CBSE Class 10 Mathematics Polynomials Worksheet Set 05

Multiple Choice Questions

Question. Sum of the zeroes of the polynomial x² + 7x + 10 are
(a) 7
(b) – 7
(c) 10
(d) – 10
Answer : B

Question. Quadratic polynomial having zeroes a and b is
(a) x² – (ab)x + (a + b)
(b) x² – (a + b)x + ab
(c) x² − (a/b) x + ab
(d) None of these
Answer : B

Question. If one zero of the quadratic polynomial x² – 5x – 6 is 6, then other zero is
(a) 0
(b) 1
(c) –1
(d) 2
Answer : C

Question. If the sum of the zeroes of the quadratic polynomial 3x² – kx + 6 is 3, then the value of k is
(a) 3
(b) 6
(c) 9
(d) 0
Answer : C

Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answers given by the students:
2x + 3, 3x² + 7x + 2, 4x³ + 3x² + 2, x3 + √3x + 7, 7x + √7 , 5x³ – 7x + 2, 2x² + 3 – 5/x, 5x – 1/2, ax³ + bx² + cx + d, x + 1/x
How many of the above ten, are not polynomials?
(a) 1
(b) 2
(c) 3
(d) 4
Answer : C

Question. How many of the above ten (in question 12), are quadratic polynomials?
(a) 0
(b) 1
(c) 2
(d) 3
Answer : B

Question. A quadratic polynomial, whose sum of zeroes is 2 and product is –8 is
(a) x² – 3x – 3
(b) x² + 2x + 8
(c) x² + 3x + 3
(d) x² – 2x – 8
Answer : D

Question. The zeores of the quadratic polynomial 6x² – 3 – 7x are
(a) 3/2, -1/3
(b) 2/3, –3
(c) 3/5, -3/7
(d) None of these
Answer : A

Question. Which of the following is not the graph of a quadratic polynomial?

Answer : D

Question. The value of p, for which (–4) is a zero of the polynomial x² – 2x – (7p + 3) is
(a) 0
(b) 2
(c) 3
(d) None of these
Answer : C

Question. The zeroes of the quadratic polynomial 4x² – 4x – 3 are
(a) 3/2, -1/3
(b) 3/2, -1/2
(c) 2/5, -2/5
(d) None of these
Answer : B

Question. What number should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the polynomial?
(a) 0
(b) 1
(c) 2
(d) 3
Answer : C

Question. A quadratic polynomial, the sum and product of whose zeroes and (–3) and 2 respectively is
(a) x² + 3x + 2
(b) x² – 3x + 2
(c) x² + 3x – 2
(d) None of these
Answer : A

Question. A quadratic polynomial whose zeroes are 5 − 3 2 and 5 + 3 2 is
(a) x² – 10x + 7
(b) x² + 10x + 7
(c) x² – 5x + 9
(d) x² + 5x – 9
Answer : A

Question. If 2 is a zero of the polynomial ax² – 2x, then the value of ‘a’ is
(a) 3
(b) 1
(c) 2
(d) 5
Answer : B

Question. The quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is
(a) x² + 5x + 6
(b) x² – 5x + 6
(c) x² – 5x – 6
(d) –x² + 5x + 6
Answer : A

Question. The zeroes of the quadratic polynomial x² + 7x + 10 are
(a) –2, –5
(b) 2, 5
(c) –3, –8
(d) 3, 8
Answer : A

Question. For what value of k, (– 4) is a zero of p(x) = x² – x – (2k – 2)?
(a) 8
(b) 11
(c) 13
(d) 15
Answer : B
 

B. Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): x² + 4x + 5 has two zeroes.
Reason (R): A quadratic polynomial can have at the most two zeroes.
Answer : D

Question. Assertion (A): A quadratic polynomial whose zeroes are 5 + √2 and 5 – √2 is x² – 10 x + 23.
Reason (R): If a and b are zeroes of the quadratic polynomial p(x), then p(x) = x² – (a + b) x + ab.
Answer : A

Question. Assertion (A): If one zero of polynomial p(x) = (k² + 4)x² + 13x + 4k is reciprocal of each other, then k = 2.
Reason (R): If (x – a) is a factor of p(x), then p(a) = 0, i.e. a is a zero of p(x).
Answer : B

Question. Assertion (A): If both zeroes of the quadratic polynomial x² –2kx + 2 are equal in magnitude but opposite in sign, then value of k is 1/2.
Reason (R): Sum of zeroes of a quadratic polynomial ax² + bx + c is − b/a
Answer : D

Question 1. Find the Zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:-
a) \( 5x^2 - 29x + 20 \)
b) \( 2\sqrt{2}x^2 - 9x + 5\sqrt{2} \)
c) \( 3\sqrt{3}x^2 - 19x + 10\sqrt{3} \)
d) \( x^2 - x - 72 \)
e) \( x^2 - 2 \)
f) \( x^2 - 5x \)
g) \( x^2 - 9 \)
Answer:

a) For \( 5x^2 - 29x + 20 \):
Splitting the middle term:
\( 5x^2 - 25x - 4x + 20 = 0 \)
\( \implies 5x(x - 5) - 4(x - 5) = 0 \)
\( \implies (5x - 4)(x - 5) = 0 \)
The zeroes are \( \alpha = \frac{4}{5} \) and \( \beta = 5 \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{4}{5} + 5 = \frac{29}{5} \). Also, \( -\frac{b}{a} = -\frac{-29}{5} = \frac{29}{5} \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{4}{5} \cdot 5 = 4 \). Also, \( \frac{c}{a} = \frac{20}{5} = 4 \). (Verified)

b) For \( 2\sqrt{2}x^2 - 9x + 5\sqrt{2} \):
Splitting the middle term:
\( 2\sqrt{2}x^2 - 4x - 5x + 5\sqrt{2} = 0 \)
\( \implies 2\sqrt{2}x(x - \sqrt{2}) - 5(x - \sqrt{2}) = 0 \)
\( \implies (2\sqrt{2}x - 5)(x - \sqrt{2}) = 0 \)
The zeroes are \( \alpha = \frac{5}{2\sqrt{2}} \) and \( \beta = \sqrt{2} \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{5}{2\sqrt{2}} + \sqrt{2} = \frac{5 + 4}{2\sqrt{2}} = \frac{9}{2\sqrt{2}} \). Also, \( -\frac{b}{a} = -\frac{-9}{2\sqrt{2}} = \frac{9}{2\sqrt{2}} \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{5}{2\sqrt{2}} \cdot \sqrt{2} = \frac{5}{2} \). Also, \( \frac{c}{a} = \frac{5\sqrt{2}}{2\sqrt{2}} = \frac{5}{2} \). (Verified)

c) For \( 3\sqrt{3}x^2 - 19x + 10\sqrt{3} \):
Splitting the middle term:
\( 3\sqrt{3}x^2 - 9x - 10x + 10\sqrt{3} = 0 \)
\( \implies 3\sqrt{3}x(x - \sqrt{3}) - 10(x - \sqrt{3}) = 0 \)
\( \implies (3\sqrt{3}x - 10)(x - \sqrt{3}) = 0 \)
The zeroes are \( \alpha = \frac{10}{3\sqrt{3}} \) and \( \beta = \sqrt{3} \).
Verification:
Sum of zeroes: \( \alpha + \beta = \frac{10}{3\sqrt{3}} + \sqrt{3} = \frac{10 + 9}{3\sqrt{3}} = \frac{19}{3\sqrt{3}} \). Also, \( -\frac{b}{a} = -\frac{-19}{3\sqrt{3}} = \frac{19}{3\sqrt{3}} \). (Verified)
Product of zeroes: \( \alpha\beta = \frac{10}{3\sqrt{3}} \cdot \sqrt{3} = \frac{10}{3} \). Also, \( \frac{c}{a} = \frac{10\sqrt{3}}{3\sqrt{3}} = \frac{10}{3} \). (Verified)

d) For \( x^2 - x - 72 \):
\( (x - 9)(x + 8) = 0 \)
The zeroes are \( \alpha = 9 \) and \( \beta = -8 \).
Verification:
Sum of zeroes: \( \alpha + \beta = 9 - 8 = 1 \). Also, \( -\frac{b}{a} = -\frac{-1}{1} = 1 \). (Verified)
Product of zeroes: \( \alpha\beta = 9 \cdot (-8) = -72 \). Also, \( \frac{c}{a} = -\frac{72}{1} = -72 \). (Verified)

e) For \( x^2 - 2 \):
\( (x - \sqrt{2})(x + \sqrt{2}) = 0 \)
The zeroes are \( \alpha = \sqrt{2} \) and \( \beta = -\sqrt{2} \).
Verification:
Sum of zeroes: \( \alpha + \beta = 0 \). Also, \( -\frac{b}{a} = 0 \). (Verified)
Product of zeroes: \( \alpha\beta = -2 \). Also, \( \frac{c}{a} = -2 \). (Verified)

f) For \( x^2 - 5x \):
\( x(x - 5) = 0 \)
The zeroes are \( \alpha = 0 \) and \( \beta = 5 \).
Verification:
Sum of zeroes: \( \alpha + \beta = 5 \). Also, \( -\frac{b}{a} = 5 \). (Verified)
Product of zeroes: \( \alpha\beta = 0 \). Also, \( \frac{c}{a} = 0 \). (Verified)

g) For \( x^2 - 9 \):
\( (x - 3)(x + 3) = 0 \)
The zeroes are \( \alpha = 3 \) and \( \beta = -3 \).
Verification:
Sum of zeroes: \( \alpha + \beta = 0 \). Also, \( -\frac{b}{a} = 0 \). (Verified)
Product of zeroes: \( \alpha\beta = -9 \). Also, \( \frac{c}{a} = -9 \). (Verified)
In simple words: For each equation, we factorise to find the individual roots. Then we confirm that the sum of these roots equals \( -b/a \) and their product equals \( c/a \), verifying the standard properties of quadratics.

Exam Tip: Always show both methods of calculation (roots summation and coefficient formulas) clearly side-by-side to secure full marks on verification questions.

Question 2. Form the Quadratic polynomials whose zeros are:-
a) \( 3 \pm \sqrt{2} \)
b) \( -\sqrt{2} \) and \( \sqrt{2} \)
c) \( \frac{1}{3} \) and \( \frac{1}{4} \)
d) \( -5 \) and \( -3 \)
e) \( 3 \) and \( \frac{1}{5} \)
f) \( \frac{1}{a} , \frac{1}{b} \)
Answer:

a) For \( 3 \pm \sqrt{2} \):
Sum of zeroes, \( S = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6 \)
Product of zeroes, \( P = (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7 \)
The polynomial is \( x^2 - Sx + P = x^2 - 6x + 7 \).

b) For \( -\sqrt{2} \) and \( \sqrt{2} \):
Sum of zeroes, \( S = -\sqrt{2} + \sqrt{2} = 0 \)
Product of zeroes, \( P = (-\sqrt{2})(\sqrt{2}) = -2 \)
The polynomial is \( x^2 - Sx + P = x^2 - 2 \).

c) For \( \frac{1}{3} \) and \( \frac{1}{4} \):
Sum of zeroes, \( S = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \)
Product of zeroes, \( P = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \)
The polynomial is \( x^2 - \frac{7}{12}x + \frac{1}{12} \), which can be written as \( 12x^2 - 7x + 1 \).

d) For \( -5 \) and \( -3 \):
Sum of zeroes, \( S = -5 + (-3) = -8 \)
Product of zeroes, \( P = (-5)(-3) = 15 \)
The polynomial is \( x^2 - Sx + P = x^2 + 8x + 15 \).

e) For \( 3 \) and \( \frac{1}{5} \):
Sum of zeroes, \( S = 3 + \frac{1}{5} = \frac{16}{5} \)
Product of zeroes, \( P = 3 \cdot \frac{1}{5} = \frac{3}{5} \)
The polynomial is \( x^2 - \frac{16}{5}x + \frac{3}{5} \), which can be written as \( 5x^2 - 16x + 3 \).

f) For \( \frac{1}{a} \) and \( \frac{1}{b} \):
Sum of zeroes, \( S = \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \)
Product of zeroes, \( P = \frac{1}{a} \cdot \frac{1}{b} = \frac{1}{ab} \)
The polynomial is \( x^2 - \left(\frac{a + b}{ab}\right)x + \frac{1}{ab} \), which can be written as \( abx^2 - (a + b)x + 1 \).
In simple words: For each pair of zeroes, we find their sum and their product. Substituting these into the standard quadratic form \( x^2 - (\text{Sum})x + \text{Product} \) gives the final polynomial.

Exam Tip: When the sum or product results in fractions, it is customary to multiply the entire polynomial by the lowest common multiple of the denominators to write it with integer coefficients.

Question 3. Find all the Zeroes of \( x^3 + 6x^2 + 11x + 6 \) if \( (x + 1) \) is a factor.
Answer: Let \( p(x) = x^3 + 6x^2 + 11x + 6 \). Since \( (x + 1) \) is a factor, \( x = -1 \) is one of its zeroes.
We perform polynomial division to find the other quadratic factor:
\[ \begin{array}{rll}
x^2 + 5x + 6 & \text{(Quotient)} \\
x + 1 \ \overline{\big) \ x^3 + 6x^2 + 11x + 6} \\
\underline{-\left(x^3 + \phantom{0}x^2\right)} \phantom{+ 11x + 6} \\
5x^2 + 11x + 6 \\
\underline{-\left(5x^2 + \phantom{0}5x\right)} \phantom{+ 6} \\
6x + 6 \\
\underline{-\left(6x + 6\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is \( x^2 + 5x + 6 \). Finding its roots by factoring:
\( x^2 + 5x + 6 = 0 \)
\( \implies (x + 2)(x + 3) = 0 \)
\( \implies x = -2 \) or \( x = -3 \).
Therefore, all the zeroes of the given polynomial are \( -1 \), \( -2 \), and \( -3 \).
In simple words: Dividing the cubic expression by the known factor \( x+1 \) leaves us with a quadratic expression \( x^2 + 5x + 6 \). Factoring this gives the remaining roots, which are \( -2 \) and \( -3 \).

Exam Tip: A cubic polynomial must have exactly three zeroes. Always list all three zeroes clearly at the end of your answer.

Question 4. Find all the Zeroes of \( x^3 - 10x^2 + 31x - 30 \) if 2 is a zero of it.
Answer: Since 2 is a zero of the polynomial \( p(x) = x^3 - 10x^2 + 31x - 30 \), we know that \( (x - 2) \) is a factor of \( p(x) \).
Dividing \( p(x) \) by \( (x - 2) \) using long division:
\[ \begin{array}{rll}
x^2 - 8x + 15 & \text{(Quotient)} \\
x - 2 \ \overline{\big) \ x^3 - 10x^2 + 31x - 30} \\
\underline{-\left(x^3 - \phantom{0}2x^2\right)} \phantom{+ 31x - 30} \\
-8x^2 + 31x - 30 \\
\underline{-\left(-8x^2 + 16x\right)} \phantom{- 30} \\
15x - 30 \\
\underline{-\left(15x - 30\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient \( x^2 - 8x + 15 \). Finding its roots:
\( x^2 - 8x + 15 = 0 \)
\( \implies (x - 3)(x - 5) = 0 \)
\( \implies x = 3 \) or \( x = 5 \).
Therefore, all the zeroes of the polynomial are \( 2 \), \( 3 \), and \( 5 \).
In simple words: Since 2 is a root, we divide the expression by \( x - 2 \). Factoring the resulting quadratic quotient \( x^2 - 8x + 15 \) gives us the other roots, which are \( 3 \) and \( 5 \).

Exam Tip: Be meticulous with sign operations during subtraction in long division to ensure the remainder is exactly zero.

Question 5. Find the values of a and b, if 2 and 3 are zeroes of \( x^3 + ax^2 + bx - 30 \).
Answer: Let \( p(x) = x^3 + ax^2 + bx - 30 \). Since 2 and 3 are zeroes of \( p(x) \):

1. For \( x = 2 \):
\( p(2) = 0 \)
\( \implies (2)^3 + a(2)^2 + b(2) - 30 = 0 \)
\( \implies 8 + 4a + 2b - 30 = 0 \)
\( \implies 4a + 2b = 22 \)
\( \implies 2a + b = 11 \) (Equation 1)

2. For \( x = 3 \):
\( p(3) = 0 \)
\( \implies (3)^3 + a(3)^2 + b(3) - 30 = 0 \)
\( \implies 27 + 9a + 3b - 30 = 0 \)
\( \implies 9a + 3b = 3 \)
\( \implies 3a + b = 1 \) (Equation 2)

Subtracting Equation 1 from Equation 2:
\( (3a + b) - (2a + b) = 1 - 11 \)
\( \implies a = -10 \)

Substituting \( a = -10 \) into Equation 1:
\( 2(-10) + b = 11 \)
\( \implies -20 + b = 11 \)
\( \implies b = 31 \).
Thus, the values are \( a = -10 \) and \( b = 31 \).
In simple words: Since 2 and 3 make the expression zero, we substitute them in to get two equations. Solving these equations together reveals that \( a = -10 \) and \( b = 31 \).

Exam Tip: Simplifying the initial equations by dividing by common factors makes simultaneous equations much faster to solve.

Question 6. Divide \( x^4 - 4x^3 + 8x^2 + 7x + 10 \) by \( (x - 2) \) and verify the division algorithm.
Answer: Let us divide \( p(x) = x^4 - 4x^3 + 8x^2 + 7x + 10 \) by \( g(x) = x - 2 \) using long division:
\[ \begin{array}{rll}
x^3 - 2x^2 + 4x + 15 & \text{(Quotient)} \\
x - 2 \ \overline{\big) \ x^4 - 4x^3 + 8x^2 + \phantom{0}7x + 10} \\
\underline{-\left(x^4 - 2x^3\right)} \phantom{+ 8x^2 + 7x + 10} \\
-2x^3 + 8x^2 + \phantom{0}7x + 10 \\
\underline{-\left(-2x^3 + 4x^2\right)} \phantom{+ 7x + 10} \\
4x^2 + \phantom{0}7x + 10 \\
\underline{-\left(4x^2 - \phantom{0}8x\right)} \phantom{+ 10} \\
15x + 10 \\
\underline{-\left(15x - 30\right)} \\
40 & \text{(Remainder)}
\end{array} \]
The quotient \( q(x) = x^3 - 2x^2 + 4x + 15 \) and the remainder \( r(x) = 40 \).

Verification of the Division Algorithm:
We must verify that \( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \).
\( \text{RHS} = (x - 2)(x^3 - 2x^2 + 4x + 15) + 40 \)
\( = x(x^3 - 2x^2 + 4x + 15) - 2(x^3 - 2x^2 + 4x + 15) + 40 \)
\( = x^4 - 2x^3 + 4x^2 + 15x - 2x^3 + 4x^2 - 8x - 30 + 40 \)
\( = x^4 - 4x^3 + 8x^2 + 7x + 10 \)
Since this is equal to the Dividend, the division algorithm is verified.
In simple words: Long division gives a quotient of \( x^3 - 2x^2 + 4x + 15 \) and a remainder of 40. Multiplying our divisor by the quotient and adding the remainder successfully returns the original expression, verifying our work.

Exam Tip: To verify the remainder quickly in your head, apply the Remainder Theorem: \( p(2) = 2^4 - 4(2)^3 + 8(2)^2 + 7(2) + 10 = 40 \).

Question 7. Find the value of k if \( (x - 2) \) is a factor of \( x^2 - kx + 10 \).
Answer: Let \( p(x) = x^2 - kx + 10 \).
Since \( (x - 2) \) is a factor of \( p(x) \), by the Factor Theorem:
\( p(2) = 0 \)
\( \implies (2)^2 - k(2) + 10 = 0 \)
\( \implies 4 - 2k + 10 = 0 \)
\( \implies 14 - 2k = 0 \)
\( \implies 2k = 14 \)
\( \implies k = 7 \).
Therefore, the value of k is 7.
In simple words: If \( x - 2 \) is a factor, substituting 2 into the equation must equal zero. This gives \( 14 - 2k = 0 \), showing that \( k = 7 \).

Exam Tip: Factor theorem problems are very straightforward. Always write down the theorem condition \( p(a) = 0 \) clearly in your steps.

Question 8. Find the value of k if 2 is zero of \( 3x^2 - 17x + k \).
Answer: Let \( p(x) = 3x^2 - 17x + k \).
Since 2 is a zero of \( p(x) \):
\( p(2) = 0 \)
\( \implies 3(2)^2 - 17(2) + k = 0 \)
\( \implies 3(4) - 34 + k = 0 \)
\( \implies 12 - 34 + k = 0 \)
\( \implies -22 + k = 0 \)
\( \implies k = 22 \).
Therefore, the value of k is 22.
In simple words: Putting 2 in place of x makes the expression equal to zero. Solving this simple equation gives us \( k = 22 \).

Exam Tip: Be precise with standard multiplication and subtraction when evaluating polynomial constants.

Question 9. Find all the zeroes of \( 4x^4 - 20x^3 + 23x^2 + 5x - 6 \) if two of its zeroes are 2 & 3.
Answer: Since 2 and 3 are zeroes of the polynomial, we know that:
\( (x - 2)(x - 3) = x^2 - 5x + 6 \) is a factor of the polynomial.

Dividing the given 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, \( 4x^2 - 1 \). Setting it equal to zero to find the zeroes:
\( 4x^2 - 1 = 0 \)
\( \implies (2x - 1)(2x + 1) = 0 \)
\( \implies x = \frac{1}{2} \) or \( x = -\frac{1}{2} \).
Thus, the complete set of zeroes is \( 2 \), \( 3 \), \( \frac{1}{2} \), and \( -\frac{1}{2} \).
In simple words: The given zeroes help us construct a dividing factor \( x^2 - 5x + 6 \). Dividing by this leaves a quotient \( 4x^2 - 1 \), which gives the final two zeroes: \( \frac{1}{2} \) and \( -\frac{1}{2} \).

Exam Tip: Factoring the quotient \( 4x^2 - 1 \) is very simple if you apply the identity \( a^2 - b^2 = (a-b)(a+b) \).

Question 10. If \( \alpha \) and \( \beta \) are the zeroes of \( x^2 + 5x + 6 \) find the value of \( \alpha^{-1} + \beta^{-1} \).
Answer: For the quadratic polynomial \( x^2 + 5x + 6 \):
\( a = 1, b = 5, c = 6 \)
Using coefficient relationships:
\( \alpha + \beta = -\frac{b}{a} = -5 \)
\( \alpha\beta = \frac{c}{a} = 6 \)

Now, rewrite the given expression:
\( \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting our values:
\( = \frac{-5}{6} \).
Thus, the value of the expression is \( -\frac{5}{6} \).
In simple words: Combining the fractions gives the sum of the roots divided by their product. Dividing \( -5 \) by \( 6 \) gives us our final answer, \( -\frac{5}{6} \).

Exam Tip: Remember that \( \alpha^{-1} = \frac{1}{\alpha} \). Always simplify the algebraic expression first before substituting the coefficient ratios.

Question 11. If \( \frac{1}{2} \) and 1 are zeroes of \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \), find the other zeroes.
Answer: Since \( \frac{1}{2} \) and \( 1 \) are zeroes of the polynomial, we know that:
\( \left(x - \frac{1}{2}\right)(x - 1) = x^2 - \frac{3}{2}x + \frac{1}{2} \)
Multiplying by 2, we can use \( 2x^2 - 3x + 1 \) as a factor.

Dividing our polynomial by \( 2x^2 - 3x + 1 \):
\[ \begin{array}{rll}
x^2 - 2 & \text{(Quotient)} \\
2x^2 - 3x + 1 \ \overline{\big) \ 2x^4 - 3x^3 - 3x^2 + 6x - 2} \\
\underline{-\left(2x^4 - 3x^3 + \phantom{0}x^2\right)} \phantom{+ 6x - 2} \\
-4x^2 + 6x - 2 \\
\underline{-\left(-4x^2 + 6x - 2\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is the quotient, \( x^2 - 2 \). Setting it equal to zero:
\( x^2 - 2 = 0 \)
\( \implies x = \pm\sqrt{2} \).
Therefore, the other zeroes of the polynomial are \( \sqrt{2} \) and \( -\sqrt{2} \).
In simple words: Using the given zeroes, we build a dividing quadratic \( 2x^2 - 3x + 1 \). Dividing our polynomial by it leaves the quotient \( x^2 - 2 \), which gives us the remaining roots: \( \sqrt{2} \) and \( -\sqrt{2} \).

Exam Tip: Clear fraction denominators from the factor before starting the long division to simplify the arithmetic significantly.

Question 12. If \( -5 \) and 7 are zeroes of \( x^4 - 6x^3 - 26x^2 + 138x - 35 \) find the other zeroes.
Answer: Since \( -5 \) and \( 7 \) are zeroes, we know that:
\( (x + 5)(x - 7) = x^2 - 2x - 35 \) is a factor of the polynomial.

Dividing the given polynomial by \( x^2 - 2x - 35 \):
\[ \begin{array}{rll}
x^2 - 4x + 1 & \text{(Quotient)} \\
x^2 - 2x - 35 \ \overline{\big) \ x^4 - 6x^3 - 26x^2 + 138x - 35} \\
\underline{-\left(x^4 - 2x^3 - 35x^2\right)} \phantom{+ 138x - 35} \\
-4x^3 + \phantom{0}9x^2 + 138x - 35 \\
\underline{-\left(-4x^3 + \phantom{0}8x^2 + 140x\right)} \phantom{- 35} \\
x^2 - \phantom{0}2x - 35 \\
\underline{-\left(x^2 - \phantom{0}2x - 35\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other quadratic factor is the quotient, \( x^2 - 4x + 1 \). Setting it equal to zero to find the remaining zeroes:
\( x^2 - 4x + 1 = 0 \)
Using the quadratic formula:
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2} \)
\( \implies x = \frac{4 \pm \sqrt{16 - 4}}{2} \)
\( \implies x = \frac{4 \pm \sqrt{12}}{2} \)
\( \implies x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \).
Therefore, the other zeroes of the polynomial are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \).
In simple words: The roots \( -5 \) and \( 7 \) give us the divisor \( x^2 - 2x - 35 \). Dividing by this factor leaves us with \( x^2 - 4x + 1 \), which we solve using the quadratic formula to get \( 2 \pm \sqrt{3} \).

Exam Tip: When factorisation of the quotient is not possible with rational numbers, use the quadratic formula to solve for irrational zeroes.

Question 13. If one of the zeroes of the polynomial \( 5z^2 + 13z - p \) is the reciprocal of the other, find p.
Answer: Let the zeroes of the quadratic polynomial be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of the zeroes is:
\( \text{Product} = \alpha \cdot \frac{1}{\alpha} = 1 \)

According to the relationship between coefficients and zeroes:
\( \text{Product of zeroes} = \frac{\text{Constant term}}{\text{Coefficient of } z^2} \)
\( \implies \frac{-p}{5} = 1 \)
\( \implies -p = 5 \)
\( \implies p = -5 \).
Thus, the value of p is -5.
In simple words: Since the zeroes are reciprocals, their product is exactly 1. Comparing this to the formula \( c/a \) tells us that \( -p/5 = 1 \), giving \( p = -5 \).

Exam Tip: Remember that whenever one zero of a quadratic polynomial is the reciprocal of the other, the coefficient of the squared term must equal the constant term (\( a = c \)).

Question 14. On dividing the polynomial \( 4x^4 - 3x^3 - 42x^2 - 55x - 17 \) by the polynomial g(x) the quotient is \( x^2 - 3x - 5 \) and the remainder is \( 5x + 8 \). Find g(x).
Answer: According to the division algorithm:
\( p(x) = g(x) \cdot q(x) + r(x) \)
\( \implies g(x) \cdot q(x) = p(x) - r(x) \)
\( \implies g(x) = \frac{p(x) - r(x)}{q(x)} \)

First, let us find the numerator \( p(x) - r(x) \):
\( p(x) - r(x) = (4x^4 - 3x^3 - 42x^2 - 55x - 17) - (5x + 8) \)
\( = 4x^4 - 3x^3 - 42x^2 - 60x - 25 \)

Now, we divide this resulting expression by \( q(x) = x^2 - 3x - 5 \) to find \( g(x) \):
\[ \begin{array}{rll}
4x^2 + 9x + 5 & \text{(g(x))} \\
x^2 - 3x - 5 \ \overline{\big) \ 4x^4 - 3x^3 - 42x^2 - 60x - 25} \\
\underline{-\left(4x^4 - 12x^3 - 20x^2\right)} \phantom{- 60x - 25} \\
9x^3 - 22x^2 - 60x - 25 \\
\underline{-\left(9x^3 - 27x^2 - 45x\right)} \phantom{- 25} \\
5x^2 - 15x - 25 \\
\underline{-\left(5x^2 - 15x - 25\right)} \\
0 & \text{(Remainder)}
\end{array} \]
Thus, the polynomial \( g(x) = 4x^2 + 9x + 5 \).
In simple words: We subtract the remainder from our main polynomial first. Then we divide that result by the given quotient using long division to find \( g(x) = 4x^2 + 9x + 5 \).

Exam Tip: Performing the subtraction step accurately is essential before starting the long division, as any minor error will prevent the division from having a remainder of zero.

Question 15. If \( \frac{1}{2} \) and 1 are zeroes of \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \), find the other zeroes.
Answer: Since \( \frac{1}{2} \) and \( 1 \) are zeroes of the polynomial, we know that:
\( \left(x - \frac{1}{2}\right)(x - 1) = x^2 - \frac{3}{2}x + \frac{1}{2} \)
We multiply by 2 to clear fractions, using \( 2x^2 - 3x + 1 \) as our factor.

Dividing the given polynomial by \( 2x^2 - 3x + 1 \):
\[ \begin{array}{rll}
x^2 - 2 & \text{(Quotient)} \\
2x^2 - 3x + 1 \ \overline{\big) \ 2x^4 - 3x^3 - 3x^2 + 6x - 2} \\
\underline{-\left(2x^4 - 3x^3 + \phantom{0}x^2\right)} \phantom{+ 6x - 2} \\
-4x^2 + 6x - 2 \\
\underline{-\left(-4x^2 + 6x - 2\right)} \\
0 & \text{(Remainder)}
\end{array} \]
The other factor is \( x^2 - 2 \). Setting it equal to zero to solve:
\( x^2 - 2 = 0 \)
\( \implies x = \pm\sqrt{2} \).
Therefore, the other zeroes of the polynomial are \( \sqrt{2} \) and \( -\sqrt{2} \).
In simple words: This is a repeat of Question 11 from the worksheet. Using the given roots, we divide the polynomial by \( 2x^2 - 3x + 1 \) and find the remaining roots to be \( \sqrt{2} \) and \( -\sqrt{2} \).

Exam Tip: Be consistent with your methods across repeated or highly similar questions on the same worksheet to ensure accuracy.

Question 16. Verify that 1, \( -2 \) and \( \frac{1}{2} \) are zeroes of \( 2x^3 + x^2 - 5x + 2 \). Also verify the relationship between the zeroes and the coefficients.
Answer: Let \( p(x) = 2x^3 + x^2 - 5x + 2 \).
First, let us verify each number by substituting it in place of x:
1. For \( x = 1 \):
\( p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0 \). (Verified)

2. For \( x = -2 \):
\( p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0 \). (Verified)

3. For \( x = \frac{1}{2} \):
\( p\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + \frac{1}{4} - 5\left(\frac{1}{2}\right) + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = -2 + 2 = 0 \). (Verified)

Verification of Relationships:
Let \( \alpha = 1 \), \( \beta = -2 \), and \( \gamma = \frac{1}{2} \). Comparing the polynomial with the standard cubic form, we have \( A = 2, B = 1, C = -5, D = 2 \).

1. Sum of zeroes:
\( \alpha + \beta + \gamma = 1 + (-2) + \frac{1}{2} = -\frac{1}{2} \)
Also, \( -\frac{B}{A} = -\frac{1}{2} \). (Verified)

2. Sum of products of zeroes taken two at a time:
\( \alpha\beta + \beta\gamma + \gamma\alpha = (1)(-2) + (-2)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)(1) = -2 - 1 + \frac{1}{2} = -\frac{5}{2} \)
Also, \( \frac{C}{A} = -\frac{5}{2} \). (Verified)

3. Product of zeroes:
\( \alpha\beta\gamma = (1)(-2)\left(\frac{1}{2}\right) = -1 \)
Also, \( -\frac{D}{A} = -\frac{2}{2} = -1 \). (Verified)
In simple words: First we plug 1, -2, and 1/2 into the expression to show they all make it zero. Then, we calculate their sum, pairwise sum, and product, showing they match the ratios of the coefficients.

Exam Tip: Be precise when evaluating \( p(1/2) \) with fractions. Show the step-by-step simplification to avoid any calculation slip-ups.

Question 17. If \( \alpha \) and \( \beta \) are the zeroes of quadratic polynomial \( x^2 - kx + 15 \) such that \( (\alpha + \beta)^2 - 2\alpha\beta = 34 \), find k.
Answer: For the quadratic polynomial \( x^2 - kx + 15 \):
Sum of zeroes, \( \alpha + \beta = -\frac{-k}{1} = k \)
Product of zeroes, \( \alpha\beta = \frac{15}{1} = 15 \)

We are given the condition:
\( (\alpha + \beta)^2 - 2\alpha\beta = 34 \)
Substituting our sum and product values:
\( (k)^2 - 2(15) = 34 \)
\( \implies k^2 - 30 = 34 \)
\( \implies k^2 = 64 \)
\( \implies k = \pm 8 \).
Therefore, the value of k is \( \pm 8 \).
In simple words: The sum of the roots is 'k' and their product is 15. Substituting these into the given formula yields \( k^2 - 30 = 34 \), which simplifies to \( k = \pm 8 \).

Exam Tip: Remember to write both positive and negative signs when taking the square root of a number, as both \( +8 \) and \( -8 \) are valid mathematical solutions.

Question 18. If one zero of polynomial \( 2x^2 - 3x + p \) is 3, then find the other root(zero). Also find the value of p.
Answer: Let the zeroes of the quadratic polynomial be \( \alpha = 3 \) and \( \beta \).
The coefficients are \( a = 2, b = -3, c = p \).

Using the sum of zeroes relationship:
\( \alpha + \beta = -\frac{b}{a} \)
\( \implies 3 + \beta = -\frac{-3}{2} \)
\( \implies 3 + \beta = \frac{3}{2} \)
\( \implies \beta = \frac{3}{2} - 3 = -\frac{3}{2} \).
Thus, the other zero is \( -\frac{3}{2} \).

Now, using the product of zeroes relationship to find p:
\( \alpha\beta = \frac{c}{a} \)
\( \implies (3)\left(-\frac{3}{2}\right) = \frac{p}{2} \)
\( \implies -\frac{9}{2} = \frac{p}{2} \)
\( \implies p = -9 \).
Therefore, the other root is \( -\frac{3}{2} \) and the value of p is -9.
In simple words: Since the sum of the roots is \( 3/2 \) and one root is 3, the other root must be \( -3/2 \). Multiplying these two roots together and comparing with \( p/2 \) tells us that \( p = -9 \).

Exam Tip: Using the sum of zeroes formula first is a very elegant way to find the other root without needing to calculate the value of \( p \) beforehand.

Question 19. If one zero of polynomial \( 2x^2 + px + 4 \) is 2, find the other zero. Also find p.
Answer: Let the zeroes of the polynomial be \( \alpha = 2 \) and \( \beta \).
Comparing with standard quadratic coefficients: \( a = 2, b = p, c = 4 \).

Using the product of zeroes relationship:
\( \alpha\beta = \frac{c}{a} \)
\( \implies 2\beta = \frac{4}{2} \)
\( \implies 2\beta = 2 \)
\( \implies \beta = 1 \).
Thus, the other zero is 1.

Now, using the sum of zeroes relationship to find p:
\( \alpha + \beta = -\frac{b}{a} \)
\( \implies 2 + 1 = -\frac{p}{2} \)
\( \implies 3 = -\frac{p}{2} \)
\( \implies p = -6 \).
Therefore, the other zero is 1 and the value of p is -6.
In simple words: Since the product of the roots is 2 and one root is 2, the other root must be 1. Adding these roots together shows their sum is 3, which helps us calculate that \( p = -6 \).

Exam Tip: When the constant term is known, use the product of zeroes relationship first to find the other root directly.

Question 20. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( ax^2 + bx + c \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \).
Answer: For the quadratic polynomial \( ax^2 + bx + c \):
Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} \)
Product of zeroes, \( \alpha\beta = \frac{c}{a} \)

Combining the fractions:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting our sum and product formulas:
\( = \frac{-b/a}{c/a} = -\frac{b}{c} \).
Thus, the value of the expression is \( -\frac{b}{c} \).
In simple words: Writing the expression with a common denominator gives sum over product. Dividing \( -b/a \) by \( c/a \) simplifies down to the clean formula \( -\frac{b}{c} \).

Exam Tip: This is a standard algebraic result. Memorising that \( \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{b}{c} \) is highly useful for quick calculations in other problems.

Question 21. If one zero of the polynomial \( (a^2 + 9)x^2 + 13x + 6a \) is the reciprocal of the other, find a.
Answer: Let the zeroes be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of the zeroes is:
\( \text{Product} = \alpha \cdot \frac{1}{\alpha} = 1 \)

According to the product of zeroes relationship \( \frac{c}{a} \):
\( \frac{6a}{a^2 + 9} = 1 \)
\( \implies a^2 + 9 = 6a \)
\( \implies a^2 - 6a + 9 = 0 \)
\( \implies (a - 3)^2 = 0 \)
\( \implies a = 3 \).
Thus, the value of a is 3.
In simple words: Reciprocal roots multiply to exactly 1. Comparing this to the formula \( c/a \) tells us that \( a^2 + 9 = 6a \), which solves to give \( a = 3 \).

Exam Tip: This question is identical to Question 21 on the previous page. Always follow the same consistent step-by-step logic.

Question 22. If \( \alpha \) and \( \beta \) are the zeroes of \( 2x^2 - 9x + 10 \), form the polynomial whose zeroes are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).
Answer: For the polynomial \( 2x^2 - 9x + 10 \):
Sum of zeroes, \( \alpha + \beta = -\frac{-9}{2} = \frac{9}{2} \)
Product of zeroes, \( \alpha\beta = \frac{10}{2} = 5 \)

For the new polynomial, let the zeroes be \( \alpha' = \frac{1}{\alpha} \) and \( \beta' = \frac{1}{\beta} \).
Sum of new zeroes:
\( S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{9/2}{5} = \frac{9}{10} \)

Product of new zeroes:
\( P' = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{5} \)

The required quadratic polynomial is:
\( P(x) = x^2 - S'x + P' \)
\( = x^2 - \frac{9}{10}x + \frac{1}{5} \)
We can multiply by 10 to write it with integer coefficients:
\( P(x) = 10x^2 - 9x + 2 \).
In simple words: From the original polynomial, we find that the sum of new zeroes is \( 9/10 \) and their product is \( 1/5 \). Placing these into our standard formula gives us the quadratic polynomial \( 10x^2 - 9x + 2 \).

Exam Tip: To find the polynomial with reciprocal zeroes quickly, simply swap the coefficients of the first and last terms: \( ax^2 + bx + c \implies cx^2 + bx + a \).

Question 23. Divide \( 2x^2 + 4x^3 + 5x - 6 \) by \( 2x^2 + 1 + 3x \) and verify the division algorithm.
Answer: Let us write both polynomials in standard descending order of their exponents:
Dividend, \( p(x) = 4x^3 + 2x^2 + 5x - 6 \)
Divisor, \( g(x) = 2x^2 + 3x + 1 \)

Dividing \( p(x) \) by \( g(x) \) using long division:
\[ \begin{array}{rll}
2x - 2 & \text{(Quotient)} \\
2x^2 + 3x + 1 \ \overline{\big) \ 4x^3 + 2x^2 + 5x - 6} \\
\underline{-\left(4x^3 + 6x^2 + 2x\right)} \phantom{- 6} \\
-4x^2 + 3x - 6 \\
\underline{-\left(-4x^2 - 6x - 2\right)} \\
9x - 4 & \text{(Remainder)}
\end{array} \]
The quotient \( q(x) = 2x - 2 \) and the remainder \( r(x) = 9x - 4 \).

Verification of the Division Algorithm:
We must verify that \( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \).
\( \text{RHS} = (2x^2 + 3x + 1)(2x - 2) + (9x - 4) \)
\( = 2x^2(2x - 2) + 3x(2x - 2) + 1(2x - 2) + 9x - 4 \)
\( = 4x^3 - 4x^2 + 6x^2 - 6x + 2x - 2 + 9x - 4 \)
\( = 4x^3 + 2x^2 + 5x - 6 \)
Since this is equal to the Dividend, the division algorithm is verified.
In simple words: First we rearrange the equations in standard order. Then we divide to find a quotient of \( 2x - 2 \) and a remainder of \( 9x - 4 \). Multiplying and adding them back together confirms our division is correct.

Exam Tip: This is identical to Question 2 from the previous page. Always rearrange the terms of polynomials in descending order of powers before starting division.

Question 24. The curve which represents a quadratic polynomial meets the x axis at \( (2, 0) \) and \( (-2, 0) \). Form the quadratic polynomial.
Answer: The points where the curve meets the x-axis represent the zeroes of the quadratic polynomial. Therefore, the zeroes are \( \alpha = 2 \) and \( \beta = -2 \).

Sum of zeroes, \( S = \alpha + \beta = 2 + (-2) = 0 \)
Product of zeroes, \( P = \alpha\beta = (2)(-2) = -4 \)

The quadratic polynomial is given by:
\( p(x) = x^2 - Sx + P \)
Substituting the values:
\( p(x) = x^2 - 4 \).
Thus, the quadratic polynomial is \( x^2 - 4 \).
In simple words: The graph crossing the x-axis at 2 and -2 means these are the roots. Using our sum (0) and product (-4) formulas, we get the polynomial \( x^2 - 4 \).

Exam Tip: Since the roots are symmetric about zero, the linear term of the quadratic polynomial is 0, leaving only \( x^2 - a^2 \).

Question 25. What must be subtracted from \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \), so that the difference is exactly divisible by \( 4x^2 + 3x - 2 \)?
Answer: We divide \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \) by \( 4x^2 + 3x - 2 \) using long division to find the remainder:
\[ \begin{array}{rll}
2x^2 + 2x - 1 & \text{(Quotient)} \\
4x^2 + 3x - 2 \ \overline{\big) \ 8x^4 + 14x^3 - 2x^2 + 7x - 8} \\
\underline{-\left(8x^4 + \phantom{0}6x^3 - 4x^2\right)} \phantom{+ 7x - 8} \\
8x^3 + \phantom{0}2x^2 + 7x - 8 \\
\underline{-\left(8x^3 + \phantom{0}6x^2 - 4x\right)} \phantom{- 8} \\
-4x^2 + 11x - 8 \\
\underline{-\left(-4x^2 - \phantom{0}3x + 2\right)} \\
14x - 10 & \text{(Remainder)}
\end{array} \]
The remainder is \( 14x - 10 \).
Therefore, \( 14x - 10 \) must be subtracted from the given polynomial so that the result is exactly divisible.
In simple words: Dividing the expression leaves us with a remainder of \( 14x - 10 \). Subtracting this remainder from the original expression makes it perfectly divisible.

Exam Tip: Always state clearly that "the polynomial to be subtracted is equal to the remainder" before writing down the final answer.

Question 26. Find the values of a and b such that \( x^4 + x^3 + 8x^2 + ax + b \) is exactly divisible by \( x^2 + 1 \)?
Answer: Since the polynomial is exactly divisible by \( x^2 + 1 \), the remainder must be 0.
Let us perform polynomial division:
\[ \begin{array}{rll}
x^2 + x + 7 & \text{(Quotient)} \\
x^2 + 1 \ \overline{\big) \ x^4 + x^3 + 8x^2 + ax + b} \\
\underline{-\left(x^4 \phantom{+ x^3} + x^2\right)} \phantom{+ ax + b} \\
x^3 + 7x^2 + ax + b \\
\underline{-\left(x^3 \phantom{+ 7x^2} + x\right)} \phantom{+ b} \\
7x^2 + (a - 1)x + b \\
\underline{-\left(7x^2 \phantom{+ (a - 1)x} + 7\right)} \\
(a - 1)x + (b - 7) & \text{(Remainder)}
\end{array} \]
The remainder is \( (a - 1)x + (b - 7) \).
Since the remainder must be 0:
1. \( a - 1 = 0 \implies a = 1 \)
2. \( b - 7 = 0 \implies b = 7 \)
Thus, the values are \( a = 1 \) and \( b = 7 \).
In simple words: We divide the polynomial by \( x^2 + 1 \). Setting the leftover remainder terms to zero gives us \( a = 1 \) and \( b = 7 \).

Exam Tip: Align your divisor terms correctly; since \( x^2 + 1 \) has no linear term, leave a gap when multiplying to keep subtraction steps clean.

Question 27. If the polynomial \( P(x) = x^4 - 6x^3 + 16x^2 - 25x + 10 \) divided by \( x^2 - 2x + k \), the remainder is \( x + a \). 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: This is a repeat of Question 6 from the previous page. We carry out division with the variable 'k' included. Comparing our variable remainder with the given \( x+a \) lets us solve for \( k=5 \) and \( a=-5 \).

Exam Tip: Review standard division properties to make sure you expand perfect squares like \( (8-k) \) correctly during coefficient comparisons.

Question 28. The zeroes of \( x^2 - kx + 6 \) are in the ratio 3: 2, find k.
Answer: Let the zeroes of the quadratic polynomial be \( 3m \) and \( 2m \).
Comparing with the standard quadratic form \( x^2 - kx + 6 \), we have:
Sum of zeroes, \( S = 3m + 2m = 5m = k \)
Product of zeroes, \( P = (3m)(2m) = 6m^2 = 6 \)

From the product relationship:
\( 6m^2 = 6 \)
\( \implies m^2 = 1 \)
\( \implies m = \pm 1 \)

Now, substitute the value of m into the sum relationship:
For \( m = 1 \):
\( k = 5(1) = 5 \)

For \( m = -1 \):
\( k = 5(-1) = -5 \).
Therefore, the value of k is \( \pm 5 \).
In simple words: We assume the roots are \( 3m \) and \( 2m \). Their product is \( 6m^2 = 6 \), which gives \( m = \pm 1 \). Substituting this back into the sum formula shows that \( k = \pm 5 \).

Exam Tip: Be sure to write \( \pm 5 \) because both positive and negative values are mathematically correct solutions for the variable \( k \).