CBSE Class 10 Mathematics Polynomials Worksheet Set 07

Polynomials

Question 1. Show that \( x^2 - 3 \) is a factor of \( 2x^4 + 3x^3 - 2x^2 - 9x - 12 \)
Answer: Let \( p(x) = 2x^4 + 3x^3 - 2x^2 - 9x - 12 \) and \( g(x) = x^2 - 3 \).
Using the polynomial long division method to divide \( p(x) \) by \( g(x) \):
\[ \begin{array}{rll}
2x^2 + 3x + 4 & \text{(Quotient)} \\
x^2 - 3 \ \overline{\big) \ 2x^4 + 3x^3 - 2x^2 - 9x - 12} \\
\underline{-\left(2x^4 \phantom{+ 3x^3} - 6x^2\right)} \phantom{- 9x - 12} \\
3x^3 + 4x^2 - 9x - 12 \\
\underline{-\left(3x^3 \phantom{+ 4x^2} - 9x\right)} \phantom{- 12} \\
4x^2 \phantom{- 9x} - 12 \\
\underline{-\left(4x^2 \phantom{- 9x} - 12\right)} \\
0 & \text{(Remainder)}
\end{array} \ forensics \]
Since the remainder obtained is 0, we can conclude that \( x^2 - 3 \) is a factor of \( 2x^4 + 3x^3 - 2x^2 - 9x - 12 \).
In simple words: Dividing the larger expression by \( x^2 - 3 \) leaves no remainder, which proves that it divides the expression perfectly and is a factor.

Exam Tip: You can also prove this by substituting the roots of \( x^2 - 3 = 0 \) (which are \( \pm\sqrt{3} \)) into the polynomial f(x). If both substitution values yield zero, it is a factor.

Question 2. Divide: \( 4x^3 + 2x^2 + 5x - 6 \) by \( 2x^2 + 3x + 1 \)
Answer: We divide \( 4x^3 + 2x^2 + 5x - 6 \) by \( 2x^2 + 3x + 1 \) using polynomial long division:
\[ \begin{array}{rll}
2x - 2 & \text{(Quotient)} \\
2x^2 + 3x + 1 \ \overline{\big) \ 4x^3 + 2x^2 + 5x - 6} \\
\underline{-\left(4x^3 + 6x^2 + 2x\right)} \phantom{- 6} \\
-4x^2 + 3x - 6 \\
\underline{-\left(-4x^2 - 6x - 2\right)} \\
9x - 4 & \text{(Remainder)}
\end{array} \]
Therefore, the quotient is \( 2x - 2 \) and the remainder is \( 9x - 4 \).
In simple words: Performing long division gives us a main result of \( 2x - 2 \) and a leftover remainder of \( 9x - 4 \).

Exam Tip: Pay close attention to signs when subtracting negative values during each subtraction step.

Question 3. Find other zeroes of the polynomial \( p(x) = 2x^4 + 7x^3 - 19x^2 - 14x + 30 \) if two of its zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \)
Answer: Since the two given zeroes of \( p(x) \) are \( \sqrt{2} \) and \( -\sqrt{2} \), their corresponding product is:
\( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \), which must be a factor of \( p(x) \).

Dividing \( p(x) \) by \( x^2 - 2 \):
\[ \begin{array}{rll}
2x^2 + 7x - 15 & \text{(Quotient)} \\
x^2 - 2 \ \overline{\big) \ 2x^4 + 7x^3 - 19x^2 - 14x + 30} \\
\underline{-\left(2x^4 \phantom{+ 7x^3} - 4x^2\right)} \phantom{- 14x + 30} \\
7x^3 - 15x^2 - 14x + 30 \\
\underline{-\left(7x^3 \phantom{- 15x^2} - 14x\right)} \phantom{+ 30} \\
-15x^2 \phantom{- 14x} + 30 \\
\underline{-\left(-15x^2 \phantom{- 14x} + 30\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor of the polynomial is the quotient, \( 2x^2 + 7x - 15 \).
Setting the quotient equal to zero and splitting the middle term to find the zeroes:
\( 2x^2 + 7x - 15 = 0 \)
\( \implies 2x^2 + 10x - 3x - 15 = 0 \)
\( \implies 2x(x + 5) - 3(x + 5) = 0 \)
\( \implies (2x - 3)(x + 5) = 0 \)
\( \implies x = \frac{3}{2} \) or \( x = -5 \).
Thus, the other zeroes are \( \frac{3}{2} \) and \( -5 \).
In simple words: The given zeroes let us build a dividing quadratic \( x^2 - 2 \). Dividing our polynomial by it leaves another quadratic equation, which we factorise to find the remaining zeroes: \( \frac{3}{2} \) and \( -5 \).

Exam Tip: Since the polynomial is of degree 4, it must have 4 zeroes. Ensure you state the other 2 zeroes explicitly at the end of your answer.

Question 4. Find all the zeroes of the polynomial \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 \), if two of its zeroes are \( \sqrt{5/3} \) and \( -\sqrt{5/3} \)
Answer: Given that the two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \), the corresponding quadratic factor is:
\( \left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3} \)
We can clear the fraction by multiplying by 3, giving \( 3x^2 - 5 \) as our factor.

Dividing the given polynomial by \( 3x^2 - 5 \):
\[ \begin{array}{rll}
x^2 + 2x + 1 & \text{(Quotient)} \\
3x^2 - 5 \ \overline{\big) \ 3x^4 + 6x^3 - 2x^2 - 10x - 5} \\
\underline{-\left(3x^4 \phantom{+ 6x^3} - 5x^2\right)} \phantom{- 10x - 5} \\
6x^3 + 3x^2 - 10x - 5 \\
\underline{-\left(6x^3 \phantom{+ 3x^2} - 10x\right)} \phantom{- 5} \\
3x^2 \phantom{- 10x} - 5 \\
\underline{-\left(3x^2 \phantom{- 10x} - 5\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient, \( x^2 + 2x + 1 \). Setting it to zero to find the remaining roots:
\( x^2 + 2x + 1 = 0 \)
\( \implies (x + 1)^2 = 0 \)
\( \implies x = -1, -1 \).
Therefore, all the zeroes are \( \sqrt{\frac{5}{3}} \), \( -\sqrt{\frac{5}{3}} \), \( -1 \), and \( -1 \).
In simple words: The given zeroes form a divisor \( 3x^2 - 5 \). Dividing our main polynomial by it leaves a quotient \( x^2 + 2x + 1 \), which has a repeating root of \( -1 \).

Exam Tip: Be sure to write out all four zeroes at the end to satisfy the prompt "Find all the zeroes".

Question 5. Find all the zeroes of \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \), if it is known that two of its zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \)
Answer: Given that the two zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \), we have the quadratic factor:
\( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \).

Dividing the given polynomial by \( x^2 - 2 \):
\[ \begin{array}{rll}
2x^2 - 3x + 1 & \text{(Quotient)} \\
x^2 - 2 \ \overline{\big) \ 2x^4 - 3x^3 - 3x^2 + 6x - 2} \\
\underline{-\left(2x^4 \phantom{- 3x^3} - 4x^2\right)} \phantom{+ 6x - 2} \\
-3x^3 + \phantom{0}x^2 + 6x - 2 \\
\underline{-\left(-3x^3 \phantom{+ \phantom{0}x^2} + 6x\right)} \phantom{- 2} \\
x^2 \phantom{+ 6x} - 2 \\
\underline{-\left(x^2 \phantom{+ 6x} - 2\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other quadratic factor is the quotient, \( 2x^2 - 3x + 1 \). Factoring it by splitting the middle term:
\( 2x^2 - 3x + 1 = 0 \)
\( \implies 2x^2 - 2x - x + 1 = 0 \)
\( \implies 2x(x - 1) - 1(x - 1) = 0 \)
\( \implies (2x - 1)(x - 1) = 0 \)
\( \implies x = \frac{1}{2} \) or \( x = 1 \).
Therefore, all the zeroes are \( \sqrt{2} \), \( -\sqrt{2} \), \( 1 \), and \( \frac{1}{2} \).
In simple words: The square-root zeroes give us a quadratic divisor \( x^2 - 2 \). After dividing the polynomial by this factor, we solve the remaining quadratic equation to find the final zeroes: \( 1 \) and \( \frac{1}{2} \).

Exam Tip: Factoring quadratic expressions like \( 2x^2 - 3x + 1 \) requires splitting the middle term into two numbers that multiply to 2 and add to -3, which are -2 and -1.

Question 6. If the polynomial \( f(x) = x^4 - 6x^3 + 16x^2 - 25x + 10 \), is divided by another polynomial \( x^2 - 2x + k \) the remainder Comes out to be x + a, find k and a
Answer: Let us divide \( x^4 - 6x^3 + 16x^2 - 25x + 10 \) by \( x^2 - 2x + k \) using polynomial division:
\[ \begin{array}{rll}
x^2 - 4x + (8 - k) & \text{(Quotient)} \\
x^2 - 2x + k \ \overline{\big) \ x^4 - 6x^3 + 16x^2 - 25x + 10} \\
\underline{-\left(x^4 - 2x^3 + kx^2\right)} \phantom{- 25x + 10} \\
-4x^3 + (16 - k)x^2 - 25x + 10 \\
\underline{-\left(-4x^3 + 8x^2 - 4kx\right)} \phantom{+ 10} \\
(8 - k)x^2 + (4k - 25)x + 10 \\
\underline{-\left[(8 - k)x^2 - 2(8 - k)x + k(8 - k)\right]} \\
\left[2k - 9\right]x + \left[k^2 - 8k + 10\right] & \text{(Remainder)}
\end{array} \]
The remainder is \( (2k - 9)x + (k^2 - 8k + 10) \).
Since the remainder is given as \( x + a \), we equate the coefficients of like terms:
1. Coefficient of x:
\( 2k - 9 = 1 \)
\( \implies 2k = 10 \)
\( \implies k = 5 \)

2. Constant term:
\( a = k^2 - 8k + 10 \)
Substituting \( k = 5 \):
\( a = (5)^2 - 8(5) + 10 \)
\( = 25 - 40 + 10 = -5 \).
Thus, \( k = 5 \) and \( a = -5 \).
In simple words: We perform division keeping 'k' as a variable. Comparing our variable remainder with the given \( x+a \) lets us solve for \( k=5 \) and \( a=-5 \).

Exam Tip: This is a standard high-weightage textbook problem. Meticulously align like-power terms to keep track of the variables during subtraction.

Question 7. Find the polynomial, whose zeroes are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \)
Answer: Let the zeroes of the polynomial be \( \alpha = 2 + \sqrt{3} \) and \( \beta = 2 - \sqrt{3} \).
Sum of zeroes, \( \alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
Product of zeroes, \( \alpha\beta = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)

The quadratic polynomial is given by the formula:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)
Substituting the values:
\( p(x) = x^2 - 4x + 1 \).
In simple words: We find the sum (4) and product (1) of the given roots and plug them into the quadratic template to get \( x^2 - 4x + 1 \).

Exam Tip: Stating the general formula \( x^2 - (\text{sum})x + \text{product} \) before putting in values is a good habit that secures partial marks even if a calculation error occurs.

Question 8. Form a quadratic polynomial, one of whose zero is \( 2 + \sqrt{5} \) and the sum of zeroes is 4
Answer: Let the zeroes of the polynomial be \( \alpha \) and \( \beta \).
We are given:
One zero, \( \alpha = 2 + \sqrt{5} \)
Sum of zeroes, \( \alpha + \beta = 4 \)

First, find the other zero \( \beta \):
\( \beta = 4 - \alpha \)
\( = 4 - (2 + \sqrt{5}) \)
\( = 2 - \sqrt{5} \)

Now, calculate the product of the zeroes:
\( \alpha\beta = (2 + \sqrt{5})(2 - \sqrt{5}) \)
\( = 2^2 - (\sqrt{5})^2 \)
\( = 4 - 5 = -1 \)

The quadratic polynomial is:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)
\( = x^2 - 4x - 1 \).
In simple words: Since the sum is 4, the second root must be \( 2 - \sqrt{5} \). Multiplying the roots gives -1, which leads to the final quadratic \( x^2 - 4x - 1 \).

Exam Tip: Remember that irrational roots of quadratic polynomials with rational coefficients always occur in conjugate pairs (\( a \pm \sqrt{b} \)).

Question 9. If \( \alpha \) and \( \beta \) are zeroes of the polynomial \( x^2 - 2x - 15 \), then form a quadratic polynomial whose zeroes are \( 2\alpha \) and \( 2\beta \)
Answer: For the given polynomial \( x^2 - 2x - 15 \):
Sum of zeroes, \( \alpha + \beta = -\frac{-2}{1} = 2 \)
Product of zeroes, \( \alpha\beta = -\frac{15}{1} = -15 \)

For the new polynomial, let the zeroes be \( a = 2\alpha \) and \( b = 2\beta \).
Sum of new zeroes:
\( a + b = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(2) = 4 \)

Product of new zeroes:
\( ab = (2\alpha)(2\beta) = 4\alpha\beta = 4(-15) = -60 \)

The required quadratic polynomial is:
\( P(x) = x^2 - (a + b)x + ab \)
\( = x^2 - 4x - 60 \).
In simple words: The original roots sum to 2 and multiply to -15. Since the new roots are twice as large, their sum becomes 4 and their product is -60, resulting in the equation \( x^2 - 4x - 60 \).

Exam Tip: It is much faster to relate the new sum and product to the old sum and product instead of factorising to find the exact values of \( \alpha \) and \( \beta \).

Question 10. Write a quadratic polynomial, the sum and product of whose zeroes are 3 and -2
Answer: We are given:
Sum of zeroes, \( S = 3 \)
Product of zeroes, \( P = -2 \)

The formula for a quadratic polynomial is:
\( p(x) = x^2 - Sx + P \)
Substituting the given values:
\( p(x) = x^2 - 3x - 2 \).
In simple words: Putting the sum (3) and product (-2) directly into the standard template gives us \( x^2 - 3x - 2 \).

Exam Tip: Be careful with signs. The term containing the sum must have a negative sign in front of it: \( x^2 - Sx + P \).

Question 11. Find the zeroes of the polynomial and verify the relationship between the zeroes and the coefficient
a) \( 4x^2 - 4x + 1 \)
b) \( x^2 - 3 \)
c) \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \)
Answer:

a) For \( 4x^2 - 4x + 1 \):
Setting the polynomial equal to zero: \( (2x - 1)^2 = 0 \implies x = \frac{1}{2}, \frac{1}{2} \).
The zeroes are \( \alpha = \frac{1}{2} \) and \( \beta = \frac{1}{2} \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1 \). Also, \( -\frac{b}{a} = -\frac{-4}{4} = 1 \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \). Also, \( \frac{c}{a} = \frac{1}{4} \). (Verified)

b) For \( x^2 - 3 \):
Setting the polynomial to zero: \( x^2 - 3 = 0 \implies x^2 = 3 \implies x = \pm\sqrt{3} \).
The zeroes are \( \alpha = \sqrt{3} \) and \( \beta = -\sqrt{3} \).
Verification:
Sum of zeroes: \( \alpha + \beta = \sqrt{3} - \sqrt{3} = 0 \). Also, \( -\frac{b}{a} = -\frac{0}{1} = 0 \). (Verified)
Product of zeroes: \( \alpha\beta = (\sqrt{3})(-\sqrt{3}) = -3 \). Also, \( \frac{c}{a} = -\frac{3}{1} = -3 \). (Verified)

c) For \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \):
Splitting the middle term: \( \sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0 \)
\( \implies \sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0 \)
\( \implies (\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0 \)
The zeroes are \( \alpha = \frac{2}{\sqrt{3}} \) and \( \beta = 2\sqrt{3} \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{2}{\sqrt{3}} + 2\sqrt{3} = \frac{2 + 6}{\sqrt{3}} = \frac{8}{\sqrt{3}} \). Also, \( -\frac{b}{a} = \frac{8}{\sqrt{3}} \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{2}{\sqrt{3}} \cdot 2\sqrt{3} = 4 \). Also, \( \frac{c}{a} = \frac{4\sqrt{3}}{\sqrt{3}} = 4 \). (Verified)
In simple words: We find the roots of each equation by factoring. Then, we check that their sum equals \( -b/a \) and their product equals \( c/a \) to confirm the coefficients match.

Exam Tip: Show both the calculation of the roots and the coefficient-based sum/product formulas to secure full marks for the verification step.

Question 12. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( 2y^2 + 7y + 5 \), write the value of \( \alpha + \beta + \alpha\beta \)
Answer: For the quadratic polynomial \( 2y^2 + 7y + 5 \):
\( a = 2, b = 7, c = 5 \)
Using the relationship between coefficients and roots:
\( \alpha + \beta = -\frac{b}{a} = -\frac{7}{2} \)
\( \alpha\beta = \frac{c}{a} = \frac{5}{2} \)

Substituting these values into the given expression:
\( \alpha + \beta + \alpha\beta = -\frac{7}{2} + \frac{5}{2} \)
\( = \frac{-7 + 5}{2} = -\frac{2}{2} = -1 \).
Thus, the value of the expression is -1.
In simple words: The sum of the roots is \( -7/2 \) and their product is \( 5/2 \). Adding these two values together gives us -1.

Exam Tip: Never spend time finding the individual values of \( \alpha \) and \( \beta \) when you can substitute the sum and product formulas directly.

Question 13. If one root of the polynomial \( 5x^3 + 13x + k \) is reciprocal of the other, then find the value of k?
Answer: Let us consider the quadratic form of the polynomial, \( 5x^2 + 13x + k \). Let its zeroes be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of the zeroes is:
\( \text{Product} = \alpha \cdot \frac{1}{\alpha} = 1 \)

Using the coefficient relationship for the product of zeroes \( \frac{c}{a} \):
\( \frac{k}{5} = 1 \)
\( \implies k = 5 \).
Therefore, the value of k is 5.
In simple words: Since the roots are reciprocals, their product is exactly 1. Equating \( c/a \) to 1 gives \( k/5 = 1 \), from which we find \( k = 5 \).

Exam Tip: When one root is the reciprocal of the other, the leading coefficient must always equal the constant term (\( a = c \)).

Question 14. What must be subtracted from \( 2x^4 - 11x^3 + 29x^2 - 40x + 29 \), so that the resulting polynomial is exactly divisible By \( x^2 - 3x + 4 \)
Answer: We divide the dividend by \( x^2 - 3x + 4 \) to find the remainder:
\[ \begin{array}{rll}
2x^2 - 5x + 6 & \text{(Quotient)} \\
x^2 - 3x + 4 \ \overline{\big) \ 2x^4 - 11x^3 + 29x^2 - 40x + 29} \\
\underline{-\left(2x^4 - 6x^3 + 8x^2\right)} \phantom{- 40x + 29} \\
-5x^3 + 21x^2 - 40x + 29 \\
\underline{-\left(-5x^3 + 15x^2 - 20x\right)} \phantom{+ 29} \\
6x^2 - 20x + 29 \\
\underline{-\left(6x^2 - 18x + 24\right)} \\
-2x + 5 & \text{(Remainder)}
\end{array} \]
The remainder is \( -2x + 5 \).
Therefore, \( -2x + 5 \) must be subtracted from the polynomial so that the result is exactly divisible.
In simple words: Carrying out long division leaves us with a remainder of \( -2x + 5 \). Subtracting this leftover part makes the division perfect with no remainder.

Exam Tip: The polynomial that must be "subtracted" is equal to the remainder itself, whereas the polynomial that must be "added" is the negative of the remainder.

Question 15. If the polynomial \( 6x^4 + 8x^3 - 5x^2 + ax + b \) is exactly divisible by the polynomial \( 2x^2 - 5 \), then find the values of a and b
Answer: Since the polynomial is exactly divisible by \( 2x^2 - 5 \), the remainder must be 0.
Let us divide the polynomial by \( 2x^2 - 5 \):
\[ \begin{array}{rll}
3x^2 + 4x + 5 & \text{(Quotient)} \\
2x^2 - 5 \ \overline{\big) \ 6x^4 + 8x^3 - 5x^2 + ax + b} \\
\underline{-\left(6x^4 \phantom{+ 8x^3} - 15x^2\right)} \phantom{+ ax + b} \\
8x^3 + 10x^2 + ax + b \\
\underline{-\left(8x^3 \phantom{+ 10x^2} - 20x\right)} \phantom{+ b} \\
10x^2 + (a + 20)x + b \\
\underline{-\left(10x^2 \phantom{+ (a + 20)x} - 25\right)} \\
(a + 20)x + (b + 25) & \text{(Remainder)}
\end{array} \]
The remainder is \( (a + 20)x + (b + 25) \).
Setting this remainder to 0:
1. \( a + 20 = 0 \implies a = -20 \)
2. \( b + 25 = 0 \implies b = -25 \)
Thus, \( a = -20 \) and \( b = -25 \).
In simple words: We divide the expression by \( 2x^2 - 5 \). Since there shouldn't be any remainder, we set the leftover terms to zero, which gives us \( a = -20 \) and \( b = -25 \).

Exam Tip: Align like-power terms carefully during subtraction steps to avoid combining terms with different powers of \( x \).

Question 16. If the zeroes of the polynomial \( x^3 - 3x^2 + x + 1 \) are a - b, a, a + b, find a and b
Answer: Let the zeroes of the cubic polynomial be \( \alpha = a - b \), \( \beta = a \), and \( \gamma = a + b \).
Comparing with the standard cubic form \( Ax^3 + Bx^2 + Cx + D \):
\( A = 1, B = -3, C = 1, D = 1 \)

Using the relationship of zeroes:
1. Sum of zeroes:
\( \alpha + \beta + \gamma = -\frac{B}{A} \)
\( \implies (a - b) + a + (a + b) = -\frac{-3}{1} \)
\( \implies 3a = 3 \)
\( \implies a = 1 \)

2. Product of zeroes:
\( \alpha\beta\gamma = -\frac{D}{A} \)
\( \implies (a - b)(a)(a + b) = -\frac{1}{1} \)
\( \implies a(a^2 - b^2) = -1 \)
Substituting \( a = 1 \):
\( 1(1 - b^2) = -1 \)
\( \implies 1 - b^2 = -1 \)
\( \implies b^2 = 2 \)
\( \implies b = \pm\sqrt{2} \).
Thus, \( a = 1 \) and \( b = \pm\sqrt{2} \).
In simple words: Adding the three roots lets us solve for \( a = 1 \). Multiplying them together and putting \( a = 1 \) into the equation gives us \( b = \pm\sqrt{2} \).

Exam Tip: Always list both the positive and negative roots when taking a square root during polynomial calculations.

Question 17. On dividing \( x^3 - 3x^2 + x + 2 \) by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively Find g(x)
Answer: According to the division algorithm:
\( p(x) = g(x) \cdot q(x) + r(x) \)
Substituting the given values:
\( x^3 - 3x^2 + x + 2 = g(x)(x - 2) + (-2x + 4) \)
\( \implies g(x)(x - 2) = (x^3 - 3x^2 + x + 2) - (-2x + 4) \)
\( \implies g(x)(x - 2) = x^3 - 3x^2 + 3x - 2 \)
\( \implies g(x) = \frac{x^3 - 3x^2 + 3x - 2}{x - 2} \)

Dividing \( x^3 - 3x^2 + 3x - 2 \) by \( x - 2 \):
\[ \begin{array}{rll}
x^2 - x + 1 & \text{(g(x))} \\
x - 2 \ \overline{\big) \ x^3 - 3x^2 + 3x - 2} \\
\underline{-\left(x^3 - 2x^2\right)} \phantom{+ 3x - 2} \\
-x^2 + 3x - 2 \\
\underline{-\left(-x^2 + 2x\right)} \phantom{- 2} \\
x - 2 \\
\underline{-\left(x - 2\right)} \\
0 & \text{(Remainder)}
\end{array} \]
Thus, the polynomial \( g(x) = x^2 - x + 1 \).
In simple words: We subtract the remainder from the original expression and divide the result by \( x - 2 \). This division gives us our final answer, \( g(x) = x^2 - x + 1 \).

Exam Tip: Ensure that you do the subtraction step correctly before starting the division process, as any mistake here will prevent the remainder from becoming zero.

Question 18. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( f(x) = 6x^2 + x - 2 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta \)
Answer: For the polynomial \( 6x^2 + x - 2 \):
\( a = 6, b = 1, c = -2 \)
Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} = -\frac{1}{6} \)
Product of zeroes, \( \alpha\beta = \frac{c}{a} = -\frac{2}{6} = -\frac{1}{3} \)

Rewriting the given expression:
\( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta \)
Substituting the sum and product values:
\( = \frac{-1/6}{-1/3} - \left(-\frac{1}{3}\right) \)
\( = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \).
Therefore, the value of the expression is \( \frac{5}{6} \).
In simple words: Combining the fractions gives the sum over the product. Plugging in our values of \( -1/6 \) and \( -1/3 \) yields the final answer of \( \frac{5}{6} \).

Exam Tip: Simplify the fraction \( \frac{\alpha+\beta}{\alpha\beta} \) carefully to avoid reciprocal errors when dividing fraction terms.

Question 19. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( 2x^2 + 3x - 5 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \)
Answer: For the quadratic polynomial \( 2x^2 + 3x - 5 \):
\( a = 2, b = 3, c = -5 \)
Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} = -\frac{3}{2} \)
Product of zeroes, \( \alpha\beta = \frac{c}{a} = -\frac{5}{2} \)

Combining the fractions:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the values:
\( = \frac{-3/2}{-5/2} = \frac{3}{5} \).
Thus, the value is \( \frac{3}{5} \).
In simple words: Combining the fractions gives the sum divided by the product. Dividing \( -3/2 \) by \( -5/2 \) simplifies to \( \frac{3}{5} \).

Exam Tip: When dividing two fractions with the same denominator, you can cancel out the denominators directly to simplify the calculation.

Question 20. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( f(x) = x^2 - 5x + k \) such that \( \alpha - \beta = 1 \), find k
Answer: For the polynomial \( x^2 - 5x + k \):
Sum of zeroes, \( \alpha + \beta = 5 \) (Equation 1)
We are given: \( \alpha - \beta = 1 \) (Equation 2)

Adding Equation 1 and Equation 2:
\( 2\alpha = 6 \implies \alpha = 3 \)
Subtracting Equation 2 from Equation 1:
\( 2\beta = 4 \implies \beta = 2 \)

The product of the zeroes is given by \( \alpha\beta = k \):
\( k = (3)(2) = 6 \).
Thus, the value of k is 6.
In simple words: The sum of roots is 5, and we are told their difference is 1. This means the roots must be 3 and 2. Multiplying them gives our answer, \( k = 6 \).

Exam Tip: You can also use the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \) to solve for \( k \) directly without finding individual roots.

Question 21. If the product of zeroes of the polynomial \( ax^2 - 6x - 6 \) is 4, find the value of a
Answer: For the polynomial \( ax^2 - 6x - 6 \), the product of zeroes is \( \frac{c}{a} = \frac{-6}{a} \).

We are given that the product is 4:
\( \frac{-6}{a} = 4 \)
\( \implies 4a = -6 \)
\( \implies a = -\frac{6}{4} = -\frac{3}{2} \).
Thus, the value of a is \( -\frac{3}{2} \).
In simple words: The product formula is \( c/a \), which translates to \( -6/a = 4 \). Solving this equation gives \( a = -\frac{3}{2} \).

Exam Tip: Always reduce fractions to their simplest form to avoid losing marks on presentation.

Question 22. If \( \alpha,\beta \) are the zeroes of quadratic polynomial \( 2x^2 + 5x + k \), find the value of k such that \( (\alpha + \beta)^2 - \alpha\beta = 24 \)
Answer: For the polynomial \( 2x^2 + 5x + k \):
Sum of zeroes, \( \alpha + \beta = -\frac{5}{2} \)
Product of zeroes, \( \alpha\beta = \frac{k}{2} \)

Substituting these values into the given equation:
\( \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = 24 \)
\( \implies \frac{25}{4} - \frac{k}{2} = 24 \)
Multiply the entire equation by 4 to clear denominators:
\( 25 - 2k = 96 \)
\( \implies -2k = 71 \)
\( \implies k = -\frac{71}{2} \).
Thus, the value of k is \( -\frac{71}{2} \).
In simple words: Placing our values into the expression gives \( 25/4 - k/2 = 24 \). Multiplying through to clear the fractions allows us to find \( k = -\frac{71}{2} \).

Exam Tip: Multiplying the entire equation by the lowest common multiple of the denominators is an excellent way to prevent algebraic errors.