implies \) \( a = 3 \)
Therefore, the value of 'a' is 3. This problem shows how polynomial remainders can be used to solve for unknown coefficients.
In simple words: We find what's left over when we divide each polynomial by \( x+3 \). Since these leftovers are equal, we set the two remainder expressions equal to each other. Solving this equation tells us the value of 'a'.

🎯 Exam Tip: Be very careful with signs when substituting negative values into polynomials, especially with powers. A mistake in a negative sign can change the entire result.

Question 18. Using remainder theorem, find the value of k if on dividing \( 2x^3 + 3x^2 - kx + 5 \) by \( x - 2 \), leaves a remainder 7.
Answer: Let the polynomial be \( f(x) = 2x^3 + 3x^2 - kx + 5 \).
When \( f(x) \) is divided by \( x - 2 \), the remainder is 7. According to the Remainder Theorem, if a polynomial \( f(x) \) is divided by \( (x - a) \), the remainder is \( f(a) \).
Here, \( a = 2 \), so \( f(2) = 7 \).
Substitute \( x = 2 \) into the polynomial expression:
\( f(2) = 2(2)^3 + 3(2)^2 - k(2) + 5 = 7 \)
\( 2(8) + 3(4) - 2k + 5 = 7 \)
\( 16 + 12 - 2k + 5 = 7 \)
\( 33 - 2k = 7 \)
\( \implies \) \( 33 - 7 = 2k \)
\( \implies \) \( 26 = 2k \)
\( \implies \) \( k = 13 \)
Thus, the value of \( k \) is 13. Knowing \( k \) allows us to fully define the polynomial.
In simple words: We put the number 2 into the expression because dividing by \( x-2 \) means \( x \) is 2. The problem tells us the leftover number is 7. So, we set the result of putting 2 into the expression equal to 7 and then solve for \( k \).

🎯 Exam Tip: The Remainder Theorem is a powerful tool to find unknown coefficients. Ensure you correctly identify the value of 'a' from the divisor \( (x-a) \) and equate the polynomial's value at 'a' to the given remainder.

Self Evaluation And Revision (Latest ICSE Questions)

Question 2. Find the value of the constants a and b if \( (x - 2) \) and \( (x + 3) \) are both factors of the expression \( x^3 + ax^2 + bx - 12 \).
Answer: Let the polynomial be \( f(x) = x^3 + ax^2 + bx - 12 \).
Since \( (x - 2) \) is a factor, by the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into the polynomial:
\( f(2) = (2)^3 + a(2)^2 + b(2) - 12 = 0 \)
\( 8 + 4a + 2b - 12 = 0 \)
\( 4a + 2b - 4 = 0 \)
Dividing by 2 gives: \( 2a + b - 2 = 0 \)
\( \implies \) \( 2a + b = 2 \) ... (i)
Since \( (x + 3) \) is also a factor, by the Factor Theorem, \( f(-3) = 0 \).
Substitute \( x = -3 \) into the polynomial:
\( f(-3) = (-3)^3 + a(-3)^2 + b(-3) - 12 = 0 \)
\( -27 + 9a - 3b - 12 = 0 \)
\( 9a - 3b - 39 = 0 \)
Dividing by 3 gives: \( 3a - b - 13 = 0 \)
\( \implies \) \( 3a - b = 13 \) ... (ii)
Now we have a system of two linear equations:
1. \( 2a + b = 2 \)
2. \( 3a - b = 13 \)
Add equation (i) and equation (ii):
\( (2a + b) + (3a - b) = 2 + 13 \)
\( 5a = 15 \)
\( \implies \) \( a = 3 \)
Substitute the value of \( a = 3 \) into equation (i):
\( 2(3) + b = 2 \)
\( 6 + b = 2 \)
\( \implies \) \( b = 2 - 6 \)
\( \implies \) \( b = -4 \)
Therefore, the values of the constants are \( a = 3 \) and \( b = -4 \). This systematic approach helps solve for multiple unknowns in polynomial problems.
In simple words: We know that if \( (x-2) \) and \( (x+3) \) are factors, then the polynomial must be zero when \( x=2 \) and when \( x=-3 \). We use these two facts to create two separate equations involving 'a' and 'b'. Then, we solve these two equations together to find what 'a' and 'b' are.

🎯 Exam Tip: When given multiple factors, each factor provides an equation. Remember to solve the resulting system of linear equations accurately, as a single calculation error can lead to incorrect constant values.

Question 3. Using factor theorem, show that \( (x - 3) \) is a factor of \( x^3 - 7x^2 + 15x - 9 \). Hence factorise the given expression completely.
Answer: Let the polynomial be \( f(x) = x^3 - 7x^2 + 15x - 9 \).
To show that \( (x - 3) \) is a factor, we need to show that \( f(3) = 0 \), according to the Factor Theorem.
Substitute \( x = 3 \) into the polynomial:
\( f(3) = (3)^3 - 7(3)^2 + 15(3) - 9 \)
\( = 27 - 7(9) + 45 - 9 \)
\( = 27 - 63 + 45 - 9 \)
\( = (27 + 45) - (63 + 9) \)
\( = 72 - 72 = 0 \)
Since \( f(3) = 0 \), \( (x - 3) \) is indeed a factor of \( x^3 - 7x^2 + 15x - 9 \).
Now, we perform long division of \( f(x) \) by \( (x - 3) \) to find the other factor:
\[ \begin{array}{r} x^2 - 4x + 3 \\ x-3\overline{|x^3 - 7x^2 + 15x - 9} \\ - \underline{(x^3 - 3x^2)} \\ -4x^2 + 15x \\ - \underline{(-4x^2 + 12x)} \\ 3x - 9 \\ - \underline{(3x - 9)} \\ 0 \\ \end{array} \] So, \( f(x) = (x - 3)(x^2 - 4x + 3) \).
Next, we factorise the quadratic expression \( x^2 - 4x + 3 \). We look for two numbers that multiply to 3 and add up to -4. These numbers are -1 and -3.
\( x^2 - 4x + 3 = x^2 - x - 3x + 3 \)
\( = x(x - 1) - 3(x - 1) \)
\( = (x - 1)(x - 3) \)
Thus, the complete factorisation of \( x^3 - 7x^2 + 15x - 9 \) is \( (x - 3)(x - 1)(x - 3) \), which can also be written as \( (x - 3)^2(x - 1) \). This shows how one factor can repeat in the factorization.
In simple words: First, we put \( x=3 \) into the expression to check if it becomes zero. If it does, then \( (x-3) \) is a factor. Then, we divide the big expression by \( (x-3) \) to get a smaller, simpler quadratic expression. Finally, we break down this quadratic into two parts to get all the factors.

🎯 Exam Tip: When showing a factor, explicitly state that \( f(a) = 0 \) implies \( (x-a) \) is a factor. For complete factorisation, always factor the quadratic quotient if it's reducible.

Question 4. Find the value of a, if \( (x- a) \) is a factor of \( x^3 - a^2x + x + 2 \).
Answer: Let the polynomial be \( f(x) = x^3 - a^2x + x + 2 \).
Given that \( (x - a) \) is a factor of \( f(x) \). By the Factor Theorem, if \( (x - a) \) is a factor, then \( f(a) = 0 \).
Substitute \( x = a \) into the polynomial expression:
\( f(a) = (a)^3 - a^2(a) + (a) + 2 = 0 \)
\( a^3 - a^3 + a + 2 = 0 \)
\( a + 2 = 0 \)
\( \implies \) \( a = -2 \)
Therefore, the value of 'a' is -2. This result is straightforward once the Factor Theorem is applied correctly.
In simple words: If \( (x-a) \) is a factor, it means if we put 'a' in place of 'x', the whole expression should become zero. So, we put 'a' into the expression, make it equal to zero, and then solve to find what 'a' must be.

🎯 Exam Tip: Be careful with signs and exponents when substituting variables into a polynomial. \( a^2(a) \) simplifies to \( a^3 \), which cancels out with \( a^3 \) in this problem.

Question 5. Use the factor theorem to factorise completely: \( x^3 + x^2 - 4x - 4 \).
Answer: Let the polynomial be \( f(x) = x^3 + x^2 - 4x - 4 \).
We look for a value of \( x \) that makes \( f(x) = 0 \). We test integer factors of the constant term, 4: \( \pm 1, \pm 2, \pm 4 \).
Let's try \( x = -1 \):
\( f(-1) = (-1)^3 + (-1)^2 - 4(-1) - 4 \)
\( = -1 + 1 + 4 - 4 \)
\( = 0 \)
Since \( f(-1) = 0 \), by the Factor Theorem, \( (x - (-1)) = (x + 1) \) is a factor of \( f(x) \).
Now, we perform long division of \( f(x) \) by \( (x + 1) \) to find the other factor:
\[ \begin{array}{r} x^2 - 4 \\ x+1\overline{|x^3 + x^2 - 4x - 4} \\ - \underline{(x^3 + x^2)} \\ -4x - 4 \\ - \underline{(-4x - 4)} \\ 0 \\ \end{array} \] So, \( f(x) = (x + 1)(x^2 - 4) \).
The quadratic expression \( x^2 - 4 \) is a difference of squares, which can be factored as \( (x)^2 - (2)^2 = (x - 2)(x + 2) \).
Thus, the complete factorisation of \( x^3 + x^2 - 4x - 4 \) is \( (x + 1)(x - 2)(x + 2) \). This highlights a common pattern for cubic polynomial factorization.
In simple words: First, we find a number that makes the expression equal to zero, which gives us the first factor. Then, we divide the original expression by this factor to get a simpler quadratic part. We notice this quadratic is a special type (difference of squares), so we break it into two more factors easily.

🎯 Exam Tip: Always look for common algebraic identities like the difference of squares \( (a^2 - b^2) = (a-b)(a+b) \) or perfect square trinomials when factorising quadratic expressions, as they can simplify the process.

Question 6. The expression \( x^3 + ax^2 + bx + 6 \). If \( (x - 2) \) is a factor and when this expression is divided by \( (x - 3) \), it leaves a remainder 3. Find the values of a and b.
Answer: Let the polynomial be \( f(x) = x^3 + ax^2 + bx + 6 \).
Given that \( (x - 2) \) is a factor of \( f(x) \). By the Factor Theorem, \( f(2) = 0 \).
Substitute \( x = 2 \) into \( f(x) \):
\( f(2) = (2)^3 + a(2)^2 + b(2) + 6 = 0 \)
\( 8 + 4a + 2b + 6 = 0 \)
\( 4a + 2b + 14 = 0 \)
Dividing by 2 gives: \( 2a + b + 7 = 0 \)
\( \implies \) \( 2a + b = -7 \) ... (i)
When \( f(x) \) is divided by \( (x - 3) \), the remainder is 3. By the Remainder Theorem, \( f(3) = 3 \).
Substitute \( x = 3 \) into \( f(x) \):
\( f(3) = (3)^3 + a(3)^2 + b(3) + 6 = 3 \)
\( 27 + 9a + 3b + 6 = 3 \)
\( 9a + 3b + 33 = 3 \)
\( \implies \) \( 9a + 3b = 3 - 33 \)
\( \implies \) \( 9a + 3b = -30 \)
Dividing by 3 gives: \( 3a + b = -10 \) ... (ii)
Now we have a system of two linear equations:
1. \( 2a + b = -7 \)
2. \( 3a + b = -10 \)
Subtract equation (i) from equation (ii):
\( (3a + b) - (2a + b) = -10 - (-7) \)
\( 3a - 2a + b - b = -10 + 7 \)
\( a = -3 \)
Substitute the value of \( a = -3 \) into equation (i):
\( 2(-3) + b = -7 \)
\( -6 + b = -7 \)
\( \implies \) \( b = -7 + 6 \)
\( \implies \) \( b = -1 \)
Therefore, the values are \( a = -3 \) and \( b = -1 \). These values satisfy both conditions given in the problem.
In simple words: We get two facts from the problem: one factor and one remainder. Each fact gives us an equation that includes 'a' and 'b'. We then solve these two equations together to find the specific values for 'a' and 'b'.

🎯 Exam Tip: Always clearly label your equations (i) and (ii) when solving simultaneous equations. This helps keep your working organised and easier to follow, reducing the chance of errors.

Question 7. Show that \( 2x + 7 \) is a factor of \( 2x^3 + 5x^2 - 11x - 14 \). Hence factorise the given expression completely, using factor theorem.
Answer: Let the polynomial be \( f(x) = 2x^3 + 5x^2 - 11x - 14 \).
To show that \( (2x + 7) \) is a factor, we need to show that \( f(-\frac{7}{2}) = 0 \), according to the Factor Theorem. This is because if \( 2x + 7 = 0 \), then \( 2x = -7 \), so \( x = -\frac{7}{2} \).
Substitute \( x = -\frac{7}{2} \) into the polynomial:
\( f(-\frac{7}{2}) = 2(-\frac{7}{2})^3 + 5(-\frac{7}{2})^2 - 11(-\frac{7}{2}) - 14 \)
\( = 2(-\frac{343}{8}) + 5(\frac{49}{4}) + \frac{77}{2} - 14 \)
\( = -\frac{343}{4} + \frac{245}{4} + \frac{154}{4} - \frac{56}{4} \) (using a common denominator of 4)
\( = \frac{-343 + 245 + 154 - 56}{4} \)
\( = \frac{399 - 399}{4} \)
\( = \frac{0}{4} = 0 \)
Since \( f(-\frac{7}{2}) = 0 \), \( (2x + 7) \) is indeed a factor of \( 2x^3 + 5x^2 - 11x - 14 \).
Now, we perform long division of \( f(x) \) by \( (2x + 7) \) to find the other factor:
\[ \begin{array}{r} x^2 - x - 2 \\ 2x+7\overline{|2x^3 + 5x^2 - 11x - 14} \\ - \underline{(2x^3 + 7x^2)} \\ -2x^2 - 11x \\ - \underline{(-2x^2 - 7x)} \\ -4x - 14 \\ - \underline{(-4x - 14)} \\ 0 \\ \end{array} \] So, \( f(x) = (2x + 7)(x^2 - x - 2) \).
Next, we factorise the quadratic expression \( x^2 - x - 2 \). We look for two numbers that multiply to -2 and add up to -1. These numbers are -2 and 1.
\( x^2 - x - 2 = x^2 - 2x + x - 2 \)
\( = x(x - 2) + 1(x - 2) \)
\( = (x + 1)(x - 2) \)
Thus, the complete factorisation of \( 2x^3 + 5x^2 - 11x - 14 \) is \( (2x + 7)(x + 1)(x - 2) \). This shows the full set of roots for the polynomial.
In simple words: First, we prove \( (2x+7) \) is a factor by putting \( x = -\frac{7}{2} \) into the expression and checking if it turns to zero. Then, we divide the original expression by \( (2x+7) \) to get a simpler quadratic. Finally, we break down this quadratic into two more factors to get all the factors.

🎯 Exam Tip: When the divisor is of the form \( (ax+b) \), the value to substitute into the polynomial is \( x = -\frac{b}{a} \). Be extra careful with fractions and signs during the substitution and calculation.

Question 8. Show that \( (x - 1) \) is a factor of \( x^3 - 7x^2 + 14x - 8 \). Hence completely factorise the above expression.
Answer: Let the polynomial be \( f(x) = x^3 - 7x^2 + 14x - 8 \).
To show that \( (x - 1) \) is a factor, we need to show that \( f(1) = 0 \), by the Factor Theorem.
Substitute \( x = 1 \) into the polynomial:
\( f(1) = (1)^3 - 7(1)^2 + 14(1) - 8 \)
\( = 1 - 7 + 14 - 8 \)
\( = (1 + 14) - (7 + 8) \)
\( = 15 - 15 = 0 \)
Since \( f(1) = 0 \), \( (x - 1) \) is indeed a factor of \( x^3 - 7x^2 + 14x - 8 \).
Now, we perform long division of \( f(x) \) by \( (x - 1) \) to find the other factor:
\[ \begin{array}{r} x^2 - 6x + 8 \\ x-1\overline{|x^3 - 7x^2 + 14x - 8} \\ - \underline{(x^3 - x^2)} \\ -6x^2 + 14x \\ - \underline{(-6x^2 + 6x)} \\ 8x - 8 \\ - \underline{(8x - 8)} \\ 0 \\ \end{array} \] So, \( f(x) = (x - 1)(x^2 - 6x + 8) \).
Next, we factorise the quadratic expression \( x^2 - 6x + 8 \). We look for two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4.
\( x^2 - 6x + 8 = x^2 - 2x - 4x + 8 \)
\( = x(x - 2) - 4(x - 2) \)
\( = (x - 2)(x - 4) \)
Thus, the complete factorisation of \( x^3 - 7x^2 + 14x - 8 \) is \( (x - 1)(x - 2)(x - 4) \). This polynomial has three distinct linear factors.
In simple words: First, we check if \( (x-1) \) is a factor by seeing if the expression becomes zero when \( x=1 \). Since it does, we then divide the main expression by \( (x-1) \) to get a quadratic part. Finally, we break this quadratic into two simpler factors to get all the factors of the original expression.

🎯 Exam Tip: When factorising, ensure you double-check your arithmetic, especially in long division and when identifying the correct pair of numbers to factorise the quadratic term. Mistakes often happen in these steps.