CBSE Class 10 Mathematics Polynomials Worksheet Set 06

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \). Find the value of
(i) \( \alpha - \beta \)
(ii) \( \alpha^2 + \beta^2 \)
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \).
\( \therefore \quad \alpha + \beta = -\frac{b}{a}; \quad \alpha\beta = \frac{c}{a} \)
(i) \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \left(-\frac{b}{a}\right)^2 - \frac{4c}{a} = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \)
\( \alpha - \beta = \frac{\sqrt{b^2 - 4ac}}{a} \)
(ii) \( \alpha^2 + \beta^2 = \alpha^2 + \beta^2 + 2\alpha\beta - 2\alpha\beta = (\alpha + \beta)^2 - 2\alpha\beta \)
\( = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2 - 2ac}{a^2} \) Ans.

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( ax^2 + bx + c \). Find the value of
(i) \( \alpha^2 - \beta^2 \)
(ii) \( \alpha^3 + \beta^3 \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of \( ax^2 + bx + c \)
\( \therefore \quad \alpha + \beta = \frac{-b}{a}, \quad \alpha\beta = \frac{c}{a} \)
(i) \( \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) = -\frac{b}{a}\sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = -\frac{b}{a}\sqrt{\left(\frac{-b}{a}\right)^2 - 4\frac{c}{a}} = -\frac{b}{a}\sqrt{\frac{b^2 - 4ac}{a^2}} = -\frac{b\sqrt{b^2 - 4ac}}{a^2} \)
(ii) \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (\alpha + \beta)[(\alpha^2 + \beta^2 + 2\alpha\beta) - 3\alpha\beta] \)
\( = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] = \frac{-b}{a} \left[ \left(\frac{-b}{a}\right)^2 - \frac{3c}{a} \right] = \frac{-b}{a} \left[ \frac{b^2}{a^2} - \frac{3c}{a} \right] = \frac{-b}{a} \left( \frac{b^2 - 3ac}{a^2} \right) = \frac{-b^3 + 3abc}{a^3} \) Ans.

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \) then form the polynomial whose zeroes are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of \( ax^2 + bx + c \)
\( \alpha + \beta = -\frac{b}{a}; \quad \alpha\beta = \frac{c}{a} \)
Sum of the zeroes = \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{\frac{-b}{a}}{\frac{c}{a}} = \frac{-b}{c} \)
Product of the zeroes = \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{\frac{c}{a}} = \frac{a}{c} \)
But required polynomial is
\( x^2 - (\text{sum of the zeroes}) x + \text{Product of the zeroes} \)

\( \implies \) \( x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} \) or \( x^2 + \frac{b}{c}x + \frac{a}{c} \) or \( c\left(x^2 + \frac{b}{c}x + \frac{a}{c}\right) \implies cx^2 + bx + a \) Ans.

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 4x + 3 \), form the polynomial whose zeroes are \( 1 + \frac{\beta}{\alpha} \) and \( 1 + \frac{\alpha}{\beta} \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the cubic polynomial \( x^2 + 4x + 3 \)
So, \( \alpha + \beta = -4, \alpha\beta = 3 \) [using sum of zeroes and product of zeroes]
Sum of the zeroes = \( 1 + \frac{\beta}{\alpha} + 1 + \frac{\alpha}{\beta} = \frac{\alpha\beta + \beta^2 + \alpha\beta + \alpha^2}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + 2\alpha\beta}{\alpha\beta} = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(-4)^2}{3} = \frac{16}{3} \)
Product of the zeroes = \( \left(1 + \frac{\beta}{\alpha}\right)\left(1 + \frac{\alpha}{\beta}\right) = 1 + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + \frac{\alpha\beta}{\alpha\beta} = \frac{\alpha\beta + \alpha^2 + \beta^2 + \alpha\beta}{\alpha\beta} = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(-4)^2}{3} = \frac{16}{3} \)
But required polynomial is: \( x^2 - (\text{sum of the zeroes}) x + \text{product the zeroes} \)
or \( x^2 - \frac{16}{3}x + \frac{16}{3} \) or \( k \left(x^2 - \frac{16}{3}x + \frac{16}{3}\right) \)
or \( 3 \left(x^2 - \frac{16}{3}x + \frac{16}{3}\right) \) (if \( k=3 \))

\( \implies \) \( 3x^2 - 16x + 16. \) Ans.

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = kx^2 + 4x + 4 \) such that \( \alpha^2 + \beta^2 = 24 \), find the values of \( k \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = kx^2 + 4x + 4 \)
\( \therefore \alpha + \beta = -\frac{4}{k} \) and \( \alpha\beta = \frac{4}{k} \)
Now, \( \alpha^2 + \beta^2 = 24 \)

\( \implies \) \( (\alpha + \beta)^2 - 2\alpha\beta = 24 \)

\( \implies \) \( \left(-\frac{4}{k}\right)^2 - 2 \times \frac{4}{k} = 24 \)

\( \implies \) \( \frac{16}{k^2} - \frac{8}{k} = 24 \)

\( \implies \) \( 16 - 8k = 24k^2 \)

\( \implies \) \( 3k^2 + k - 2 = 0 \)

\( \implies \) \( 3k^2 + 3k - 2k - 2 = 0 \)

\( \implies \) \( 3k(k + 1) - 2(k + 1) = 0 \)

\( \implies \) \( (k + 1)(3k - 2) = 0 \)

\( \implies \) \( k + 1 = 0 \) or \( 3k - 2 = 0 \)

\( \implies \) \( k = -1 \) or \( k = \frac{2}{3} \)
Hence, \( k = -1 \) or \( k = \frac{2}{3} \)

 

Question. *Verify that: \( \frac{1}{2}, 1, -2 \) are zeroes of cubic polynomial \( 2x^3 + x^2 - 5x + 2 \). Also verify the relationship between the zeroes and their coefficients.
Answer: \( p(x) = 2x^3 + x^2 - 5x + 2 \)
\( p\left(\frac{1}{2}\right) = 2 \times \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 5 \times \frac{1}{2} + 2 \)
\( = \frac{1}{4} + \frac{1}{4} - \frac{5}{2} + 2 = 0 \)
\( p(1) = 2 \times (1)^3 + (1)^2 - 5 \times (1) + 2 = 0 \)
\( p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = 0 \)
Therefore, \( \frac{1}{2}, 1 \) and \( -2 \) are zeroes.
Now, comparing \( p(x) \) with \( ax^3 + bx^2 + cx + d \)
\( a = 2, b = 1, c = -5, d = 2 \)
Sum of zeroes = \( \frac{1}{2} + 1 - 2 = \frac{-1}{2} = \frac{-b}{a} \)
Sum of product of zeroes taken two at a time
\( = \frac{1}{2} \times 1 + \frac{1}{2} \times -2 + 1 \times -2 = \frac{-5}{2} = \frac{c}{a} \)
Product of zeroes = \( \frac{1}{2} \times 1 \times -2 = \frac{-2}{2} = \frac{-d}{a} \) Verified.

 

Question. If \( \alpha, \beta \) are the zeroes of the polynomial \( f(x) = 2x^2 + 5x + k \). If it satisfying the relation \( \alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4} \), then find the value of \( k \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( f(x) = 2x^2 + 5x + k \).
\( \therefore \alpha + \beta = -\frac{5}{2} \) and \( \alpha\beta = \frac{k}{2} \)
Now, \( \alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4} \)
adding and subtracting \( \alpha\beta \)

\( \implies \) \( (\alpha^2 + \beta^2 + 2\alpha\beta) - \alpha\beta = \frac{21}{4} \)

\( \implies \) \( (\alpha + \beta)^2 - \alpha\beta = \frac{21}{4} \)

\( \implies \) \( \frac{25}{4} - \frac{k}{2} = \frac{21}{4} \quad \left[\because \alpha + \beta = -\frac{5}{2} \text{ and } \alpha\beta = \frac{k}{2}\right] \)

\( \implies \) \( \frac{k}{2} = -1 \)

\( \implies \) \( k = 2. \)

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = 2x^2 - 5x + 7 \), find a polynomial whose zeroes are \( 2\alpha + 3\beta \) and \( 3\alpha + 2\beta \).
Answer: Since \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = 2x^2 - 5x + 7 \).
\( \therefore \alpha + \beta = -\left(-\frac{5}{2}\right) = \frac{5}{2} \) and \( \alpha\beta = \frac{7}{2} \)
Let \( S \) and \( P \) denote respectively the sum and product of the zeroes of the required polynomial.
Then, \( S = (2\alpha + 3\beta) + (3\alpha + 2\beta) = 5(\alpha + \beta) = 5 \times \frac{5}{2} = \frac{25}{2} \)
and \( P = (2\alpha + 3\beta)(3\alpha + 2\beta) \)

\( \implies \) \( P = 6(\alpha^2 + \beta^2) + 13\alpha\beta = 6\alpha^2 + 6\beta^2 + 12\alpha\beta + \alpha\beta = 6(\alpha + \beta)^2 + \alpha\beta \)

\( \implies \) \( P = 6 \times \left(\frac{5}{2}\right)^2 + \frac{7}{2} = \frac{75}{2} + \frac{7}{2} = 41 \)
Hence, the required polynomial is given by \( = k(x^2 - Sx + P) \)

\( \implies \) \( k\left(x^2 - \frac{25}{2}x + 41\right) \), where \( k \) is any non-zero real number.

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x) = 3x^2 - 4x + 1 \), find a quadratic polynomial whose zeroes are \( \frac{\alpha^2}{\beta} \) and \( \frac{\beta^2}{\alpha} \).
Answer: Sum of zeroes of given polynomial \( 3x^2 - 4x + 1 \) is \( \alpha + \beta = \frac{4}{3} \)
Product of zeroes of given polynomial \( 3x^2 - 4x + 1 \) is \( \alpha\beta = \frac{1}{3} \)
The zeroes of required polynomial are \( \frac{\alpha^2}{\beta} \) and \( \frac{\beta^2}{\alpha} \) so the sum of zeroes of required polynomial
\( = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta} = \frac{\left(\frac{4}{3}\right)^3 - 3 \times \frac{1}{3} \times \frac{4}{3}}{\left(\frac{1}{3}\right)} = \frac{28}{9} \)
Product of zeroes = \( \frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha} = \alpha\beta = \frac{1}{3} \)
Hence required polynomial = \( x^2 - \frac{28}{9}x + \frac{1}{3} = 9x^2 - 28x + 3. \) Ans.

 

Question. *Example 10. If \( \alpha, \beta, \gamma \) are zeroes of polynomial \( 6x^3 + 3x^2 - 5x + 1 \), then find the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \).
Answer: \( p(x) = 6x^3 + 3x^2 - 5x + 1 \)
\( a = 6, b = 3, c = -5, d = 1 \)
\( \alpha, \beta \) and \( \gamma \) are zeroes.
\( \therefore \alpha + \beta + \gamma = \frac{-b}{a} = \frac{-3}{6} = \frac{-1}{2} \)
\( \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} = \frac{-5}{6} \)
\( \alpha\beta\gamma = \frac{-d}{a} = \frac{-1}{6} \)
Now, \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} = \frac{-5/6}{-1/6} = 5 \)

EXERCISE 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( 2x^2 - 4x + 5 \), then find the value of
(i) \( \alpha^2 + \beta^2 \)
(ii) \( \frac{1}{\alpha} + \frac{1}{\beta} \)
(iii) \( (\alpha - \beta)^2 \)
(iv) \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)
Answer: (i) -1, (ii) 4/5, (iii) -6, (iv) -4/25

 

Question. If \( \alpha \) and \( \beta \) be the zeroes of the polynomial \( ax^2 + bx + c \), then find the value of
(i) \( \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} \)
(ii) \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)
Answer: (i) \( \frac{-b}{\sqrt{ac}} \), (ii) \( \frac{b^2 - 2ac}{c^2} \)

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 8x + 6 \). Form a polynomial whose zeroes are
(i) \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \)
(ii) \( \frac{1}{\alpha^2} \) and \( \frac{1}{\beta^2} \)
(iii) \( 1 + \frac{\beta}{\alpha}, 1 + \frac{\alpha}{\beta} \)
(iv) \( \alpha - \beta, \alpha + \beta \)
Answer: (i) \( 6x^2 + 8x + 1 \), (ii) \( 36x^2 - 52x + 1 \), (iii) \( 3x^2 - 32x + 32 \), (iv) \( x^2 + (8 \pm 2\sqrt{10})x - 16\sqrt{10} \)

 

Question. If \( \alpha \) and \( \frac{1}{\alpha} \) are the zeroes of polynomial \( 4x^2 - 2x + 1(k - 4) \). Find the value of \( k \).
Answer: \( k = 8 \)

 

Question. For what value of \( k \) the zeroes of polynomial \( x^2 - 5x + k \) are differ by 1 i.e. \( \alpha - \beta = 1 \).
Answer: \( k = 6 \)

 

Question. If sum of squares of zeroes of a quadratic polynomial: \( p(x) = x^2 - 8x + k \) is find the value of \( k \).
Answer: \( k = 12 \)

 

Question. If \( \alpha \) & \( \beta \) are zeroes of quadratic polynomial \( x^2 - ax + b \) then find value of \( \alpha^2 + \beta^2 \).
Answer: \( a^2 - 2b \)

 

Question. If one of the zeroes of polynomial \( p(x) : x^2 - 8x + k \) is seven times of other then find the value of \( k \).
Answer: \( k = 7 \)

 

Question. If zeroes of polynomial: \( 2x^2 - 3x + \lambda \) and \( ax^2 - 6x + \mu \) are same then find the value of: \( \frac{\lambda \times a}{\mu} \).
Answer: 2

 

Question. If zeroes of polynomial: \( 1x^2 - 2x - 3 \) are half of the zeroes of polynomial \( ax^2 + bx + c \), then find the value of \( (a - b - c) \).
Answer: \( a - b - c = 17. \)

 

Question. What must be added to \( f(x) = 4x^4 + 2x^3 - 2x^2 + x - 1 \) so that the resulting polynomial is divisible by \( g(x) = x^2 + 2x - 3 \).
Answer:
By division algorithm, we have \( f(x) = g(x) \times q(x) + r(x) \)

\( \implies \) \( f(x) - r(x) = g(x) \times q(x) \implies f(x) + \{-r(x)\} = g(x) \times q(x) \)
Clearly, RHS is divisible by \( g(x) \). Therefore, LHS is also divisible by \( g(x) \). Thus, if we add \( -r(x) \) to \( f(x) \), then the resulting polynomial is divisible by \( g(x) \). Let us now find the remainder when \( f(x) \) is divided by \( g(x) \).
By long division of \( 4x^4 + 2x^3 - 2x^2 + x - 1 \) by \( x^2 + 2x - 3 \), we get:
\( r(x) = -61x + 65 \)
Therefore, \( -r(x) = 61x - 65 \).
Hence, we should add \( 61x - 65 \) to \( f(x) \) so that the resulting polynomial is divisible by \( g(x) \). Ans.

 

Question. What must be subtracted from \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \) so that the resulting polynomial is exactly divisible by \( 4x^2 + 3x - 2 \)?
Answer:
We know that Dividend = Quotient \( \times \) Divisor + Remainder

\( \implies \) Dividend - Remainder = Quotient \( \times \) Division
Clearly, RHS of above result is divisible by the divisor.
Therefore, LHS is also divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor.
Dividing \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \) by \( 4x^2 + 3x - 2 \), we get:
Remainder = \( 14x - 10 \).
Thus, if we subtract the remainder i.e., \( 14x - 10 \) from \( 8x^4 + 14x^3 - 2x^2 + 7x - 8 \), it will be exactly divisible by \( 4x^2 + 3x - 2 \). Ans.

 

Question. Obtain all the zeroes of \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 \), if two of its zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \).
Answer:
Since two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \)

\( \implies \) \( \left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3} \) or \( 3x^2 - 5 \) is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial and \( 3x^2 - 5 \).
On dividing \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 \) by \( 3x^2 - 5 \), we get quotient \( x^2 + 2x + 1 \).
\( \therefore 3x^4 + 6x^3 - 2x^2 - 10x - 5 = (3x^2 - 5)(x^2 + 2x + 1) = (3x^2 - 5)(x + 1)^2 \)
Zeroes of \( (x + 1)^2 \) are -1, -1.
Hence, all its zeroes are \( \sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}, -1, -1 \). Ans.

 

Question. If two zeroes of the polynomial \( x^4 - 6x^3 - 26x^2 + 138x - 35 \) are \( 2 \pm \sqrt{3} \), find other zeroes.
Answer:
We have: \( 2 \pm \sqrt{3} \) are two zeroes of the polynomial \( p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35 \).
Let \( x = 2 \pm \sqrt{3} \). So, \( x - 2 = \pm \sqrt{3} \)
Squaring, we get \( x^2 - 4x + 4 = 3 \), i.e., \( x^2 - 4x + 1 = 0 \)
Let us divide \( p(x) \) by \( x^2 - 4x + 1 \) to obtain other zeroes.
On dividing, we get quotient \( x^2 - 2x - 35 \).
\( \therefore p(x) = (x^2 - 4x + 1)(x^2 - 2x - 35) \)
\( = (x^2 - 4x + 1)(x^2 - 7x + 5x - 35) \)
\( = (x^2 - 4x + 1)[x(x - 7) + 5(x - 7)] \)
\( = (x^2 - 4x + 1)(x + 5)(x - 7) \)
So, \( (x + 5) \) and \( (x - 7) \) are other factors of \( p(x) \).
\( \therefore -5 \) and 7 are other zeroes of the given polynomial. Ans.

 

Question. Find all the zeroes of \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \), if you know that two of its zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \).
Answer:
Since two zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \), \( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \) is a factor of given polynomial.
Now, we divide the given polynomial by \( x^2 - 2 \).
On dividing \( 2x^4 - 3x^3 - 3x^2 + 6x - 2 \) by \( x^2 - 2 \), we get quotient \( 2x^2 - 3x + 1 \).
To find other zeroes, factorize the quotient:
\( 2x^2 - 3x + 1 = 2x^2 - 2x - x + 1 = 2x(x - 1) - 1(x - 1) = (2x - 1)(x - 1) \).
Zeroes are \( \frac{1}{2} \) and 1. Ans.

 

Question. Obtain all zeroes of polynomial \( p(x) = x^3 + 13x^2 + 32x + 20 \), if one of its zeroes is -2.
Answer:
As (-2) is its one of zeroes so \( p(x) \) is divisible by \( (x + 2) \) so on dividing \( x^3 + 13x^2 + 32x + 20 \) by \( (x + 2) \), we get quotient \( x^2 + 11x + 10 \).
Now, to find zeroes of \( x^2 + 11x + 10 \) we can write \( x^2 + 11x + 10 = (x + 10)(x + 1) \)

\( \implies \) Zeroes are -1, -10
Hence zeroes of \( x^3 + 13x^2 + 32x + 20 \) are -2, -1 and -10. Ans.

EXERCISE 

 

Question. Apply the division algorithm to find the quotient \( q(x) \) and remainder \( r(x) \) on dividing \( p(x) \) by \( g(x) \).
(i) \( p(x) = x^4 + 4 \); \( g(x) = x^2 + x + 1 \)
(ii) \( p(x) = 6x^4 + 3x^2 + 2 \); \( g(x) = x^2 + 1 \)
(iii) \( p(x) = 8x^4 + 4x^2 + 4x + 4 \); \( g(x) = x^2 - x + 1 \)
(iv) \( p(x) = 10x^4 + 8x^2 + 6x + 3 \); \( g(x) = x^2 - 5x + 6 \)
Answer:
(i) \( q(x) = x^2 - x \), \( r(x) = x + 4 \)
(ii) \( q(x) = 6x^2 - 3 \), \( r(x) = 5 \)
(iii) \( q(x) = 8x^2 + 8x + 4 \), \( r(x) = 0 \)
(iv) \( q(x) = 10x^2 + 50x + 198 \), \( r(x) = 696x - 1185 \).

 

Question. Check whether the polynomial \( g(x) \) is a factor of the poly-nomial \( p(x) \) by division algorithm, in the following cases.
(i) \( p(x) = 2x^4 - 3x^3 + 4x^2 - 7x + 4 \); \( g(x) = x^2 - 2x + 1 \)
(ii) \( p(x) = 6x^4 + x^3 + 10x^2 + 3x + 4 \); \( g(x) = 2x^2 + x + 1 \)
(iii) \( p(x) = 9x^4 + 4x^3 - 2x^2 + x + 10 \); \( g(x) = 3x^2 + 2x + 5 \)
(iv) \( p(x) = 8x^4 + 6x^3 - 3x^2 + x + 6 \); \( g(x) = 4x^2 + 2x + 3 \)
Answer:
(i) Yes, second factor is \( 2x^2 + x + 4 \).
(ii) Yes, second factor is \( 3x^2 - x + 4 \).
(iii) No, \( q(x) = 3x^2 - \frac{2}{3}x - \frac{47}{9} \) and \( r(x) = \frac{133}{9}x + \frac{325}{9} \).
(iv) No, \( q(x) = 2x^2 + \frac{1}{2}x - \frac{5}{2} \) and \( r(x) = \frac{9}{2}x + \frac{27}{2} \).

 

Question. Obtain all zeroes of polynomial \( p(x) = x^4 - 5x^3 - x^2 + 9x - 6 \), if two of its zeroes are \( (-\sqrt{3}) \) and \( (\sqrt{3}) \).
Answer: \( -\sqrt{3}, \sqrt{3}, 1 \) and 2. Ans.

 

Question. If two zeroes of the polynomial \( x^4 + x^3 - 15x^2 - 29x - 6 \) are \( 2 \pm \sqrt{5} \). Find other zeroes.
Answer: Other zeroes are -3 and 2. Ans.

 

Question. Given that \( \sqrt{2} \) is a zero of the cubic polynomial \( 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \), find its other two zeroes.
Answer: \( \frac{-\sqrt{2}}{2}, \frac{-2\sqrt{2}}{3} \). Ans.

 

Question. Find k so that \( x^2 + 2x + k \) is a factor of \( 2x^4 + x^3 - 14x^2 + 5x + 6 \). Also find all the zeroes of the two polynomials.
Answer: \( k = -3 \). Zeroes of \( 2x^4 + x^3 - 14x^2 + 5x + 6 \) are 1, -3, 2, \( -\frac{1}{2} \). Zeroes of \( x^2 + 2x - 3 \) are 1, -3. Ans.

 

Question. Given that \( x - \sqrt{5} \) is a factor of the cubic polynomial \( x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \), find all the zeroes of the polynomial.
Answer: \( \sqrt{5}, \sqrt{5} + \sqrt{2}, \sqrt{5} - \sqrt{2} \). Ans.

 

Question. For which values of a and b, are the zeroes of \( q(x) = x^3 + 2x^2 + a \) also the zeroes of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \)? Which zeroes of \( p(x) \) are not the zeroes of \( q(x) \)?
Answer: \( a = -1, b = -2 \). 1 and 2 are the zeroes of \( q(x) \) which are not the zeroes of \( p(x) \). Ans.

 

Short Answer Type Questions

Question. If \( \alpha \) and \( \beta \) are zeroes of the quadratic polynomial \( x^2 - 6x + a \); find the value of 'a' if \( 3\alpha + 2\beta = 20 \).
Answer: \( a = -16 \)

 

Question. Check whether the polynomial \( g(x) = x^3 - 3x + 1 \) is the factor of polynomial \( p(x) = x^5 - 4x^3 + x^2 + 3x + 1 \).
Answer: No

 

Question. Obtain all zeroes of \( f(x) = x^4 - 3x^3 - x^2 + 9x - 6 \) if two of its zeroes are \( (-\sqrt{3}) \) and \( \sqrt{3} \).
Answer: 1, 2

 

Question. If \( \alpha, \beta, \gamma \) are zeroes of polynomial \( 6x^3 + 3x^2 - 5x + 1 \), then find the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \).
Answer: 5

 

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 - 5x + k \) such that \( \alpha - \beta = 1 \), find the value of \( k \).
Answer: \( k = 6 \)

 

Question. If \( \alpha, \beta \) are the two zeroes of the polynomial \( 25p^2 - 15p + 2 \), find a quadratic polynomial whose zeroes are \( \frac{1}{2\alpha} \) and \( \frac{1}{2\beta} \).
Answer: \( 8x^2 - 30x + 25 \)

 

 

Please click on below link to download CBSE Class 10 Mathematics Probability Worksheet Set A