CBSE Class 10 Mathematics Polynomials Worksheet Set 06

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 \( 5x^2 - 29x + 20 \):
\( 5x^2 - 25x - 4x + 20 = 5x(x - 5) - 4(x - 5) = (5x - 4)(x - 5) \)
Zeroes are \( \alpha = \frac{4}{5} \) and \( \beta = 5 \).
Verification: Sum \( \alpha + \beta = \frac{4}{5} + 5 = \frac{29}{5} = -\frac{b}{a} \). Product \( \alpha\beta = \frac{4}{5} \times 5 = 4 = \frac{c}{a} \). (Verified)

b) For \( 2\sqrt{2}x^2 - 9x + 5\sqrt{2} \):
\( 2\sqrt{2}x^2 - 4x - 5x + 5\sqrt{2} = 2\sqrt{2}x(x - \sqrt{2}) - 5(x - \sqrt{2}) = (2\sqrt{2}x - 5)(x - \sqrt{2}) \)
Zeroes are \( \alpha = \frac{5}{2\sqrt{2}} = \frac{5\sqrt{2}}{4} \) and \( \beta = \sqrt{2} \).
Verification: Sum \( \alpha + \beta = \frac{5}{2\sqrt{2}} + \sqrt{2} = \frac{9}{2\sqrt{2}} = -\frac{b}{a} \). Product \( \alpha\beta = \frac{5}{2\sqrt{2}} \times \sqrt{2} = \frac{5}{2} = \frac{5\sqrt{2}}{2\sqrt{2}} = \frac{c}{a} \). (Verified)

c) For \( 3\sqrt{3}x^2 - 19x + 10\sqrt{3} \):
\( 3\sqrt{3}x^2 - 9x - 10x + 10\sqrt{3} = 3\sqrt{3}x(x - \sqrt{3}) - 10(x - \sqrt{3}) = (3\sqrt{3}x - 10)(x - \sqrt{3}) \)
Zeroes are \( \alpha = \frac{10}{3\sqrt{3}} = \frac{10\sqrt{3}}{9} \) and \( \beta = \sqrt{3} \).
Verification: Sum \( \alpha + \beta = \frac{10}{3\sqrt{3}} + \sqrt{3} = \frac{19}{3\sqrt{3}} = -\frac{b}{a} \). Product \( \alpha\beta = \frac{10}{3\sqrt{3}} \times \sqrt{3} = \frac{10}{3} = \frac{c}{a} \). (Verified)

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

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

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

g) For \( x^2 - 9 \):
\( (x - 3)(x + 3) = 0 \)
Zeroes are \( \alpha = 3 \) and \( \beta = -3 \).
Verification: Sum \( \alpha + \beta = 0 = -\frac{b}{a} \). Product \( \alpha\beta = -9 = \frac{c}{a} \). (Verified)
In simple words: Factorize each quadratic expression to find the values of x that make it zero. Verify your answers by checking if their sum equals \( -b/a \) and their product equals \( c/a \).

Exam Tip: Keep your calculations organized. Always write down the values of the coefficients a, b, and c first to avoid simple sign errors in verification.

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) Let \( \alpha = 3 + \sqrt{2} \), \( \beta = 3 - \sqrt{2} \).
Sum \( S = 6 \), Product \( P = (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7 \).
The polynomial is \( x^2 - 6x + 7 \).

b) Let \( \alpha = -\sqrt{2} \), \( \beta = \sqrt{2} \).
Sum \( S = 0 \), Product \( P = -2 \).
The polynomial is \( x^2 - 2 \).

c) Let \( \alpha = \frac{1}{3} \), \( \beta = \frac{1}{4} \).
Sum \( S = \frac{7}{12} \), Product \( P = \frac{1}{12} \).
The polynomial is \( 12x^2 - 7x + 1 \).

d) Let \( \alpha = -5 \), \( \beta = -3 \).
Sum \( S = -8 \), Product \( P = 15 \).
The polynomial is \( x^2 + 8x + 15 \).

e) Let \( \alpha = 3 \), \( \beta = \frac{1}{5} \).
Sum \( S = \frac{16}{5} \), Product \( P = \frac{3}{5} \).
The polynomial is \( 5x^2 - 16x + 3 \).

f) Let \( \alpha = \frac{1}{a} \), \( \beta = \frac{1}{b} \).
Sum \( S = \frac{a+b}{ab} \), Product \( P = \frac{1}{ab} \).
The polynomial is \( abx^2 - (a+b)x + 1 \).
In simple words: To form a quadratic polynomial, use the formula \( x^2 - Sx + P \), where S is the sum of the zeroes and P is their product. Multiply by a constant to clear any fractions.

Exam Tip: If the zeroes are fractions, choose a suitable constant multiplier to write the polynomial with integer coefficients.

Question 3. Find all the Zeroes of \( x^3 + 6x^2 + 11x + 6 \) if \( (x + 1) \) is a factor.
Answer: Since \( (x + 1) \) is a factor, \( x = -1 \) is one of the zeroes.
Dividing \( x^3 + 6x^2 + 11x + 6 \) by \( (x + 1) \) using polynomial division:
\( x^3 + 6x^2 + 11x + 6 = (x + 1)(x^2 + 5x + 6) \)
Now, factorize the quadratic quotient:
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)
Therefore, the zeroes of the polynomial are \( -1, -2, \) and \( -3 \).
In simple words: Divide the cubic polynomial by the given factor to find a quadratic equation. Factoring this quadratic gives the other two zeroes, which are \( -2 \) and \( -3 \).

Exam Tip: Use synthetic division or standard long division to quickly find the quadratic quotient without making algebraic mistakes.

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, \( (x - 2) \) is a factor of the polynomial.
Dividing the polynomial by \( (x - 2) \):
\( x^3 - 10x^2 + 31x - 30 = (x - 2)(x^2 - 8x + 15) \)
Factorizing the remaining quadratic term:
\( x^2 - 8x + 15 = (x - 3)(x - 5) \)
Thus, the zeroes of the cubic polynomial are \( 2, 3, \) and \( 5 \).
In simple words: Since 2 is a zero, divide the expression by \( x-2 \). Factor the remaining part to find that the other two zeroes are 3 and 5.

Exam Tip: Check your final factorization by ensuring the product of the constant terms of the factors is equal to the constant term of the original polynomial (\( -2 \times -3 \times -5 = -30 \)).

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

Since 3 is a zero of \( p(x) \):
\( p(3) = 3^3 + a(3)^2 + b(3) - 30 = 0 \)
\( \implies 27 + 9a + 3b - 30 = 0 \implies 9a + 3b = 3 \implies 3a + b = 1 \) - (Equation 2)

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

Substituting \( a = -10 \) back into Equation 1:
\( 2(-10) + b = 11 \implies -20 + b = 11 \implies b = 31 \)
Therefore, \( a = -10 \) and \( b = 31 \).
In simple words: Plug 2 and 3 into the polynomial to get two linear equations. Solving these equations gives \( a = -10 \) and \( b = 31 \).

Exam Tip: Substituting known zeroes into a polynomial always leads to a system of linear equations, which can be solved easily by elimination.

Question 6. Divide \( x^4 - 4x^3 + 8x^2 + 7x + 10 \) by \( (x - 2) \) and verify the division algorithm.
Answer: Performing polynomial division:
\( \frac{x^4 - 4x^3 + 8x^2 + 7x + 10}{x - 2} \)
Using long division:
- First term: \( x^4 / x = x^3 \). Subtract \( x^3(x-2) \) to get remainder \( -2x^3 + 8x^2 + 7x + 10 \).
- Second term: \( -2x^3 / x = -2x^2 \). Subtract \( -2x^2(x-2) \) to get remainder \( 4x^2 + 7x + 10 \).
- Third term: \( 4x^2 / x = 4x \). Subtract \( 4x(x-2) \) to get remainder \( 15x + 10 \).
- Fourth term: \( 15x / x = 15 \). Subtract \( 15(x-2) \) to get remainder \( 40 \).

Thus, the Quotient \( q(x) = x^3 - 2x^2 + 4x + 15 \) and Remainder \( r(x) = 40 \).

Verification:
\( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \)
\( \text{LHS} = (x - 2)(x^3 - 2x^2 + 4x + 15) + 40 \)
\( = (x^4 - 2x^3 + 4x^2 + 15x) - (2x^3 - 4x^2 + 8x + 30) + 40 \)
\( = x^4 - 4x^3 + 8x^2 + 7x + 10 = \text{Dividend} \). (Hence Verified)
In simple words: Divide the polynomial by \( x-2 \) to get a quotient of \( x^3 - 2x^2 + 4x + 15 \) and a remainder of 40. Multiplying them back verifies the division algorithm.

Exam Tip: When performing long division, ensure you write down terms of all degrees in descending order and keep the signs correct during subtraction.

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, by the Factor Theorem, \( p(2) = 0 \):
\( 2^2 - k(2) + 10 = 0 \)
\( \implies 4 - 2k + 10 = 0 \)
\( \implies 14 - 2k = 0 \implies 2k = 14 \implies k = 7 \)
Thus, the value of \( k \) is 7.
In simple words: Plug \( x = 2 \) into the expression and set it equal to 0. Solving this gives \( k = 7 \).

Exam Tip: The Factor Theorem states that if \( x - a \) is a factor of \( p(x) \), then \( p(a) \) must equal 0. Use this direct substitution to solve for unknown variables quickly.

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 the polynomial, \( p(2) = 0 \):
\( 3(2)^2 - 17(2) + k = 0 \)
\( \implies 12 - 34 + k = 0 \)
\( \implies -22 + k = 0 \implies k = 22 \)
Thus, the value of \( k \) is 22.
In simple words: Substitute 2 into the expression and solve for \( k \), which gives \( k = 22 \).

Exam Tip: Zero of a polynomial means the value of x that makes the polynomial equal to zero. Simply substitute and equate to zero.

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, \( (x - 2)(x - 3) = x^2 - 5x + 6 \) is a factor of the given polynomial.
Dividing the polynomial by \( x^2 - 5x + 6 \):
- First term: \( 4x^4 / x^2 = 4x^2 \). Subtract \( 4x^2(x^2 - 5x + 6) \) to get remainder \( -x^2 + 5x - 6 \).
- Second term: \( -x^2 / x^2 = -1 \). Subtract \( -1(x^2 - 5x + 6) \) to get remainder 0.

So, the other factor is \( 4x^2 - 1 \).
To find the remaining zeroes, set \( 4x^2 - 1 = 0 \):
\( \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \)
Therefore, all the zeroes of the polynomial are \( 2, 3, \frac{1}{2}, \) and \( -\frac{1}{2} \).
In simple words: Multiply the two known factors to get \( x^2 - 5x + 6 \). Divide the big expression by this to get \( 4x^2 - 1 \), which yields the other two zeroes: \( 1/2 \) and \( -1/2 \).

Exam Tip: When dividing by a quadratic factor, make sure the remainder is exactly zero to confirm your division was correct.

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 \( x^2 + 5x + 6 \):
Sum of zeroes: \( \alpha + \beta = -5 \)
Product of zeroes: \( \alpha\beta = 6 \)

We need to find the value of:
\( \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the values:
\( \frac{\alpha + \beta}{\alpha\beta} = \frac{-5}{6} \)
Thus, the value of \( \alpha^{-1} + \beta^{-1} \) is \( -\frac{5}{6} \).
In simple words: Write \( \frac{1}{\alpha} + \frac{1}{\beta} \) as a single fraction: \( \frac{\alpha + \beta}{\alpha\beta} \). Since the sum of zeroes is \( -5 \) and the product is 6, the answer is \( -\frac{5}{6} \).

Exam Tip: Do not waste time calculating the individual zeroes \( \alpha \) and \( \beta \). Simplify the target expression first and use the relationships directly.

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, \( \left(x - \frac{1}{2}\right)(x - 1) = \left(2x - 1\right)(x - 1) = 2x^2 - 3x + 1 \) is a factor.
Let us divide the polynomial by \( 2x^2 - 3x + 1 \) using long division:
- First term: \( 2x^4 / 2x^2 = x^2 \). Subtract \( x^2(2x^2 - 3x + 1) \) to get remainder \( -4x^2 + 6x - 2 \).
- Second term: \( -4x^2 / 2x^2 = -2 \). Subtract \( -2(2x^2 - 3x + 1) \) to get remainder 0.

The other factor is \( x^2 - 2 \).
To find the other zeroes, set this factor to 0:
\( x^2 - 2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2} \)
Thus, the other zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \).
In simple words: Multiply the factors of the known zeroes to get \( 2x^2 - 3x + 1 \). Divide the original expression by this to get \( x^2 - 2 \). Setting this to zero gives the other zeroes: \( \sqrt{2} \) and \( -\sqrt{2} \).

Exam Tip: Be comfortable working with irrational numbers. Do not convert \( \sqrt{2} \) into decimal form unless requested.

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, \( (x + 5)(x - 7) = x^2 - 2x - 35 \) is a factor.
Let us divide the polynomial by \( x^2 - 2x - 35 \):
- First term: \( x^4 / x^2 = x^2 \). Subtract \( x^2(x^2 - 2x - 35) \) to get remainder \( -4x^3 + 9x^2 + 138x - 35 \).
- Second term: \( -4x^3 / x^2 = -4x \). Subtract \( -4x(x^2 - 2x - 35) \) to get remainder \( x^2 - 2x - 35 \).
- Third term: \( x^2 / x^2 = 1 \). Subtract \( 1(x^2 - 2x - 35) \) to get remainder 0.

So, the other factor is \( x^2 - 4x + 1 \).
To find the other zeroes, set \( x^2 - 4x + 1 = 0 \):
Using the quadratic formula:
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \)
Therefore, the other zeroes of the polynomial are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \).
In simple words: Divide the polynomial by the product of the known factors, \( x^2 - 2x - 35 \), to get \( x^2 - 4x + 1 \). Solving this gives the other zeroes as \( 2 \pm \sqrt{3} \).

Exam Tip: Use the quadratic formula when the quadratic factor cannot be easily factored by splitting the middle term.

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 polynomial be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of these zeroes is:
\( \text{Product} = \alpha \times \frac{1}{\alpha} = 1 \)

From the given polynomial \( 5z^2 + 13z - p \):
\( \text{Product of zeroes} = \frac{\text{Constant term}}{\text{Coefficient of } z^2} = \frac{-p}{5} \)

Equating the two:
\( \frac{-p}{5} = 1 \implies -p = 5 \implies p = -5 \)
Thus, the value of \( p \) is \( -5 \).
In simple words: Since the zeroes are reciprocals, their product must be 1. According to the formula, the product is \( -p/5 \), which gives \( p = -5 \).

Exam Tip: Whenever a problem states that one zero is the reciprocal of the other, use the product of zeroes relationship, which immediately simplifies to \( a = c \).

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: By the division algorithm:
\( p(x) = g(x) \cdot q(x) + r(x) \)
\( \implies 4x^4 - 3x^3 - 42x^2 - 55x - 17 = g(x) \cdot (x^2 - 3x - 5) + (5x + 8) \)
\( \implies g(x) \cdot (x^2 - 3x - 5) = (4x^4 - 3x^3 - 42x^2 - 55x - 17) - (5x + 8) \)
\( \implies g(x) \cdot (x^2 - 3x - 5) = 4x^4 - 3x^3 - 42x^2 - 60x - 25 \)
\( \implies g(x) = \frac{4x^4 - 3x^3 - 42x^2 - 60x - 25}{x^2 - 3x - 5} \)

Let us perform polynomial division:
- First term: \( 4x^4 / x^2 = 4x^2 \). Subtract \( 4x^2(x^2 - 3x - 5) \) to get remainder \( 9x^3 - 22x^2 - 60x - 25 \).
- Second term: \( 9x^3 / x^2 = 9x \). Subtract \( 9x(x^2 - 3x - 5) \) to get remainder \( 5x^2 - 15x - 25 \).
- Third term: \( 5x^2 / x^2 = 5 \). Subtract \( 5(x^2 - 3x - 5) \) to get remainder 0.

So, \( g(x) = 4x^2 + 9x + 5 \).
In simple words: Subtract the remainder from the dividend first, and then divide the result by the quotient to get \( g(x) = 4x^2 + 9x + 5 \).

Exam Tip: Be careful to change the signs of the remainder terms when subtracting them from the main dividend.

Question 15. If \( \frac{1}{2} \) and 1 are zeroes of \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \), find the other zeroes.
Answer: This question is identical to Q11.
The other zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \).
In simple words: The division of the polynomial by the factors of \( 1/2 \) and 1 yields the other zeroes as \( \pm \sqrt{2} \).

Exam Tip: Always make sure to check if a question is repeated on a worksheet to save valuable time.

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 \( p(x) = 2x^3 + x^2 - 5x + 2 \).
- For \( x = 1 \): \( p(1) = 2(1)^3 + 1^2 - 5(1) + 2 = 2 + 1 - 5 + 2 = 0 \) (Verified)
- For \( x = -2 \): \( p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = -16 + 4 + 10 + 2 = 0 \) (Verified)
- For \( x = \frac{1}{2} \): \( p\left(\frac{1}{2}\right) = 2\left(\frac{1}{8}\right) + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = \frac{1}{2} - \frac{5}{2} + 2 = 0 \) (Verified)

Relationship Verification:
Here, \( a = 2, b = 1, c = -5, d = 2 \).
Let \( \alpha = 1, \beta = -2, \gamma = \frac{1}{2} \).
1. Sum of zeroes:
\( \alpha + \beta + \gamma = 1 - 2 + \frac{1}{2} = -\frac{1}{2} \)
Formula: \( -\frac{b}{a} = -\frac{1}{2} \). (Verified)
2. Sum of product of zeroes taken two at a time:
\( \alpha\beta + \beta\gamma + \gamma\alpha = (1)(-2) + (-2)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)(1) = -2 - 1 + \frac{1}{2} = -\frac{5}{2} \)
Formula: \( \frac{c}{a} = -\frac{5}{2} \). (Verified)
3. Product of zeroes:
\( \alpha\beta\gamma = (1)(-2)\left(\frac{1}{2}\right) = -1 \)
Formula: \( -\frac{d}{a} = -\frac{2}{2} = -1 \). (Verified)
In simple words: Substitute each number into the expression to check that they all make the equation zero. Then verify the sum and product formulas for the zeroes of a cubic polynomial.

Exam Tip: For cubic verification, you must verify all three relations: sum, sum of products, and product of zeroes.

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: From the polynomial \( x^2 - kx + 15 \):
Sum of zeroes: \( \alpha + \beta = k \)
Product of zeroes: \( \alpha\beta = 15 \)

Given condition:
\( (\alpha + \beta)^2 - 2\alpha\beta = 34 \)
Substitute the values:
\( k^2 - 2(15) = 34 \)
\( \implies k^2 - 30 = 34 \implies k^2 = 64 \implies k = \pm 8 \).
In simple words: Sum of zeroes is \( k \) and their product is 15. Substituting these into the given condition gives \( k^2 - 30 = 34 \), which simplifies to \( k = \pm 8 \).

Exam Tip: Don't forget that taking the square root of 64 gives both positive and negative values, \( \pm 8 \).

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: Since 3 is a zero of \( 2x^2 - 3x + p \):
\( 2(3)^2 - 3(3) + p = 0 \implies 18 - 9 + p = 0 \implies 9 + p = 0 \implies p = -9 \)

Let the other zero be \( \beta \).
Sum of zeroes:
\( 3 + \beta = -\frac{b}{a} = \frac{3}{2} \implies \beta = \frac{3}{2} - 3 = -\frac{3}{2} \)
Thus, the other zero is \( -\frac{3}{2} \), and the value of \( p \) is \( -9 \).
In simple words: Plug 3 in to find \( p = -9 \). Then use the sum of zeroes relation to get the other zero, which is \( -1.5 \).

Exam Tip: Using the sum of zeroes relationship is faster than factoring the polynomial to find the second zero.

Question 19. If one zero of polynomial \( 2x^2 + px + 4 \) is 2, find the other zero. Also find \( p \).
Answer: Since 2 is a zero of \( 2x^2 + px + 4 \):
\( 2(2)^2 + p(2) + 4 = 0 \implies 8 + 2p + 4 = 0 \implies 2p = -12 \implies p = -6 \)

Let the other zero be \( \beta \).
Product of zeroes:
\( 2 \times \beta = \frac{c}{a} = \frac{4}{2} = 2 \implies \beta = 1 \)
Thus, the other zero is 1, and the value of \( p \) is \( -6 \).
In simple words: Substitute 2 to get \( p = -6 \). Then use the product of zeroes relation to find that the other zero is 1.

Exam Tip: Using the product of zeroes relationship is a great shortcut to find the second zero when the constant term and first term are both known.

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 \( ax^2 + bx + c \):
\( \alpha + \beta = -\frac{b}{a} \)
\( \alpha\beta = \frac{c}{a} \)

We need to find the value of:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the values:
\( \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \)
Thus, the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) is \( -\frac{b}{c} \).
In simple words: Combining the terms into a single fraction gives \( \frac{\alpha + \beta}{\alpha\beta} \). Since sum is \( -b/a \) and product is \( c/a \), the result is \( -b/c \).

Exam Tip: Learn this general formula \( \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{b}{c} \) as a shortcut, as it is a very common question in exams.

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 \( \frac{1}{\alpha} \).
Product of zeroes:
\( \text{Product} = \alpha \times \frac{1}{\alpha} = 1 \)

From the polynomial \( (a^2 + 9)x^2 + 13x + 6a \):
\( \text{Product of zeroes} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{6a}{a^2 + 9} \)

Equating the two:
\( \frac{6a}{a^2 + 9} = 1 \implies a^2 + 9 = 6a \implies a^2 - 6a + 9 = 0 \)
\( \implies (a - 3)^2 = 0 \implies a = 3 \).
Thus, the value of \( a \) is 3.
In simple words: Since the zeroes are reciprocals, their product is 1. Setting \( \frac{6a}{a^2+9} = 1 \) and simplifying gives \( (a-3)^2 = 0 \), which means \( a = 3 \).

Exam Tip: Use algebraic identity \( x^2 - 2xy + y^2 = (x-y)^2 \) to factorize the quadratic equation quickly.

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 \( 2x^2 - 9x + 10 \):
\( \alpha + \beta = \frac{9}{2} \)
\( \alpha\beta = \frac{10}{2} = 5 \)

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

The new quadratic polynomial is:
\( x^2 - Sx + P = x^2 - \frac{9}{10}x + \frac{1}{5} \)
Multiplying by 10 to clear the fraction:
\( 10x^2 - 9x + 2 \)
In simple words: Calculate the new sum and product of zeroes as \( 9/10 \) and \( 1/5 \). Using \( x^2 - Sx + P \) gives the new polynomial as \( 10x^2 - 9x + 2 \).

Exam Tip: Alternatively, when zeroes are reciprocated, we can get the new polynomial simply by reversing the coefficients: \( ax^2 + bx + c \implies cx^2 + bx + a \).

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

Let us divide \( p(x) by g(x) \):
- First term: \( 4x^3 / 2x^2 = 2x \). Subtract \( 2x(2x^2 + 3x + 1) = 4x^3 + 6x^2 + 2x \) to get remainder \( -4x^2 + 3x - 6 \).
- Second term: \( -4x^2 / 2x^2 = -2 \). Subtract \( -2(2x^2 + 3x + 1) = -4x^2 - 6x - 2 \) to get remainder \( 9x - 4 \).

Thus, the Quotient \( q(x) = 2x - 2 \) and Remainder \( r(x) = 9x - 4 \).

Verification:
\( \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \)
\( \text{RHS} = (2x^2 + 3x + 1)(2x - 2) + (9x - 4) \)
\( = (4x^3 - 4x^2 + 6x^2 - 6x + 2x - 2) + 9x - 4 \)
\( = 4x^3 + 2x^2 + 5x - 6 = \text{Dividend} \). (Hence Verified)
In simple words: Rearrange the expressions first. Divide to find a quotient of \( 2x - 2 \) and a remainder of \( 9x - 4 \). Multiply divisor and quotient and add remainder to verify.

Exam Tip: Never start dividing polynomials without writing them down in standard descending power order first.

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 \( x \)-intercepts of the curve are the zeroes of the quadratic polynomial.
Thus, the zeroes are \( \alpha = 2 \) and \( \beta = -2 \).
Sum of zeroes: \( S = 2 + (-2) = 0 \)
Product of zeroes: \( P = 2 \times (-2) = -4 \)
The quadratic polynomial is:
\( x^2 - Sx + P = x^2 - 0x - 4 = x^2 - 4 \)
Thus, the polynomial is \( x^2 - 4 \).
In simple words: The points where the curve crosses the x-axis are its zeroes: 2 and -2. Sum is 0 and product is -4, giving the polynomial \( x^2 - 4 \).

Exam Tip: Remember that the \( x \)-coordinates of the points where a graph cuts the \( x \)-axis are the real zeroes of the polynomial.

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: The expression that must be subtracted is the remainder obtained when \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \) is divided by \( 4x^2 + 3x - 2 \).
Let us divide:
- First term: \( 8x^4 / 4x^2 = 2x^2 \). Subtract \( 2x^2(4x^2 + 3x - 2) = 8x^4 + 6x^3 - 4x^2 \) to get remainder \( 8x^3 + 2x^2 + 7x - 8 \).
- Second term: \( 8x^3 / 4x^2 = 2x \). Subtract \( 2x(4x^2 + 3x - 2) = 8x^3 + 6x^2 - 4x \) to get remainder \( -4x^2 + 11x - 8 \).
- Third term: \( -4x^2 / 4x^2 = -1 \). Subtract \( -1(4x^2 + 3x - 2) = -4x^2 - 3x + 2 \) to get remainder \( 14x - 10 \).

Since the remainder is \( 14x - 10 \), the expression to be subtracted is \( 14x - 10 \).
In simple words: Divide the dividend by the divisor. The leftover remainder is the term that must be subtracted to make it completely divisible, which is \( 14x - 10 \).

Exam Tip: Be careful with sign changes during subtraction in long division; for example, subtracting \( -4x \) is equivalent to adding \( 4x \).

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: Let us divide the polynomial by \( x^2 + 1 \):
- First term: \( x^4 / x^2 = x^2 \). Subtract \( x^2(x^2 + 1) = x^4 + x^2 \) to get remainder \( x^3 + 7x^2 + ax + b \).
- Second term: \( x^3 / x^2 = x \). Subtract \( x(x^2 + 1) = x^3 + x \) to get remainder \( 7x^2 + (a - 1)x + b \).
- Third term: \( 7x^2 / x^2 = 7 \). Subtract \( 7(x^2 + 1) = 7x^2 + 7 \) to get remainder \( (a - 1)x + (b - 7) \).

For exact divisibility, the remainder must be 0:
\( (a - 1)x + (b - 7) = 0 \cdot x + 0 \)
Comparing coefficients on both sides:
\( a - 1 = 0 \implies a = 1 \)
\( b - 7 = 0 \implies b = 7 \)
Thus, \( a = 1 \) and \( b = 7 \).
In simple words: Perform division to find the remainder \( (a - 1)x + (b - 7) \). For exact division, set this remainder to 0, which yields \( a = 1 \) and \( b = 7 \).

Exam Tip: When a polynomial is exactly divisible by another, always set the remainder equal to zero and solve for the unknown coefficients.

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 \( x^2 - 2x + k \):
- First term: \( x^4 / x^2 = x^2 \). Subtract \( x^2(x^2 - 2x + k) = x^4 - 2x^3 + kx^2 \) to get remainder \( -4x^3 + (16 - k)x^2 - 25x + 10 \).
- Second term: \( -4x^3 / x^2 = -4x \). Subtract \( -4x(x^2 - 2x + k) = -4x^3 + 8x^2 - 4kx \) to get remainder \( (8 - k)x^2 + (4k - 25)x + 10 \).
- Third term: \( (8 - k)x^2 / x^2 = (8 - k) \). Subtract \( (8 - k)(x^2 - 2x + k) = (8 - k)x^2 - 2(8 - k)x + k(8 - k) = (8 - k)x^2 + (2k - 16)x + (8k - k^2) \).
The remainder is:
\( [(4k - 25) - (2k - 16)]x + [10 - (8k - k^2)] = (2k - 9)x + (k^2 - 8k + 10) \)

Given that the remainder is \( x + a \). Comparing the coefficients:
\( 2k - 9 = 1 \implies 2k = 10 \implies k = 5 \)
Substituting \( k = 5 \) into the constant term:
\( a = k^2 - 8k + 10 = 5^2 - 8(5) + 10 = 25 - 40 + 10 = -5 \)
Thus, \( k = 5 \) and \( a = -5 \).
In simple words: Divide the polynomial to get a remainder in terms of \( k \). Comparing it with \( x + a \) gives \( k = 5 \) and \( a = -5 \).

Exam Tip: Keep your algebraic expressions grouped carefully inside parentheses to avoid errors when comparing multi-term coefficients.

Question 28. The zeroes of \( x^2 - kx + 6 \) are in the ratio 3: 2, find \( k \).
Answer: Let the zeroes be \( 3m \) and \( 2m \).
From the polynomial \( x^2 - kx + 6 \):
Product of zeroes:
\( (3m)(2m) = 6 \implies 6m^2 = 6 \implies m^2 = 1 \implies m = \pm 1 \)

Sum of zeroes:
\( 3m + 2m = k \implies 5m = k \)
- If \( m = 1 \): \( k = 5 \).
- If \( m = -1 \): \( k = -5 \).
Therefore, the value of \( k \) is \( \pm 5 \).
In simple words: Let the zeroes be \( 3m \) and \( 2m \). Since their product is 6, we get \( m = \pm 1 \). The sum is \( 5m = k \), so \( k = \pm 5 \).

Exam Tip: Always consider both the positive and negative cases when taking the square root of a variable during intermediate steps.