CBSE Class 10 Mathematics Polynomials Worksheet Set 08

Question 1. Show that \( x^2 - 3 \) is a factor of \( 2x^4 + 3x^3 - 2x^2 - 9x - 12 \)
Answer: Let the given polynomial be \( p(x) = 2x^4 + 3x^3 - 2x^2 - 9x - 12 \) and the divisor be \( g(x) = x^2 - 3 \).
To show that \( g(x) \) is a factor of \( p(x) \), we divide \( p(x) \) by \( g(x) \) using the long division method:
\[ \begin{array}{rll}
2x^2 + 3x + 4 & \text{(Quotient)} \\
x^2 - 3 \ \overline{\big) \ 2x^4 + 3x^3 - 2x^2 - 9x - 12} \\
\underline{-\left(2x^4 \phantom{+ 3x^3} - 6x^2\right)} \phantom{- 9x - 12} \\
3x^3 + 4x^2 - 9x - 12 \\
\underline{-\left(3x^3 \phantom{+ 4x^2} - 9x\right)} \phantom{- 12} \\
4x^2 \phantom{- 9x} - 12 \\
\underline{-\left(4x^2 \phantom{- 9x} - 12\right)} \\
0 & \text{(Remainder)}
\end{array} \]
Since the remainder obtained is 0, \( x^2 - 3 \) is a factor of \( 2x^4 + 3x^3 - 2x^2 - 9x - 12 \).
In simple words: When we divide the larger expression by \( x^2 - 3 \), the division is perfect with no leftover remainder, proving it is indeed a factor.

Exam Tip: Alternatively, you can substitute the zeroes of \( x^2 - 3 = 0 \) (which are \( \pm\sqrt{3} \)) into the polynomial. If both \( p(\sqrt{3}) = 0 \) and \( p(-\sqrt{3}) = 0 \), then the polynomial is divisible by \( x^2 - 3 \).

Question 2. Divide: \( 4x^3 + 2x^2 + 5x - 6 \) by \( 2x^2 + 3x + 1 \)
Answer: Let us divide \( p(x) = 4x^3 + 2x^2 + 5x - 6 \) by \( g(x) = 2x^2 + 3x + 1 \) 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: Dividing the cubic expression by the quadratic one gives us a quotient of \( 2x - 2 \) and leaves a remainder of \( 9x - 4 \).

Exam Tip: Be extra careful with signs when subtracting negative terms during polynomial long division steps.

Question 3. Divide \( (6 + 19 x + x^2 - 6x^3) \) by \( (2 + 5x - 3x^2) \) and verify the division algorithm
Answer: Let us rewrite both polynomials in standard descending order of their exponents:
Dividend, \( p(x) = -6x^3 + x^2 + 19x + 6 \)
Divisor, \( g(x) = -3x^2 + 5x + 2 \)

Performing long division:
\[ \begin{array}{rll}
2x + 3 & \text{(Quotient)} \\
-3x^2 + 5x + 2 \ \overline{\big) \ -6x^3 + \phantom{0}x^2 + 19x + 6} \\
\underline{-\left(-6x^3 + 10x^2 + \phantom{0}4x\right)} \phantom{+ 6} \\
-9x^2 + 15x + 6 \\
\underline{-\left(-9x^2 + 15x + 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
Thus, the quotient \( q(x) = 2x + 3 \) and the remainder \( r(x) = 0 \).

Verification of the Division Algorithm:
We need to verify that \( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \).
\( \text{LHS} = -6x^3 + x^2 + 19x + 6 \)
\( \text{RHS} = (-3x^2 + 5x + 2)(2x + 3) + 0 \)
\( = -3x^2(2x + 3) + 5x(2x + 3) + 2(2x + 3) \)
\( = -6x^3 - 9x^2 + 10x^2 + 15x + 4x + 6 \)
\( = -6x^3 + x^2 + 19x + 6 \)
Since \( \text{LHS} = \text{RHS} \), the division algorithm is verified.
In simple words: First we rearrange the equations in order. Then we divide to find a quotient of \( 2x+3 \) with no remainder. Multiplying the divisor by our quotient brings us back to the original expression, which confirms our division is correct.

Exam Tip: Always arrange the terms of both polynomials in standard form (highest to lowest power of \( x \)) before initiating the division.

Question 4. 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 \( -\sqrt{2} \), we know that:
\( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \) is a factor of \( p(x) \).

Dividing \( p(x) \) by \( x^2 - 2 \) to find the remaining factors:
\[ \begin{array}{rll}
2x^2 + 7x - 15 & \text{(Quotient)} \\
x^2 - 2 \ \overline{\big) \ 2x^4 + 7x^3 - 19x^2 - 14x + 30} \\
\underline{-\left(2x^4 \phantom{+ 7x^3} - 4x^2\right)} \phantom{- 14x + 30} \\
7x^3 - 15x^2 - 14x + 30 \\
\underline{-\left(7x^3 \phantom{- 15x^2} - 14x\right)} \phantom{+ 30} \\
-15x^2 \phantom{- 14x} + 30 \\
\underline{-\left(-15x^2 \phantom{- 14x} + 30\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient, \( 2x^2 + 7x - 15 \). We can find its zeroes by splitting the middle term:
\( 2x^2 + 7x - 15 = 0 \)
\( \implies 2x^2 + 10x - 3x - 15 = 0 \)
\( \implies 2x(x + 5) - 3(x + 5) = 0 \)
\( \implies (2x - 3)(x + 5) = 0 \)
\( \implies x = \frac{3}{2} \) or \( x = -5 \).
Therefore, the other zeroes of the polynomial are \( \frac{3}{2} \) and \( -5 \).
In simple words: Since we know two of the roots, we build a quadratic factor \( x^2 - 2 \) and divide our original polynomial by it. Solving the resulting quadratic equation gives us the other two roots: \( \frac{3}{2} \) and \( -5 \).

Exam Tip: The degree of the polynomial is 4, so it must have exactly 4 zeroes in total. Always state all remaining zeroes clearly at the end.

Question 5. 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 two zeroes of the polynomial are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \), the corresponding factor is:
\( \left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3} \)
Multiplying by 3, we can use \( 3x^2 - 5 \) as our divisor.

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 \( x^2 + 2x + 1 \). Finding its zeroes:
\( x^2 + 2x + 1 = 0 \)
\( \implies (x + 1)^2 = 0 \)
\( \implies x = -1, -1 \).
Therefore, all the zeroes of the given polynomial are \( \sqrt{\frac{5}{3}} \), \( -\sqrt{\frac{5}{3}} \), \( -1 \), and \( -1 \).
In simple words: We create a quadratic divisor \( 3x^2 - 5 \) from the two known roots. Dividing the polynomial by this factor gives a remaining quadratic \( x^2 + 2x + 1 \), which has duplicate roots at \( -1 \).

Exam Tip: Be sure to write "all the zeroes" in your final statement, listing both the given ones and the newly calculated ones.

Question 6. 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:
\( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \).

We divide \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \) 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 quotient obtained is \( 2x^2 - 3x + 1 \). We can find its roots by factorisation:
\( 2x^2 - 3x + 1 = 0 \)
\( \implies 2x^2 - 2x - x + 1 = 0 \)
\( \implies 2x(x - 1) - 1(x - 1) = 0 \)
\( \implies (2x - 1)(x - 1) = 0 \)
\( \implies x = \frac{1}{2} \) or \( x = 1 \).
Thus, the complete set of zeroes for this polynomial is \( \sqrt{2} \), \( -\sqrt{2} \), \( 1 \), and \( \frac{1}{2} \).
In simple words: The two given square-root zeroes give us the dividing factor \( x^2 - 2 \). Dividing our polynomial by it leaves us with \( 2x^2 - 3x + 1 \), which splits to give the remaining answers: \( 1 \) and \( \frac{1}{2} \).

Exam Tip: Splitting the middle term of \( 2x^2 - 3x + 1 \) requires finding two numbers that multiply to 2 and add up to -3, which are -2 and -1.

Question 7. Find all the zeroes of \( 2x^4 - 9x^3 + 5x^2 + 3x - 1 \), if two of its zeroes are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \)
Answer: Let \( \alpha = 2 + \sqrt{3} \) and \( \beta = 2 - \sqrt{3} \) be the given zeroes.
Sum of the zeroes, \( \alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
Product of the zeroes, \( \alpha\beta = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)
The quadratic factor is given by:
\( x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - 4x + 1 \).

Dividing our polynomial by \( x^2 - 4x + 1 \):
\[ \begin{array}{rll}
2x^2 - x - 1 & \text{(Quotient)} \\
x^2 - 4x + 1 \ \overline{\big) \ 2x^4 - 9x^3 + 5x^2 + 3x - 1} \\
\underline{-\left(2x^4 - 8x^3 + 2x^2\right)} \phantom{+ 3x - 1} \\
-x^3 + 3x^2 + 3x - 1 \\
\underline{-\left(-x^3 + 4x^2 - x\right)} \phantom{- 1} \\
-x^2 + 4x - 1 \\
\underline{-\left(-x^2 + 4x - 1\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other quadratic factor is \( 2x^2 - x - 1 \). Finding its roots:
\( 2x^2 - x - 1 = 0 \)
\( \implies 2x^2 - 2x + x - 1 = 0 \)
\( \implies 2x(x - 1) + 1(x - 1) = 0 \)
\( \implies (2x + 1)(x - 1) = 0 \)
\( \implies x = -\frac{1}{2} \) or \( x = 1 \).
Therefore, all the zeroes are \( 2 + \sqrt{3} \), \( 2 - \sqrt{3} \), \( 1 \), and \( -\frac{1}{2} \).
In simple words: We combine the two given roots to find the quadratic factor \( x^2 - 4x + 1 \). After dividing the main polynomial by this, we factor the resulting quadratic expression to find the final two roots: 1 and \( -\frac{1}{2} \).

Exam Tip: When given zeroes of the form \( a \pm \sqrt{b} \), you can also set \( x = a \pm \sqrt{b} \implies (x - a)^2 = b \) to directly find the quadratic factor.

Question 8. Find all the zeroes of polynomial \( 4x^4 - 20x^3 + 23x^2 + 5x - 6 \) if two of its zeroes are 2 and 3
Answer: Given that 2 and 3 are zeroes of the polynomial, we know that:
\( (x - 2)(x - 3) = x^2 - 5x + 6 \) is a factor.

Dividing our polynomial by \( x^2 - 5x + 6 \):
\[ \begin{array}{rll}
4x^2 - 1 & \text{(Quotient)} \\
x^2 - 5x + 6 \ \overline{\big) \ 4x^4 - 20x^3 + 23x^2 + 5x - 6} \\
\underline{-\left(4x^4 - 20x^3 + 24x^2\right)} \phantom{+ 5x - 6} \\
-x^2 + 5x - 6 \\
\underline{-\left(-x^2 + 5x - 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient, \( 4x^2 - 1 \). Finding its roots:
\( 4x^2 - 1 = 0 \)
\( \implies (2x - 1)(2x + 1) = 0 \)
\( \implies x = \frac{1}{2} \) or \( x = -\frac{1}{2} \).
Thus, the zeroes are \( 2 \), \( 3 \), \( \frac{1}{2} \), and \( -\frac{1}{2} \).
In simple words: Using the roots 2 and 3, we get the dividing factor \( x^2 - 5x + 6 \). Dividing our polynomial by it leaves us with \( 4x^2 - 1 \), which gives the final two zeroes: \( \frac{1}{2} \) and \( -\frac{1}{2} \).

Exam Tip: The factor \( 4x^2 - 1 \) is of the form \( a^2 - b^2 \), which easily factorises as \( (2x - 1)(2x + 1) \).

Question 9. When a polynomial f(x) is divided by \( x^2 - 5 \) the quotient is \( x^2 - 2x - 3 \) and remainder is zero. Find the polynomial and all its zeroes.
Answer: According to the division algorithm for polynomials:
\( f(x) = g(x) \cdot q(x) + r(x) \)
Given:
Divisor, \( g(x) = x^2 - 5 \)
Quotient, \( q(x) = x^2 - 2x - 3 \)
Remainder, \( r(x) = 0 \)

Substituting these values:
\( f(x) = (x^2 - 5)(x^2 - 2x - 3) + 0 \)
\( = x^2(x^2 - 2x - 3) - 5(x^2 - 2x - 3) \)
\( = x^4 - 2x^3 - 3x^2 - 5x^2 + 10x + 15 \)
\( = x^4 - 2x^3 - 8x^2 + 10x + 15 \)

To find all zeroes of \( f(x) \), we set both factors equal to 0:
1. \( x^2 - 5 = 0 \implies x = \pm\sqrt{5} \)
2. \( x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0 \implies x = 3, -1 \)
So, the polynomial is \( x^4 - 2x^3 - 8x^2 + 10x + 15 \) and its zeroes are \( 3 \), \( -1 \), \( \sqrt{5} \), and \( -\sqrt{5} \).
In simple words: We multiply the divisor and the quotient to find the polynomial, which is \( x^4 - 2x^3 - 8x^2 + 10x + 15 \). Setting each part to zero gives us the roots: \( 3 \), \( -1 \), and \( \pm\sqrt{5} \).

Exam Tip: Never forget to write the final polynomial expression and clearly state all four individual zeroes.

Question 10. 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 long division:
\[ \begin{array}{rll}
x^2 - 4x + (8 - k) & \text{(Quotient)} \\
x^2 - 2x + k \ \overline{\big) \ x^4 - 6x^3 + 16x^2 - 25x + 10} \\
\underline{-\left(x^4 - 2x^3 + kx^2\right)} \phantom{- 25x + 10} \\
-4x^3 + (16 - k)x^2 - 25x + 10 \\
\underline{-\left(-4x^3 + 8x^2 - 4kx\right)} \phantom{+ 10} \\
(8 - k)x^2 + (4k - 25)x + 10 \\
\underline{-\left[(8 - k)x^2 - 2(8 - k)x + k(8 - k)\right]} \\
\left[4k - 25 + 16 - 2k\right]x + \left[10 - 8k + k^2\right] \\
\end{array} \]
The remainder simplifies to:
\( (2k - 9)x + (k^2 - 8k + 10) \)

We are given that the remainder is \( x + a \). Equating the coefficients of \( x \) and constant terms:
1. \( 2k - 9 = 1 \)
\( \implies 2k = 10 \)
\( \implies k = 5 \)

2. \( a = k^2 - 8k + 10 \)
Substituting \( k = 5 \):
\( a = 5^2 - 8(5) + 10 \)
\( = 25 - 40 + 10 \)
\( = -5 \)
Thus, the value of \( k = 5 \) and \( a = -5 \).
In simple words: We carry out the division with the variable 'k' included. This gives us a remainder containing 'k', which we compare with the given remainder \( x + a \) to solve and find \( k = 5 \) and \( a = -5 \).

Exam Tip: This is a very popular high-weightage question in board exams. Practice the coefficient comparison step carefully to avoid calculation errors.

Question 11. 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 upon division must be 0.

Performing long division:
\[ \begin{array}{rll}
3x^2 + 4x + 5 & \text{(Quotient)} \\
2x^2 - 5 \ \overline{\big) \ 6x^4 + 8x^3 - 5x^2 + ax + b} \\
\underline{-\left(6x^4 \phantom{+ 8x^3} - 15x^2\right)} \phantom{+ ax + b} \\
8x^3 + 10x^2 + ax + b \\
\underline{-\left(8x^3 \phantom{+ 10x^2} - 20x\right)} \phantom{+ b} \\
10x^2 + (a + 20)x + b \\
\underline{-\left(10x^2 \phantom{+ (a + 20)x} - 25\right)} \\
(a + 20)x + (b + 25) & \text{(Remainder)}
\end{array} \]
Since the remainder must be 0:
\( (a + 20)x + (b + 25) = 0 \)

Equating the coefficient of \( x \) and the constant term 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: Ensure that when dividing, you line up terms with identical powers of \( x \) properly to prevent errors during subtraction.

Question 12. Find the values of m and n so that \( x^4 + mx^3 + nx^2 - 3x + n \) is divisible by \( x^2 - 1 \)
Answer: Let \( p(x) = x^4 + mx^3 + nx^2 - 3x + n \).
Since \( p(x) \) is divisible by \( x^2 - 1 \), the zeroes of \( x^2 - 1 = 0 \) (which are \( x = 1 \) and \( x = -1 \)) must also be zeroes of \( p(x) \).

Using the factor theorem:
1. For \( x = 1 \):
\( p(1) = 0 \)
\( \implies (1)^4 + m(1)^3 + n(1)^2 - 3(1) + n = 0 \)
\( \implies 1 + m + n - 3 + n = 0 \)
\( \implies m + 2n = 2 \) (Equation 1)

2. For \( x = -1 \):
\( p(-1) = 0 \)
\( \implies (-1)^4 + m(-1)^3 + n(-1)^2 - 3(-1) + n = 0 \)
\( \implies 1 - m + n + 3 + n = 0 \)
\( \implies -m + 2n = -4 \) (Equation 2)

Adding Equation 1 and Equation 2:
\( (m + 2n) + (-m + 2n) = 2 + (-4) \)
\( \implies 4n = -2 \)
\( \implies n = -\frac{1}{2} \)

Substituting \( n = -\frac{1}{2} \) into Equation 1:
\( m + 2\left(-\frac{1}{2}\right) = 2 \)
\( \implies m - 1 = 2 \)
\( \implies m = 3 \)
Therefore, the values are \( m = 3 \) and \( n = -\frac{1}{2} \).
In simple words: Because the polynomial divides by \( x^2 - 1 \), substituting \( 1 \) and \( -1 \) into the equation must equal zero. This gives us two simple equations that we solve to get \( m = 3 \) and \( n = -\frac{1}{2} \).

Exam Tip: Using the factor theorem here is much faster and cleaner than doing polynomial long division with unknown coefficients.

Question 13. 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:
\( p(x) = g(x) \cdot q(x) + r(x) \)
Given:
\( p(x) = x^3 - 3x^2 + x + 2 \)
Quotient, \( q(x) = x - 2 \)
Remainder, \( r(x) = -2x + 4 \)

Substituting the values:
\( x^3 - 3x^2 + x + 2 = g(x)(x - 2) + (-2x + 4) \)
\( \implies g(x)(x - 2) = (x^3 - 3x^2 + x + 2) - (-2x + 4) \)
\( \implies g(x)(x - 2) = x^3 - 3x^2 + x + 2 + 2x - 4 \)
\( \implies g(x)(x - 2) = x^3 - 3x^2 + 3x - 2 \)
\( \implies g(x) = \frac{x^3 - 3x^2 + 3x - 2}{x - 2} \)

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

Exam Tip: Make sure to do the subtraction step correctly before starting the division process, as any mistake here will lead to a non-zero remainder.

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: To find the polynomial that must be subtracted, we need to divide \( 2x^4 - 11x^3 + 29x^2 - 40x + 29 \) by \( x^2 - 3x + 4 \) and find the remainder.

Performing long division:
\[ \begin{array}{rll}
2x^2 - 5x + 6 & \text{(Quotient)} \\
x^2 - 3x + 4 \ \overline{\big) \ 2x^4 - 11x^3 + 29x^2 - 40x + 29} \\
\underline{-\left(2x^4 - 6x^3 + 8x^2\right)} \phantom{- 40x + 29} \\
-5x^3 + 21x^2 - 40x + 29 \\
\underline{-\left(-5x^3 + 15x^2 - 20x\right)} \phantom{+ 29} \\
6x^2 - 20x + 29 \\
\underline{-\left(6x^2 - 18x + 24\right)} \\
-2x + 5 & \text{(Remainder)}
\end{array} \]
The remainder is \( -2x + 5 \).
Therefore, \( -2x + 5 \) must be subtracted from the polynomial so that it becomes exactly divisible.
In simple words: When we divide our expression by \( x^2 - 3x + 4 \), we are left with a remainder of \( -2x + 5 \). Subtraction of this remainder makes the expression completely divisible.

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

Question 15. Find the polynomial, whose zeroes are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \)
Answer: Let the zeroes 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, \( \alpha\beta = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)

A quadratic polynomial is given by:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)
Substituting the sum and product:
\( p(x) = x^2 - 4x + 1 \).
In simple words: We calculate the sum of the roots (4) and their product (1). Plugging these into the standard quadratic template gives us \( x^2 - 4x + 1 \).

Exam Tip: Remember to write the standard formula \( x^2 - (\text{sum})x + \text{product} \) to secure step-wise marks before inserting the actual values.

Question 16. 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 \( \beta \).
Given:
One zero, \( \alpha = 2 + \sqrt{5} \)
Sum of zeroes, \( \alpha + \beta = 4 \)

First, we 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 given by:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)
\( = x^2 - 4x - 1 \).
In simple words: With one root as \( 2 + \sqrt{5} \) and the sum as 4, the second root must be \( 2 - \sqrt{5} \). Multiplying them gives a product of -1, leading to the polynomial \( x^2 - 4x - 1 \).

Exam Tip: Irrational roots of quadratic polynomials with rational coefficients always occur in conjugate pairs. If one is \( a + \sqrt{b} \), the other is \( a - \sqrt{b} \).

Question 17. 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 \( \alpha' = 2\alpha \) and \( \beta' = 2\beta \).
Sum of new zeroes:
\( \alpha' + \beta' = 2\alpha + 2\beta \)
\( = 2(\alpha + \beta) \)
\( = 2(2) = 4 \)

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

The required quadratic polynomial is:
\( p(x) = x^2 - (\alpha' + \beta')x + \alpha'\beta' \)
\( = x^2 - 4x - 60 \).
In simple words: From the original equation, the sum of roots is 2 and the product is -15. Since the new roots are twice as large, their sum is 4 and their product is -60, giving the polynomial \( x^2 - 4x - 60 \).

Exam Tip: Express the new sum and product in terms of the old sum and product before doing the actual calculations to avoid mistakes.

Question 18. 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, \( P = -2 \)

A quadratic polynomial is represented as:
\( p(x) = x^2 - Sx + P \)
Substituting the given values:
\( p(x) = x^2 - 3x - 2 \).
In simple words: Substituting the given sum (3) and product (-2) directly into our template gives us \( x^2 - 3x - 2 \).

Exam Tip: This is a simple 1-mark question. Be careful not to swap the signs of the sum and product terms.

Question 19. 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) \( 4x^2 - 7 \)
d) \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \)
Answer:

a) For \( 4x^2 - 4x + 1 \):
\( 4x^2 - 4x + 1 = 0 \implies (2x - 1)^2 = 0 \implies x = \frac{1}{2}, \frac{1}{2} \).
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 \( x^2 - 3 \):
\( x^2 - 3 = 0 \implies x^2 = 3 \implies x = \pm\sqrt{3} \).
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 \( 4x^2 - 7 \):
\( 4x^2 - 7 = 0 \implies x^2 = \frac{7}{4} \implies x = \pm\frac{\sqrt{7}}{2} \).
The zeroes are \( \alpha = \frac{\sqrt{7}}{2} \) and \( \beta = -\frac{\sqrt{7}}{2} \).
Verification:
Sum of zeroes: \( \alpha + \beta = 0 \). Also, \( -\frac{b}{a} = 0 \). (Verified)
Product of zeroes: \( \alpha\beta = \left(\frac{\sqrt{7}}{2}\right)\left(-\frac{\sqrt{7}}{2}\right) = -\frac{7}{4} \). Also, \( \frac{c}{a} = -\frac{7}{4} \). (Verified)

d) For \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \):
\( \sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0 \)
\( \implies \sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0 \)
\( \implies (\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0 \)
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: For each equation, we find the roots by factoring. Then, we check that their sum equals \( -b/a \) and their product equals \( c/a \), validating the standard relationship rules.

Exam Tip: When splitting the middle term of \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \), look for numbers that multiply to 12 and add to -8, which are -6 and -2.

Question 20. 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 zeroes and coefficients:
\( \alpha + \beta = -\frac{b}{a} = -\frac{7}{2} \)
\( \alpha\beta = \frac{c}{a} = \frac{5}{2} \)

Now, calculate the value of 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 the product is \( 5/2 \). Adding these two values together gives our answer of -1.

Exam Tip: Do not solve for \( \alpha \) and \( \beta \) individually. Substitute the values of \( \alpha+\beta \) and \( \alpha\beta \) directly to save time.

Question 21. If one root of the polynomial \( 5x^2 + 13x + k \) is reciprocal of the other, then find the value of k?
Answer: Let one root of the polynomial be \( \alpha \). Therefore, the other root is \( \frac{1}{\alpha} \).

The product of the roots is given by:
\( \text{Product of roots} = \alpha \cdot \frac{1}{\alpha} = 1 \)

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

Exam Tip: Whenever the question states that one zero is the reciprocal of the other, the coefficient of \( x^2 \) must always equal the constant term (\( a = c \)).

Question 22. If one zero of the polynomial \( (a^2 + 9) x^2 + 13x + 6a \) is reciprocal of the other. Find the value of a
Answer: Let the zeroes be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of the zeroes is:
\( \text{Product} = \alpha \cdot \frac{1}{\alpha} = 1 \)

Using the relationship of coefficients for the product of zeroes \( \frac{c}{a} \):
\( \frac{6a}{a^2 + 9} = 1 \)
\( \implies a^2 + 9 = 6a \)
\( \implies a^2 - 6a + 9 = 0 \)
\( \implies (a - 3)^2 = 0 \)
\( \implies a = 3 \).
Thus, the value of a is 3.
In simple words: Reciprocal roots multiply to 1, meaning the constant term must equal the coefficient of \( x^2 \). Solving \( a^2 + 9 = 6a \) gives us \( a = 3 \).

Exam Tip: Solve the quadratic equation by identifying it as a perfect square expansion of \( (a - 3)^2 = 0 \) to reach the answer quickly.

Question 23. 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, we have coefficients:
\( A = 1, B = -3, C = 1, D = 1 \)

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 \( a = 1 \):
\( 1(1^2 - b^2) = -1 \)
\( \implies 1 - b^2 = -1 \)
\( \implies b^2 = 2 \)
\( \implies b = \pm\sqrt{2} \).
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 for \( b \) when taking the square root of 2.

Question 24. 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, \( \alpha\beta = \frac{c}{a} = -\frac{2}{6} = -\frac{1}{3} \)

Now, rewrite the given expression:
\( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta \)
Substituting the values:
\( = \frac{-1/6}{-1/3} - \left(-\frac{1}{3}\right) \)
\( = \frac{1}{2} + \frac{1}{3} \)
\( = \frac{3 + 2}{6} = \frac{5}{6} \).
Therefore, the value of the expression is \( \frac{5}{6} \).
In simple words: We combine the fraction part to read as sum over product. Plugging in our calculated values of \( -1/6 \) and \( -1/3 \) gives us the final sum of \( \frac{5}{6} \).

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

Question 25. 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, \( \alpha\beta = \frac{c}{a} = -\frac{5}{2} \)

We need to find:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the values:
\( = \frac{-3/2}{-5/2} \)
\( = \frac{3}{5} \).
Thus, the value is \( \frac{3}{5} \).
In simple words: The combined fraction simplifies to the sum of the roots divided by their product. Dividing \( -3/2 \) by \( -5/2 \) yields \( \frac{3}{5} \).

Exam Tip: Notice that when dividing fractions with identical denominators, the denominators cancel out directly.

Question 26. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial f(x) = \( x^2 - 8x + k \) such that \( \alpha^2 + \beta^2 = 40 \), find k
Answer: For the polynomial \( x^2 - 8x + k \):
Sum of zeroes, \( \alpha + \beta = 8 \)
Product of zeroes, \( \alpha\beta = k \)

Using the algebraic identity:
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
We are given \( \alpha^2 + \beta^2 = 40 \):
\( (8)^2 - 2k = 40 \)
\( \implies 64 - 2k = 40 \)
\( \implies 2k = 24 \)
\( \implies k = 12 \).
Thus, the value of k is 12.
In simple words: The sum of the roots is 8. Writing the equation for the sum of squares as \( 8^2 - 2k = 40 \) allows us to solve and find \( k = 12 \).

Exam Tip: Remember to use the standard identity \( \alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \) which is extremely common in polynomial questions.

Question 27. If \( \alpha \), \( \beta \) are the zeroes of a polynomial, such that \( \alpha + \beta = 6 \) and \( \alpha\beta = 4 \), then writes the polynomial
Answer: We are given:
Sum of zeroes, \( S = \alpha + \beta = 6 \)
Product of zeroes, \( P = \alpha\beta = 4 \)

The general formula for a quadratic polynomial is:
\( p(x) = x^2 - Sx + P \)
Substituting the given values:
\( p(x) = x^2 - 6x + 4 \).
In simple words: We place the given sum of 6 and product of 4 into our standard quadratic template, yielding \( x^2 - 6x + 4 \).

Exam Tip: Writing the polynomial as \( k(x^2 - Sx + P) \) where \( k \) is a constant is also a correct, more general representation.

Question 28. If the product of zeroes of the polynomial \( ax^2 - 6x - 6 \) is 4, find the value of a
Answer: For the quadratic polynomial \( ax^2 - 6x - 6 \):
The constant term \( c = -6 \) and the coefficient of \( x^2 \) is \( a \).

We are given that the product of zeroes is 4:
\( \text{Product} = \frac{c}{a} = 4 \)
\( \implies \frac{-6}{a} = 4 \)
\( \implies 4a = -6 \)
\( \implies a = -\frac{6}{4} = -\frac{3}{2} \).
Thus, the value of a is \( -\frac{3}{2} \).
In simple words: The product of roots equals \( c/a \), which means \( -6/a = 4 \). Solving this gives us \( a = -\frac{3}{2} \).

Exam Tip: Always simplify fractions to their lowest terms for your final answer.

Question 29. 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, \( \alpha\beta = \frac{k}{2} \)

According to the given condition:
\( (\alpha + \beta)^2 - \alpha\beta = 24 \)
Substituting our sum and product:
\( \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = 24 \)
\( \implies \frac{25}{4} - \frac{k}{2} = 24 \)
Multiply the entire equation by 4 to clear denominators:
\( 25 - 2k = 96 \)
\( \implies -2k = 96 - 25 \)
\( \implies -2k = 71 \)
\( \implies k = -\frac{71}{2} \).
Therefore, the value of k is \( -\frac{71}{2} \).
In simple words: Putting \( -5/2 \) and \( k/2 \) into the equation 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 a great way to avoid fractional errors.

Question 30. If \( \alpha \) and \( \beta \) are zeroes of \( x^2 + 5x + 5 \), find the value of \( \alpha^{-1} + \beta^{-1} \)
Answer: For the polynomial \( x^2 + 5x + 5 \):
Sum of zeroes, \( \alpha + \beta = -5 \)
Product of zeroes, \( \alpha\beta = 5 \)

We need to evaluate:
\( \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the values:
\( = \frac{-5}{5} = -1 \).
Thus, the value of \( \alpha^{-1} + \beta^{-1} \) is -1.
In simple words: The expression is just the sum of the roots divided by their product. This means dividing \( -5 \) by \( 5 \), which equals -1.

Exam Tip: Remember that \( x^{-1} \) is the same as \( \frac{1}{x} \). Combine the terms first before substituting coefficients.

Question 31. \( \alpha \), \( \beta \) are the zeroes of the quadratic polynomial \( x^2 - (k+6)x + 2 (2k - 1) \). Find the value of k if \( \alpha + \beta = \frac{1}{2} \alpha\beta \)
Answer: For the polynomial \( x^2 - (k+6)x + 2 (2k - 1) \):
Sum of zeroes, \( \alpha + \beta = k + 6 \)
Product of zeroes, \( \alpha\beta = 2(2k - 1) = 4k - 2 \)

We are given the condition:
\( \alpha + \beta = \frac{1}{2} \alpha\beta \)
Substituting our values:
\( k + 6 = \frac{1}{2}(4k - 2) \)
\( \implies k + 6 = 2k - 1 \)
\( \implies 2k - k = 6 + 1 \)
\( \implies k = 7 \).
Therefore, the value of k is 7.
In simple words: We find the sum of the roots is \( k+6 \) and the product is \( 4k-2 \). Setting the sum equal to half the product allows us to solve and find \( k = 7 \).

Exam Tip: Take care to expand the term \( \frac{1}{2}(4k - 2) \) as \( 2k - 1 \) correctly before moving terms around.

Question 32. If \( \alpha \), \( \beta \) are the zeroes of the quadratic polynomial \( x^2 - 7x + 10 \), find the value of \( \alpha^3 + \beta^3 \)
Answer: For the quadratic polynomial \( x^2 - 7x + 10 \):
Sum of zeroes, \( \alpha + \beta = 7 \)
Product of zeroes, \( \alpha\beta = 10 \)

Using the algebraic identity:
\( \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] \)
Substituting the values:
\( \alpha^3 + \beta^3 = (7)[(7)^2 - 3(10)] \)
\( = 7[49 - 30] \)
\( = 7[19] = 133 \).
Therefore, the value is 133.
In simple words: Using the sum (7) and product (10) of the roots, we apply the sum-of-cubes identity. This simplifies to \( 7 \times 19 \), giving us our answer of 133.

Exam Tip: The identity \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \) is also extremely useful and gives the same result.

Question 33. m, n are zeroes of \( ax^2 - 12x + c \). Find the value of a and c if m + n = m n = 3
Answer: For the quadratic polynomial \( ax^2 - 12x + c \):
Sum of zeroes, \( m + n = \frac{12}{a} \)
Product of zeroes, \( mn = \frac{c}{a} \)

We are given:
\( m + n = 3 \)
\( \implies \frac{12}{a} = 3 \)
\( \implies 3a = 12 \)
\( \implies a = 4 \)

We are also given:
\( mn = 3 \)
\( \implies \frac{c}{a} = 3 \)
Substituting \( a = 4 \):
\( \frac{c}{4} = 3 \)
\( \implies c = 12 \).
Thus, the values are \( a = 4 \) and \( c = 12 \).
In simple words: Knowing the sum of roots is 3 allows us to find \( a = 4 \) from \( 12/a = 3 \). Using this value of 'a' with the product equation gives \( c = 12 \).

Exam Tip: Solve for the variable in the denominator ('a') first, as you will need its value to solve for the second variable ('c').

Question 34. If \( \alpha \) and \( \beta \) are the zeroes of \( x^2 - 8x + k \), such that \( \alpha^2 + \beta^2 = 40 \), find k
Answer: For the quadratic polynomial \( x^2 - 8x + k \):
Sum of zeroes, \( \alpha + \beta = 8 \)
Product of zeroes, \( \alpha\beta = k \)

Using the identity:
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
Substituting the given values:
\( 40 = (8)^2 - 2k \)
\( \implies 40 = 64 - 2k \)
\( \implies 2k = 64 - 40 \)
\( \implies 2k = 24 \)
\( \implies k = 12 \).
Thus, the value of k is 12.
In simple words: The sum of the roots is 8. Writing the equation for the sum of squares as \( 8^2 - 2k = 40 \) allows us to solve and find \( k = 12 \).

Exam Tip: This question is identical to Question 26 on the worksheet. Double-check your steps to ensure consistency.

Question 35. Find the sum and the product of the zeroes of cubic polynomial \( 2x^3 - 5x^2 - 14x + 8 \)
Answer: Comparing with the standard cubic form \( ax^3 + bx^2 + cx + d \), we have:
\( a = 2, b = -5, c = -14, d = 8 \)

Let the zeroes be \( \alpha \), \( \beta \), and \( \gamma \).
1. Sum of zeroes:
\( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2} \)

2. Sum of products of zeroes taken two at a time:
\( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-14}{2} = -7 \)

3. Product of zeroes:
\( \alpha\beta\gamma = -\frac{d}{a} = -\frac{8}{2} = -4 \).
Thus, the sum is \( \frac{5}{2} \), the sum of product of zeroes taken two at a time is \( -7 \), and the product of the zeroes is \( -4 \).
In simple words: For this cubic equation, the sum of roots is \( 5/2 \), the sum of their pairwise products is -7, and their total product is -4.

Exam Tip: Make sure to distinguish between the three different relationships (sum, pairwise sum, and product) for cubic polynomials.

Question 36. Find the sum and product of the zeroes of quadratic polynomial \( x^2 - 3 \)
Answer: For the quadratic polynomial \( x^2 - 3 \):
Comparing with the standard quadratic form \( ax^2 + bx + c \):
\( a = 1, b = 0, c = -3 \)

Using the coefficient relationships:
Sum of zeroes \( = -\frac{b}{a} = -\frac{0}{1} = 0 \)
Product of zeroes \( = \frac{c}{a} = \frac{-3}{1} = -3 \).
Therefore, the sum is 0 and the product is -3.
In simple words: Since there is no 'x' term, the sum of the roots is 0. The constant term tells us their product is -3.

Exam Tip: You can also find the actual zeroes first (which are \( \pm\sqrt{3} \)) and then verify their sum (\( 0 \)) and product (\( -3 \)).

Question 37. If 1 is a zero of polynomial \( ax^2 - 3(a-1)x - 1 \), then find the value of a
Answer: Since 1 is a zero of the polynomial \( p(x) = ax^2 - 3(a - 1)x - 1 \), we must have \( p(1) = 0 \).

Substituting \( x = 1 \):
\( a(1)^2 - 3(a - 1)(1) - 1 = 0 \)
\( \implies a - 3a + 3 - 1 = 0 \)
\( \implies -2a + 2 = 0 \)
\( \implies 2a = 2 \)
\( \implies a = 1 \).
Thus, the value of a is 1.
In simple words: Since 1 is a root, putting \( x=1 \) into the equation must give zero. Simplifying the resulting equation shows that \( a = 1 \).

Exam Tip: Be careful when distributing the negative sign in the term \( -3(a-1) \), which expands to \( -3a+3 \).