Rs Aggarwal (2015) for Class 10 Math Chapter 2 - Polynomials

Share with your friends

Rs Aggarwal (2015) Solutions for Class 10 Math Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among Class 10 students for Math Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal (2015) Book of Class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal (2015) Solutions. All Rs Aggarwal (2015) Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 43:

Question 1:

Find the zeros of the quadratic polynomial (x² + 3x − 10) and verify the relation between its zeros and coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = x^{2} + 3 x - 10 \\ = x^{2} + 5 x - 2 x - 10 \\ = x (x + 5) - 2 (x + 5) \\ = (x - 2) (x + 5) \\ ∴ f (x) = 0 = > (x - 2) (x + 5) = 0 \\ = > x - 2 = 0 or x + 5 = 0 \\ = > x = 2 or x = - 5 \\ So, the zeroes of f (x)are 2 and - 5 . \\ Sum of the zeroes = 2+(- 5) = - 3 = \frac{- 3}{1} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes= 2× (- 5) = - 10 = \frac{- 10}{1} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 2:

Find the zeros of the quadratic polynomial (6x² − 7x − 3) and verify the relation between its zeros an coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = 6 x^{2} - 7 x - 3 \\ = 6 x^{2} - 9 x + 2 x - 3 \\ = 3 x (2 x - 3) + 1 (2 x - 3) \\ = (3 x + 1) (2 x - 3) \\ ∴ f (x) = 0 = > (3 x + 1) (2 x - 3) = 0 \\ = > 3 x + 1 = 0 or 2 x - 3 = 0 \\ = > x = \frac{- 1}{3} or x = \frac{3}{2} \\ So, the zeroes of f (x) are \frac{- 1}{3} and \frac{3}{2} . \\ Sum of the zeroes = (- \frac{1}{3}) + \frac{3}{2} = \frac{- 2 + 9}{6} = \frac{7}{6} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeros= (- \frac{1}{3}) \times \frac{3}{2} = - \frac{3}{6} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 3:

Find the zeros of the quadratic polynomial 4x² − 4x − 3 and verify the relation between the zeros and the coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = 4 x^{2} - 4 x - 3 \\ = 4 x^{2} - (6 x - 2 x) - 3 \\ = 4 x^{2} - 6 x + 2 x - 3 \\ = 2 x (2 x - 3) + 1 (2 x - 3) \\ = (2 x + 1) (2 x - 3) \\ ∴ f (x) = 0 = > (2 x + 1) (2 x - 3) = 0 \\ = > 2 x + 1 = 0 or 2 x - 3 = 0 \\ = > x = \frac{- 1}{2} or x = \frac{3}{2} \\ So, the zeros of f (x) are \frac{- 1}{2} and \frac{3}{2} . \\ Sum of the zeros = (\frac{- 1}{2}) + \frac{3}{2} = \frac{- 1 + 3}{2} = \frac{2}{2} = 1 = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeros= (\frac{- 1}{2}) \times \frac{3}{2} = \frac{- 3}{4} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 4:

Find the zeros of the quadratic polynomial 5x² − 4 − 8x and verify the relationship between the zeros and the coefficients of the given polynomial.

Answer:

$\begin{array}{l} We have: \\ f (x) = 5 x^{2} - 4 - 8 x \\ = 5 x^{2} - 8 x - 4 \\ \begin{array}{l} = 5 x^{2} - (10 x - 2 x) - 4 \\ = 5 x^{2} - 10 x + 2 x - 4 \\ = 5 x (x - 2) + 2 (x - 2) \\ = (5 x + 2) (x - 2) \\ ∴ f (x) = 0 = > (5 x + 2) (x - 2) = 0 \\ = > 5 x + 2 = 0 or x - 2 = 0 \\ = > x = \frac{- 2}{5} or x = 2 \\ So, the zeros of f (x) are \frac{- 2}{5} and 2. \end{array} \\ Sum of the zeros = (\frac{- 2}{5}) + 2 = \frac{- 2 + 10}{5} = \frac{8}{5} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeros= (\frac{- 2}{5}) \times 2 = \frac{- 4}{5} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 5:

Find the zeros of the quadratic polynomial 6x² − 3 − 7x and verify the relationship between the zeros and the coefficients of the given polynomial.

Answer:

$\begin{array}{l} We have: \\ f (x) = 6 x^{2} - 3 - 7 x \\ = 6 x^{2} - 7 x - 3 \\ \begin{array}{l} = 6 x^{2} - (9 x - 2 x) - 3 \\ = 6 x^{2} - 9 x + 2 x - 3 \\ = 3 x (2 x - 3) + 1 (2 x - 3) \\ = (3 x + 1) (2 x - 3) \\ ∴ f (x) = 0 = > (3 x + 1) (2 x - 3) = 0 \\ = > 3 x + 1 = 0 or 2 x - 3 = 0 \\ = > x = \frac{- 1}{3} or x = \frac{3}{2} \\ So, the zeroes of f (x) are \frac{- 1}{3} and \frac{3}{2} . \\ Sum of zeroes = (\frac{- 1}{3}) + \frac{3}{2} = \frac{- 2 + 9}{6} = \frac{7}{6} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of zeroes= (\frac{- 1}{3}) \times \frac{3}{2} = \frac{- 3^{1}}{6^{2}} = \frac{- 1}{2} = \frac{constant term}{(coefficient of x^{2})} . \end{array} \end{array}$

Page No 43:

Question 6:

Find the zeros of the quadratic polynomial 2x² − 11x + 15 and verify the relation between the zeros and the coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = 2 x^{2} - 11 x + 15 \\ = 2 x^{2} - (6 x + 5 x) + 15 \\ = 2 x^{2} - 6 x - 5 x + 15 \\ = 2 x (x - 3) - 5 (x - 3) \\ = (2 x - 5) (x - 3) \\ ∴ f (x) = 0 = > (2 x - 5) (x - 3) = 0 \\ = > 2 x - 5 = 0 or x - 3 = 0 \\ = > x = \frac{5}{2} or x = 3 \\ So, the zeroes of f (x) are \frac{5}{2} and 3. \\ Sum of the zeroes = \frac{5}{2} + 3 = \frac{5 + 6}{2} = \frac{11}{2} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes = \frac{5}{2} \times 3 = \frac{15}{2} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 7:

Find the zeros of the quadratic polynomial (x² − 5) and verify the relation between the zeros and the coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = x^{2} - 5 \\ It can be written as x^{2} + 0 x - 5 . \\ = (x^{2} - (\sqrt{5})^{2}) \\ = (x + \sqrt{5}) (x - \sqrt{5}) \\ ∴ f (x) = 0 = > (x + \sqrt{5}) (x - \sqrt{5}) = 0 \\ = > x + \sqrt{5} = 0 or x - \sqrt{5} = 0 \\ = > x = - \sqrt{5} or x = \sqrt{5} \\ So, the zeroes of f (x) are - \sqrt{5} and \sqrt{5} . \\ Here, the coefficient of x is 0 and the coefficient of x^{2} is 1 . \\ Sum of the zeroes = - \sqrt{5} + \sqrt{5} = \frac{0}{1} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes= - \sqrt{5} \times \sqrt{5} = \frac{- 5}{1} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 8:

Find the zeros of the quadratic polynomial (8 x² − 4) and verify the relation between the zeros and the coefficients.

Answer:

$\begin{array}{l} We have: \\ f (x) = 8 x^{2} - 4 \\ It can be written as 8 x^{2} + 0 x - 4 \\ = 4 {(\sqrt{2} x)^{2} - (1)^{2}} \\ = 4 (\sqrt{2} x + 1) (\sqrt{2} x - 1) \\ ∴ f (x) = 0 = > (\sqrt{2} x + 1) (\sqrt{2} x - 1) = 0 \\ = > \sqrt{2} x + 1 = 0 or \sqrt{2} x - 1 = 0 \\ = > x = \frac{- 1}{\sqrt{2}} or x = \frac{1}{\sqrt{2}} \\ So, the zeroes of f (x) are \frac{- 1}{\sqrt{2}} and \frac{1}{\sqrt{2}} \\ Here the coefficient of x is 0 and the coefficient of x^{2} is \sqrt{2} \\ Sum of the zeroes = \frac{- 1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{- 1 + 1}{\sqrt{2}} = \frac{0}{\sqrt{2}} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes= \frac{- 1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{- 1 \times 4}{2 \times 4} = \frac{- 4}{8} = \frac{constant term}{(coefficient of x^{2})} \end{array}$

Page No 43:

Question 9:

Find the zeros of the quadratic polynomial (5u² + 10u) and verify the relation between the zeros and the coefficients.

Answer:

$\begin{array}{l} We have: \\ f (u) = 5 u^{2} + 10 u \\ It can be written as 5 u (u + 2) \\ ∴ f (u) = 0 = > 5 u = 0 o r u + 2 = 0 \\ = > u = 0 or u =-2 \\ So, the zeroes of f (u) are - 2 and 0 . \\ Sum of the zeroes = - 2+ 0= - 2 = \frac{- 2 \times 5}{1 \times 5} = \frac{- 10}{5} = \frac{- (coefficient of u)}{(coefficient of u^{2})} \\ Product of the zeroes= - 2 \times 0 = 0 = \frac{0 \times 5}{1 \times 5} = \frac{0}{5} = \frac{constant term}{(coefficient of u^{2})} \end{array}$

Page No 43:

Question 10:

Find the quadratic polynomial whose zeros are 2 and −6. Verify the relation between the coefficients and the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α =2 and β = - 6 \\ Sum of the zeroes, (α + β) = 2 + (- 6) = - 4 \\ Product of the zeroes, α β = 2 \times (- 6) = - 12 \\ ∴ Required polynomial = x^{2} - (α + β) x + α β = x^{2} - (- 4) x - 12 \\ = x^{2} + 4 x - 12 \\ Sum of the zeroes = - 4 = \frac{- 4}{1} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes = - 12 = \frac{- 12}{1} = \frac{constant term}{coefficient of x^{2}} \end{array}$

Page No 43:

Question 11:

Find the quadratic polynomial whose zeros are $\frac{2}{3}$ and $\frac{- 1}{4}$ . Verify the relation between the coefficients and the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α = \frac{2}{3} and β = \frac{- 1}{4} . \\ Sum of the zeroes = (α + β) = \frac{2}{3} + (\frac{- 1}{4}) = \frac{8 - 3}{12} = \frac{5}{12} \\ Product of the zeroes = α β = \frac{2}{3} \times (\frac{- 1}{4}) = \frac{- 2^{1}}{1 2^{6}} = \frac{- 1}{6} \\ ∴ Required polynomial = x^{2} - (α + β) x + α β = x^{2} - \frac{5}{12} x + (\frac{- 1}{6}) \\ = x^{2} - \frac{5}{12} x - \frac{1}{6} \\ \begin{array}{l} Sum of the zeroes = \frac{5}{12} = \frac{- (coefficient of x)}{(coefficient of x^{2})} \\ Product of the zeroes = \frac{- 1}{6} = \frac{constant term}{coefficient of x^{2}} \end{array} \end{array}$

Page No 44:

Question 12:

Find the quadratic polynomial, sum of whose zeros is 8 and their product is 12. Hence, find the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α and β be the zeroes of the required polynomial f (x) . \\ Then (α + β)=8 and α β =12 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - 8 x + 12 \\ Hence, required polynomial f (x) = x^{2} - 8 x + 12 \\ ∴ f (x) = 0 = > x^{2} - 8 x + 12 = 0 \\ => x^{2} - (6 x + 2 x) + 12 = 0 \\ => x^{2} - 6 x - 2 x + 12 = 0 \\ => x (x - 6) - 2 (x - 6) = 0 \\ => (x - 2) (x - 6) = 0 \\ => (x - 2) = 0 or (x - 6) = 0 \\ => x = 2 or x = 6 \\ So, the zeroes of f (x) are 2 and 6. \end{array}$

Page No 44:

Question 13:

Find the quadratic polynomial, the sum of whose zeros is −5 and their product is 6. Hence, find the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α and β be the zeros of the required polynomial f (x) . \\ Then (α + β)= - 5 and α β =6 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - (- 5) x + 6 \\ = > f (x) = x^{2} + 5 x + 6 \\ Hence, the required polynomial is f (x) = x^{2} + 5 x + 6 . \\ ∴ f (x) = 0 = > x^{2} + 5 x + 6 = 0 \\ => x^{2} + 3 x + 2 x + 6 = 0 \\ => x (x + 3) + 2 (x + 3) = 0 \\ => (x +2) (x + 3) = 0 \\ => (x +2) = 0 or (x + 3) = 0 \\ => x = - 2 or x = - 3 \\ So, the zeros of f (x) are - 2 and - 3. \end{array}$

Page No 44:

Question 14:

Find the quadratic polynomial, the sum of whose zeros is $(\frac{5}{2})$ and their product is 1. Hence, find the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α and β be the zeros of the required polynomial f (x) . \\ Then (α + β)= \frac{5}{2} and α β =1 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - \frac{5}{2} x + 1 \\ \begin{array}{l} = > f (x) = 2 x^{2} - 5 x + 2 \\ Hence, the required polynomial is f (x) = 2 x^{2} - 5 x + 2 . \\ ∴ f (x) = 0 = > 2 x^{2} - 5 x + 2 = 0 \\ => 2 x^{2} - (4 x + x) + 2 = 0 \\ => 2 x^{2} - 4 x - x + 2 = 0 \\ => 2 x (x - 2) - 1 (x - 2) = 0 \\ => (2 x - 1) (x - 2) = 0 \\ => (2 x - 1) = 0 or (x - 2) = 0 \\ => x = \frac{1}{2} or x = 2 \\ So, the zeros of f (x) are \frac{1}{2} and 2 . \end{array} \end{array}$

Page No 44:

Question 15:

Find the quadratic polynomial, the sum of whose zeros is 0 and their product is −1. Hence, find the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α and β be the zeros of the required polynomial f (x) . \\ Then (α + β)=0 and α β = - 1 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - 0 x + (- 1) \\ = > f (x) = x^{2} - 1 \\ Hence, the required polynomial is f (x) = x^{2} - 1 . \\ ∴ f (x) = 0 = > x^{2} - 1 = 0 \\ => (x + 1) (x - 1) = 0 \\ => (x + 1) = 0 or (x - 1) = 0 \\ => x = - 1 or x = 1 \\ So, the zeros of f (x) are - 1 and 1 . \end{array}$

Page No 44:

Question 16:

Find the quadratic polynomial, the sum of whose zeros is $\sqrt{2}$ and their product is −12. Hence, find the zeros of the polynomial.

Answer:

$\begin{array}{l} Let α and β be the zeros of the required polynomial f (x) . \\ Then (α + β)= \sqrt{2} and α β = - 12 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - \sqrt{2} x + (- 12) \\ = > f (x) = x^{2} - \sqrt{2} x - 12 \\ Hence, the required polynomial is f (x) = x^{2} - \sqrt{2} x - 12 . \\ ∴ f (x) = 0 = > x^{2} - \sqrt{2} x - 12 = 0 \\ => x^{2} - 3 \sqrt{2} x + 2 \sqrt{2} x - 12 = 0 \\ => x (x - 3 \sqrt{2}) + 2 \sqrt{2} (x - 3 \sqrt{2}) = 0 \\ => (x - 3 \sqrt{2}) (x + 2 \sqrt{2}) = 0 \\ => x = 3 \sqrt{2} or x = - 2 \sqrt{2} \\ So, the zeros of f (x) are 3 \sqrt{2} and - 2 \sqrt{2} . \end{array}$

Page No 44:

Question 17:

If α, β are the zeros of a polynomial, such that α + β = 6 and αβ = 4, then write the polynomial.

Answer:

$\begin{array}{l} Given: (α + β)=6 and α β =4 \\ ∴ f (x) = x^{2} - (α + β) x + α β \\ = > f (x) = x^{2} - 6 x + 4 \\ Hence, the required polynomial is f (x) = x^{2} - 6 x + 4. \end{array}$

Page No 52:

Question 1:

Verity that 3, −2, 1 are the zeros of the cubic polynomial p(x) = x³ − 2x² − 5x + 6 and verify the relation between its zeros and coefficients.

Answer:

$\begin{array}{l} The given polynomial is p (x) = (x^{3} - 2 x^{2} - 5 x + 6) \\ ∴ p (3) = (3^{3} - 2 \times 3^{2} - 5 \times 3 + 6) = (27 - 18 - 15 + 6) = 0 \\ p (- 2) = [(- 2^{3}) - 2 \times (- 2^{2}) - 5 \times (- 2) + 6] = (- 8 - 8 + 10 + 6) = 0 \\ p (1) = (1^{3} - 2 \times 1^{2} - 5 \times 1 + 6) = (1 - 2 - 5 + 6) = 0 \\ ∴ 3, - 2 and 1 are the zeroes of p (x), \\ Let α =3, β = - 2 and γ =1. Then we have: \\ (α + β + γ) = (3 - 2 + 1) = 2 = \frac{- (coefficient of x^{2})}{(coefficient of x^{3})} \\ (α β + β γ + γ α) = (- 6 - 2 + 3) = \frac{- 5}{1} = \frac{coefficient of x}{coefficient of x^{3}} \\ α β γ = {3 \times (- 2) \times 1} = \frac{- 6}{1} = \frac{- (constant term)}{(coefficient of x^{3})} \end{array}$

Page No 52:

Question 2:

Verify that 5, −2 and $\frac{1}{3}$ are the zeros of the cubic polynomial p(x) = 3x³ − 10x² − 27x + 10 and verify the relation between its zeros and coefficients.

Answer:

$\begin{array}{l} p (x) = (3 x^{3} - 10 x^{2} - 27 x + 10) \\ p (5) = (3 \times 5^{3} - 10 \times 5^{2} - 27 \times 5 + 10) = (375 - 250 - 135 + 10) = 0 \\ p (- 2) = [3 \times (- 2^{3}) - 10 \times (- 2^{2}) - 27 \times (- 2) + 10] = (- 24 - 40 + 54 + 10) = 0 \\ p (\frac{1}{3}) = {3 \times {(\frac{1}{3})}^{3})^{3} - 10 \times {(\frac{1}{3})}^{2} - 27 \times \frac{1}{3} + 10} = (3 \times \frac{1}{27} - 10 \times \frac{1}{9} - 9 + 10) \\ = (\frac{1}{9} - \frac{10}{9} + 1) = (\frac{1 - 10 + 9}{9}) = (\frac{0}{9}) = 0 \\ ∴ 5, - 2 and \frac{1}{3} are the zeroes of p (x) . \\ Let α =5, β = - 2 and γ = \frac{1}{3} . Then we have : \\ (α + β + γ) = (5 - 2 + \frac{1}{3}) = (\frac{10}{3}) = \frac{- (coefficient of x^{2})}{(coefficient of x^{3})} \\ (α β + β γ + γ α) = (- 10 - \frac{2}{3} + \frac{5}{3}) = \frac{- 27}{3} = \frac{coefficient of x}{coefficient of x^{3}} \\ α β γ = {5 \times (- 2) \times \frac{1}{3}} = \frac{- 10}{3} = \frac{- (constant term)}{(coefficient of x^{3})} \end{array}$

Page No 52:

Question 3:

Find a cubic polynomial whose zeros are −2, −3 and −1.

Answer:

$\begin{array}{l} Let α = - 2, β = - 3 and γ = - 1 \\ Now, (α + β + γ) = (- 2 - 3 - 1) = - 6 \\ and (α β + β γ + γ α) = (6 + 3 + 2) = 11 \\ Also, α β γ = {(- 2) \times (- 3) \times (- 1)} = (- 6) \\ ∴ Required polynomial = x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ \\ i .e., x^{3} + 6 x^{2} + 11 x + 6 \end{array}$

Page No 52:

Question 4:

Find a cubic polynomial whose zeros are 3, $\frac{1}{2}$ and −1.

Answer:

$\begin{array}{l} Let α = 3, β = \frac{1}{2} and γ = - 1 \\ Now, (α + β + γ) = {3 + \frac{1}{2} + (- 1)} = \frac{6 + 1 - 2}{2} = \frac{5}{2} \\ and (α β + β γ + γ α) = {(3 \times \frac{1}{2}) + \frac{1}{2} \times (- 1) + (- 1) \times 3} = (\frac{3}{2} - \frac{1}{2} - 3) = \frac{- 4}{2} \\ Also, α β γ = {3 \times \frac{1}{2} \times (- 1)} = (- \frac{3}{2}) \\ ∴ Required polynomial = x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ \\ i .e., x^{3} - (\frac{5}{2}) x^{2} + (- \frac{4}{2}) x - (- \frac{3}{2}) = x^{3} - \frac{5}{2} x^{2} - \frac{4}{2} x + \frac{3}{2} = 2 x^{3} - 5 x^{2} - 4 x + 3 \end{array}$

Page No 52:

Question 5:

When f(x) = 4x³ − 8x² + 8x + 1 is divided by a polynomial g(x), we get (2x − 1) as quotient and (x + 3) as remainder. Find g(x).

Answer:

$\begin{array}{l} Let f (x) = (4 x^{3} - 8 x^{2} + 8 x + 1), q (x) = (2 x - 1) and r (x) = (x + 3) . \\ Then, f (x) = g (x) \times q (x) + r (x) \\ = > g (x) = \frac{f (x) - r (x)}{q (x)} \\ Now, [f (x) - r (x)] = (4 x^{3} - 8 x^{2} + 8 x + 1) - (x + 3) \\ = (4 x^{3} - 8 x^{2} + 7 x - 2) \\ On dividing (4 x^{3} - 8 x^{2} + 7 x - 2) by (2 x - 1), we get g (x) . \end{array}$

$\begin{array}{l} ∴ g (x) = 2 x^{2} - 3 x + 2 \end{array}$

Page No 52:

Question 6:

Divide (2x² + x − 15) by (x + 3) and verify the division algorithm.

Answer:

$\begin{array}{l} First, we will write the given polynomials in their standard forms in \\ decreasing order of degrees and then perform the division as shown below: \end{array}$

$\begin{array}{l} ∴ Quotient = (2 x - 5) and Remainder = 0 \\ => (Quotient \times divisor + remainder) = (2 x - 5) \times (x + 3) + 0 \\ = 2 x^{2} - 5 x + 6 x - 15 \\ = 2 x^{2} + x - 15 \\ = dividend \\ Thus, quotient \times divisor + remainder=dividend \\ Hence, the division algorithm is verified . \end{array}$

Page No 52:

Question 7:

Divide (12 − 17x − 5x²) by (3 − 5x) and verify the division algorithm.

Answer:

$\begin{array}{l} First, we will write the given polynomials in their standard forms in \\ decreasing order of degrees and then perform the division as shown below: \end{array}$

$\begin{array}{l} ∴ Quotient = (x + 4) and remainder=0 \\ ⇒ (Quotient \times divisor) + remainder= (x + 4) \times (- 5 x + 3) + 0 \\ = - 5 x^{2} - 20 x + 3 x + 12 \\ = 5 x^{2} - 17 x + 12 \\ = dividend \\ Thus, (q uotient \times divisor) + remainder = dividend \\ Hence, the division algorithm is verified . \end{array}$

Page No 52:

Question 8:

Divide (3x³ − 4x² + 7x − 2) by (x² − x + 2) and verify the division algorithm.

Answer:

$\begin{array}{l} First, we will write the given polynomials in their standard forms in \\ decreasing order of degrees and then perform the division as shown below: \end{array}$

$\begin{array}{l} ∴ Quotient = (3 x - 1) and remainder = 0 \\ \Rightarrow (Quotient \times divisor) + remainder= (3 x - 1) \times (x^{2} - x + 2) + 0 \\ = 3 x^{3} - x^{2} - 3 x^{2} + x + 6 x - 2 \\ = 3 x^{3} - 4 x^{2} + 7 x - 2 \\ = dividend \\ Thus, (quotient \times divisor) + remainder = dividend \\ Hence, the division algorithm is verified . \end{array}$

Page No 52:

Question 9:

Divide (6 + 19x + x² − 6x³) by (2+ 5x − 3x²) and verify the division algorithm.

Answer:

$\begin{array}{l} First, we will write the given polynomials in their standard forms in decreasing order of \\ degrees and then perform the division as shown below: \end{array}$

$\begin{array}{l} ∴ Quotient = (2 x + 3) and r emainder = 0 \\ \Rightarrow (Quotient \times d ivisor) + remainder = (2 x + 3) \times (2 + 5 x - 3 x^{2}) + 0 \\ = 4 x + 6 + 10 x^{2} + 15 x - 6 x^{3} - 9 x^{2} + 0 \\ = - 6 x^{3} + x^{2} + 19 x + 6 \\ = dividend \\ Thus, (q uotient \times divisor) + remainder = dividend \\ Hence, the division algorithm is verified . \end{array}$

Page No 52:

Question 10:

It being given that 2 is one of the zeros of the polynomial x³ − 4x² + x + 6. Find its other zeros.

Answer:

$\begin{array}{l} Let f (x) = x^{3} - 4 x^{2} + x + 6 \\ Since 2 is a zero of f (x), (x - 2) is a factor of f (x) . \\ On dividing f (x) by (x - 2), we get: \end{array}$

$\begin{array}{l} f (x) = (x^{3} - 4 x^{2} + x + 6) \\ = (x - 2) (x^{2} - 2 x - 3) \\ = (x - 2) (x^{2} - 3 x + x - 3) \\ = (x - 2) {x (x - 3) + 1 (x - 3)} \\ = (x - 2) (x - 3) (x + 1) \\ ∴ f (x) = 0 = > (x - 2) = 0 or (x - 3) = 0 or (x + 1) = 0 \\ = > x = 2 or x = 3 or x = - 1 \\ Thus, the other two zeroes are 3 and - 1 . \end{array}$

Page No 52:

Question 11:

It is given that −1 is one of the zeros of the polynomial x³ + 2x² − 11x − 12. Find all the given zeros of the given polynomial.

Answer:

$\begin{array}{l} Let f (x) = x^{3} + 2 x^{2} - 11 x - 12 \\ Since -1 is a zero of f (x), (x + 1) is a factor of f (x) . \\ On dividing f (x) by (x + 1), we get: \end{array}$

$\begin{array}{l} f (x) = x^{3} + 2 x^{2} - 11 x - 12 \\ = (x + 1) (x^{2} + x - 12) \\ = (x + 1) {x^{2} + 4 x - 3 x - 12} \\ = (x + 1) {x (x + 4) - 3 (x + 4)} \\ = (x + 1) (x - 3) (x + 4) \\ ∴ f (x) = 0 = > (x + 1) (x - 3) (x + 4) = 0 \\ = > (x + 1) = 0 or (x - 3) = 0 or (x + 4) = 0 \\ = > x = - 1 or x = 3 or x = - 4 \\ Thus, all the zeroes are - 1, 3 and - 4 . \end{array}$

Page No 52:

Question 12:

If 1 and −2 are two zeros of the polynomial (x³ − 4x² − 7x + 10), find its third zero.

Answer:

$\begin{array}{l} Let f (x) = x^{3} - 4 x^{2} - 7 x + 10 \\ Since 1 and - 2 are the zeroes of f (x), it follows that each one of (x - 1) and \\ (x + 2) is a factor of f (x) . \\ Consequently, (x - 1) (x + 2) = (x^{2} + x - 2) is a factor of f (x) . \\ On dividing f (x) by (x^{2} + x - 2), we get: \end{array}$

$\begin{array}{l} f (x) = 0 = > (x^{2} + x - 2) (x - 5) = 0 \\ = > (x - 1) (x + 2) (x - 5) = 0 \\ = > x = 1 or x = - 2 or x = 5 \\ Hence, the third zero is 5 . \end{array}$

Page No 52:

Question 13:

If 3 and −3 are two zeros of the polynomial (x⁴ + x³ − 11x² − 9x + 18), find all the zeros of the given polynomial.

Answer:

$\begin{array}{l} Let f (x) = x^{4} + x^{3} - 11 x^{2} - 9 x + 18 \\ Since 3 and - 3 are the zeroes of f (x), it follows that each one of (x + 3) and \\ (x - 3) is a factor of f (x) . \\ Consequently, (x - 3) (x + 3) = (x^{2} - 9) is a factor of f (x) . \\ On dividing f (x) by (x^{2} - 9), we get: \end{array}$

$\begin{array}{l} f (x) = 0 = > (x^{2} + x - 2) (x^{2} - 9) = 0 \\ = > (x^{2} + 2 x - x - 2) (x - 3) (x + 3) \\ \begin{array}{l} = > (x - 1) (x + 2) (x - 3) (x + 3) = 0 \\ = > x = 1 or x = - 2 or x = 3 or x = - 3 \\ Hence, all the zeroes are 1, - 2, 3 and - 3. \end{array} \end{array}$

Page No 52:

Question 14:

If 2 and −2 are two zeros of the polynomial (x⁴ + x³ − 34x² − 4x + 120), find all the zeros of given polynomial.

Answer:

$\begin{array}{l} Let f (x) = x^{4} + x^{3} - 34 x^{2} - 4 x + 120 \\ Since 2 and - 2 are the zeroes of f (x), it follows that each one of (x - 2) and \\ (x + 2) is a factor of f (x) . \\ Consequently, (x - 2) (x + 2) = (x^{2} - 4) is a factor of f (x) . \\ On dividing f (x) by (x^{2} - 4),we get: \end{array}$

$\begin{array}{l} ∴ f (x) = 0 \\ = > (x^{2} + x - 30) (x^{2} - 4) = 0 \\ \begin{array}{l} = > (x^{2} + 6 x - 5 x - 30) (x - 2) (x + 2) \\ \begin{array}{l} = > [x (x + 6) - 5 (x + 6)] (x - 2) (x + 2) \\ \begin{array}{l} = > (x - 5) (x + 6) (x - 2) (x + 2) = 0 \\ = > x = 5 or x = - 6 or x = 2 or x = - 2 \\ Hence, all the zeros are 2, - 2, 5 and - 6 . \end{array} \end{array} \end{array} \end{array}$

Page No 53:

Question 15:

Find all the zeros of (x⁴+ x³ − 23x² − 3x + 60), if it is given that two of its zeros are $\sqrt{3}$ and $- \sqrt{3}$ .

Answer:

$\begin{array}{l} Let f (x) = x^{4} + x^{3} - 23 x^{2} - 3 x + 60 \\ Since \sqrt{3} and - \sqrt{3} are the zeroes of f (x), it follows that each one of (x - \sqrt{3}) and (x + \sqrt{3}) is a factor of f (x) . \\ Consequently, (x - \sqrt{3}) (x + \sqrt{3}) = (x^{2} - 3) is a factor of f (x) . \\ On dividing f (x) by (x^{2} - 3), we get: \end{array}$

$\begin{array}{l} ∴ f (x) = 0 \\ = > (x^{2} + x - 20) (x^{2} - 3) = 0 \\ \begin{array}{l} = > (x^{2} + 5 x - 4 x - 20) (x^{2} - 3) \\ = > [x (x + 5) - 4 (x + 5)] (x^{2} - 3) \\ \begin{array}{l} \begin{array}{l} = > (x - 4) (x + 5) (x - \sqrt{3}) (x + \sqrt{3}) = 0 \\ = > x = 4 or x = - 5 or x = \sqrt{3} or x = - \sqrt{3} \end{array} \\ Hence, all the zeroes are \sqrt{3}, - \sqrt{3}, 4 and - 5. \end{array} \end{array} \end{array}$

Page No 53:

Question 16:

Find all the zeros of (2x⁴ − 3x³ − 5x² + 9x − 3), it being given that two of its zeros are $\sqrt{3}$ and $- \sqrt{3}$ .

Answer:

$\begin{array}{l} The given polynomial is f (x) = 2 x^{4} - 3 x^{3} - 5 x^{2} + 9 x - 3 . \\ Since \sqrt{3} and - \sqrt{3} are the zeroes of f (x), it follows that each one of (x - \sqrt{3}) and (x + \sqrt{3}) is a factor of f (x) . \\ Consequently, (x - \sqrt{3}) (x + \sqrt{3}) = (x^{2} - 3) is a factor of f (x) . \\ On dividing f (x) by (x^{2} - 3), we get: \end{array}$

$\begin{array}{l} ∴ f (x) = 0 \\ = > 2 x^{4} - 3 x^{3} - 5 x^{2} + 9 x - 3 = 0 \\ \begin{array}{l} = > (x^{2} - 3) (2 x^{2} - 3 x + 1) = 0 \\ => (x^{2} - 3) (2 x^{2} - 2 x - x + 1) = 0 \\ \begin{array}{l} = > (x - \sqrt{3}) (x + \sqrt{3}) (2 x - 1) (x - 1) = 0 \\ = > x = \sqrt{3} or x = - \sqrt{3} 3 or x = \frac{1}{2} or x = 1 \end{array} \end{array} \end{array}$

Page No 53:

Question 17:

Find all the zeros of the polynomial (2x⁴ − 11x³ + 7x² + 13x), it being given that two if its zeros are $3 + \sqrt{2}$ and $3 - \sqrt{2}$ .

Answer:

$\begin{array}{l} The given polynomial is f (x) = 2 x^{4} - 11 x^{3} + 7 x^{2} + 13 x - 7 . \\ Since (3 + \sqrt{2}) and (3 - \sqrt{2}) are the zeroes of f (x), it follows that each one of (x + 3 + \sqrt{2}) and (x + 3 - \sqrt{2}) is a factor of f (x) . \\ Consequently, [x - (3 + \sqrt{2})] [x - (3 - \sqrt{2})] = [(x - 3) - \sqrt{2}] [(x - 3) + \sqrt{2}] \\ = [(x - 3)^{2} - 2] = x^{2} - 6 x + 7, which is a factor of f (x) \end{array} On dividing f (x) by (x^{2} - 6 x + 7), we get:$

$\begin{array}{l} ∴ f (x) = 0 \\ \Rightarrow 2 x^{4} - 11 x^{3} + 7 x^{2} + 13 x - 7 = 0 \\ \begin{array}{l} = > (x^{2} - 6 x + 7) (2 x^{2} + x - 1) = 0 \\ = > (x + 3 + \sqrt{2}) (x + 3 - \sqrt{2}) (2 x - 1) (x + 1) = 0 \\ = > x = - 3 - \sqrt{2} or x = - 3 + \sqrt{2} or x = \frac{1}{2} or x = - 1 \\ Hence, all the zeros are (- 3 - \sqrt{2}), (- 3 + \sqrt{2}), \frac{1}{2} and - 1. \end{array} \end{array}$

Page No 53:

Question 18:

Obtain all other zeros of (x⁴ + 4x³ − 2x² − 20x − 15) if two of its zeros are $\sqrt{5}$ and $- \sqrt{5}$ .

Answer:

$\begin{array}{l} The given polynomial is f (x) = x^{4} + 4 x^{3} - 2 x^{2} - 20 x - 15 . \\ Since (x - \sqrt{5}) and (x + \sqrt{5}) are the zeroes of f (x), it follows that each one of (x - \sqrt{5}) and (x + \sqrt{5}) is a factor of f (x) . \\ Consequently, (x - \sqrt{5}) (x + \sqrt{5}) = (x^{2} - 5) is a factor of f (x) . \\ On dividing f (x) by (x^{2} - 5), we get: \end{array}$

$\begin{array}{l} ∴ f (x) = 0 \\ = > x^{4} + 4 x^{3} - 7 x^{2} - 20 x - 15 = 0 \\ \begin{array}{l} = > (x^{2} - 5) (x^{2} + 4 x + 3) = 0 \\ = > (x - \sqrt{5}) (x + \sqrt{5}) (x + 1) (x + 3) = 0 \\ = > x = \sqrt{5} or x = - \sqrt{5} or x = - 1 or x = - 3 \\ Hence, all the zeros are \sqrt{5}, - \sqrt{5}, - 1 and - 3. \end{array} \end{array}$

Page No 56:

Question 1:

Which of the following is a polynomial?

(a) $x^{2} - 5 x + 6 \sqrt{x} + 2$
(b) $x^{3 / 2} - x + x^{1 / 2} + 1$
(c) $\sqrt{x} + \frac{1}{\sqrt{x}}$
(d) None of these

Answer:

(d) none of these

A polynomial in x of degree n is an expression of the form p(x) =a_o+a₁x+a₂x²+...+a_n x_ⁿ, where a_n $\neq$ 0.

Page No 57:

Question 2:

Which of the following is not a polynomial?

(a) $\sqrt{3} x^{2} - 2 \sqrt{3} x + 5$
(b) $9 x^{2} - 4 x + \sqrt{2}$
(c) $\frac{3}{2} x^{3} + 6 x^{2} - \frac{1}{\sqrt{2}} x - 8$
(d) $x + \frac{3}{x}$

Answer:

$\begin{array}{l} (d) x + \frac{3}{x} is not a polynomial . \end{array}$
It is because in the second term, the degree of x is −1 and an expression with a negative degree is not a polynomial.

Page No 57:

Question 3:

The zeros of the polynomial x² − 2x − 3 are

(a) −3, 1
(b) −3, −1
(c) 3, −1
(d) 3, 1

Answer:

$\begin{array}{l} (c) 3, - 1 \\ \begin{array}{l} Let f (x) = x^{2} - 2 x - 3 = 0 \\ = x^{2} - 3 x + x - 3 = 0 \end{array} = x (x - 3) + 1 (x - 3) = 0 = (x - 3) (x + 1) = 0 = x = 3 or x = - 1 \end{array}$

Page No 57:

Question 4:

The zeros of the polynomial $x^{2} - \sqrt{2} x - 12$ are
(a) $\sqrt{2}, - \sqrt{2}$
(b) $3 \sqrt{2}, - 2 \sqrt{2}$
(c) $- 3 \sqrt{2}, 2 \sqrt{2}$
(d) $3 \sqrt{2}, 2 \sqrt{2}$

Answer:

$(b) 3 \sqrt{2}, - 2 \sqrt{2} Let f(x)= x^{2} - \sqrt{2} x - 12 = 0 = > x^{2} - 3 \sqrt{2} x + 2 \sqrt{2} x - 12 = 0 = > x (x - 3 \sqrt{2}) + 2 \sqrt{2} (x - 3 \sqrt{2}) = 0 = > (x - 3 \sqrt{2}) (x + 2 \sqrt{2}) = 0 = > x = 3 \sqrt{2} or x = - 2 \sqrt{2}$

Page No 57:

Question 5:

The zeros of the polynomial $4 x^{2} + 5 \sqrt{2} x - 3$ are
(a) $- 3 \sqrt{2}, \sqrt{2}$
(b) $- 3 \sqrt{2}, \frac{\sqrt{2}}{2}$
(c) $\frac{- 3 \sqrt{2}}{2}, \frac{\sqrt{2}}{4}$
(d) none of these

Answer:

$(c) - \frac{3}{\sqrt{2}}, \frac{\sqrt{2}}{4} Let f (x) = 4 x^{2} + 5 \sqrt{2} x - 3 = 0 = > 4 x^{2} + 6 \sqrt{2} x - \sqrt{2} x - 3 = 0 = > 2 \sqrt{2} x (\sqrt{2} x + 3) - 1 (\sqrt{2} x + 3) = 0 = > (\sqrt{2} x + 3) (2 \sqrt{2} x - 1) = 0 = > x = - \frac{3}{\sqrt{2}} or x = \frac{1}{2 \sqrt{2}} = > x = - \frac{3}{\sqrt{2}} or x = \frac{1}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$

Page No 57:

Question 6:

The zeros of the polynomial $x^{2} + \frac{1}{6} x - 2$ are
(a) −3, 4
(b) $\frac{- 3}{2}, \frac{4}{3}$
(c) $\frac{- 4}{3}, \frac{3}{2}$
(d) none of these

Answer:

$(b) \frac{- 3}{2}, \frac{4}{3} Let f (x) = x^{2} + \frac{1}{6} x - 2 = 0 = > 6 x^{2} + x - 12 = 0 = > 6 x^{2} + 9 x - 8 x - 12 = 0 = > 3 x (2 x + 3) - 4 (2 x + 3) = 0 = > (2 x + 3) (3 x - 4) = 0 ∴ x = \frac{- 3}{2} or x = \frac{4}{3}$

Page No 57:

Question 7:

The zeros of the polynomial $7 x^{2} - \frac{11}{3} x - \frac{2}{3}$ are
(a) $\frac{2}{3}, \frac{- 1}{7}$
(b) $\frac{2}{7}, \frac{- 1}{3}$
(c) $\frac{- 2}{3}, \frac{1}{7}$
(d) none of these

Answer:

$(a) \frac{2}{3}, \frac{- 1}{7} Let f (x) = 7 x^{2} - \frac{11}{3} x - \frac{2}{3} = 0 \Rightarrow 21 x^{2} - 11 x - 2 = 0 \Rightarrow 21 x^{2} - 14 x + 3 x - 2 = 0 \Rightarrow 7 x (3 x - 2) + 1 (3 x - 2) = 0 \Rightarrow (3 x - 2) (7 x + 1) = 0 \Rightarrow x = \frac{2}{3} or x = \frac{- 1}{7}$

Page No 57:

Question 8:

A quadratic polynomial whose zeros are 5 and −3, is

(a) x² + 2x − 15
(b) x² − 2x + 15
(c) x² − 2x − 15
(d) none of these

Answer:

$(c) x^{2} - 2 x - 15 Here, the zeroes are 5 and - 3 . Let α = 5 and β So, sum of the zeroes, α+β = 5+ (- 3) = 2 A lso, product of the zeroes, αβ = 5 \times (- 3) = - 15 The polynomial will be x^{2} - (α + β) x + α β . ∴ The r equired polynomial is x^{2} - 2 x - 15 .$

Page No 57:

Question 9:

A quadratic polynomial whose zeros are $\frac{3}{5}$ and $\frac{- 1}{2}$ ,is
(a) 10x² + x + 3
(b) 10x² + x − 3
(c) 10x² − x + 3
(d) $x^{2} - \frac{1}{10} x - \frac{3}{10}$

Answer:

$(d) x^{2} - \frac{1}{10} x - \frac{3}{10} Here, the zeroes are \frac{3}{5} and \frac{- 1}{2} . Let α = \frac{3}{5} a n d β = \frac{- 1}{2} So, sum of the zeroes, α+β= \frac{3}{5} + (\frac{- 1}{2}) = \frac{1}{10} \begin{array}{l} Also, product of the zeroes, αβ= \frac{3}{5} \times (\frac{- 1}{2}) = \frac{- 3}{10} \\ The polynomial will be x^{2} - (α + β) x + α β . \end{array} ∴ The r equired polynomial is x^{2} - \frac{1}{10} x - \frac{3}{10} .$

Page No 57:

Question 10:

The sum and product of the zeros of a quadratic polynomial are 3 and −10 respectively. The quadratic polynomial is

(a) x² − 3x + 10
(b) x² + 3x −10
(c) x² − 3x −10
(d) x² + 3x + 10

Answer:

$(c) x^{2} - 3 x - 10 Given: Sum of zeroes, α+β = 3 Also, product of zeroes, αβ=-10 ∴ Required polynomial = x^{2} - (α+β) + αβ = x^{2} - 3 x - 10$

Page No 57:

Question 11:

How many polynomials are there having 4 and −2 as zeros?

(a) One
(b) Two
(c) Three
(d) More than three

Answer:

$(d) More than three An i nfinite number of polynomials can be formed having zeroes 4 and -2, which will be in the form n (x - 4) (x + 2) or n (x^{2} - 2 x - 8), for any arbitrary real number n .$

Page No 57:

Question 12:

The zeros of the quadratic polynomial x² + 88x + 125 are

(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal

Answer:

$(b) both negative Let α and β be the zeroes of x^{2} + 88 x + 125 . Then α + β = - 88 and α \times β =125 This can only happen when both the zeroes are negative .$

Page No 58:

Question 13:

If α and β are the zero of x² + 5x + 8, then the value of (α + β) is

(a) 5
(b) −5
(c) 8
(d) −8

Answer:

$(b) - 5 Given: α and β are the zeroes of x^{2} + 5 x + 8 . If α + β i s t h e s u m o f t h e r o o t s a n d α β is t h e p r o d u c t, t h e n t h e r e q u i r e d p o l y n i m i a l w i l l b e x^{2} - (α + β) + α β . ∴ α + β = - 5$

Page No 58:

Question 14:

If α and β are the zero of 2x² + 5x − 8, then the value of (αβ) is

(a) $\frac{- 5}{2}$
(b) $\frac{5}{2}$
(c) $\frac{- 9}{2}$
(d) $\frac{9}{2}$

Answer:

$(c) \frac{- 9}{2} Given: α and β are the zeroes of 2 x^{2} + 5 x - 9 . If α and β are the zeroes, then x^{2} - (α + β) x + α β is the required polynomial . The polynomial will be x^{2} - \frac{5}{2} x - \frac{9}{2} . ∴ α β = \frac{- 9}{2}$

Page No 58:

Question 15:

If one zero of the quadratic polynomial kx² + 3x + k is 2, then the value of k is

(a) $\frac{5}{6}$
(b) $\frac{- 5}{6}$
(c) $\frac{6}{5}$
(d) $\frac{- 6}{5}$

Answer:

$(d) \frac{- 6}{5} \begin{array}{l} Since 2 is a zero of k x^{2} + 3 x + k, we have: \end{array} k \times {(2)}^{2} + 3 \times 2 + k = 0 = > 4 k + k + 6 = 0 = > 5 k = - 6 = > k = \frac{- 6}{5}$

Page No 58:

Question 16:

If one zero of the quadratic polynomial (k − 1) x² + kx + 1 is −4, then the value of k is

(a) $\frac{- 5}{4}$
(b) $\frac{5}{4}$
(c) $\frac{- 4}{3}$
(d) $\frac{4}{3}$

Answer:

$(b) \frac{5}{4} \begin{array}{l} Since - 4 is a zero of (k - 1) x^{2} + k x + 1, we have: \\ (k - 1) \times {(- 4)}^{2} + k \times (- 4) + 1 = 0 \end{array} = > 16 k - 16 - 4 k + 1 = 0 = > 12 k - 15 = 0 = > k = \frac{1 5^{5}}{1 2^{4}} = > k = \frac{5}{4}$

Page No 58:

Question 17:

If −2 and 3 are the zeros of the quadratic polynomial x² + (a + 1) x + b, then

(a) a = −2, b = 6
(b) a = 2, b = −6
(c) a = −2, b = −6
(d) a = 2, b = 6

Answer:

$(c) a=-2, b= -6 Given: - 2 and 3 are the zeroes of x^{2} + (a + 1) x + b . \begin{array}{l} {Now, (- 2)}^{2} + (a + 1) \times (- 2) + b = 0 = > 4 - 2 a - 2 + b = 0 \\ => b - 2 a = - 2 ... (1) \end{array} Also, 3^{2} + (a + 1) \times 3 + b = 0 = > 9 + 3 a + 3 + b = 0 => b + 3 a = - 12 ... (2) On subtracting (1) from (2), we get a = - 2 ∴ b = - 2 - 4 = - 6 [From (1)]$

Page No 58:

Question 18:

If one of the zeroes of the quadratic polynomial x² + bx + c is negative of the other, then

(a) b = 0 and c is positive
(b) b = 0 and c is negative
(c) b ≠ 0 and c is positive
(d) b ≠ 0 and c is negative

Answer:

$(b) b =0 and c is negative Let α and - α be the zeroes of x^{2} + b x + c . Then α + (- α) = - b = > - b = 0 = > b = 0 Also, α \times (- α) = c = > - α^{2} = c = > c is negative, as α^{2} > 0 ∴ b = 0 and c is negative .$

Page No 58:

Question 19:

If the zeros of the quadratic polynomial ax² + bx + c, where a ≠ 0 and c ≠ 0, are equal then

(a) c and a have the same sign
(b) c and a have opposite signs
(c) c and b have the same sign
(d) c and b have opposite sign

Answer:

$(a) c and a have the same sign Let α and α be the zeroes of a x^{2} + b x + c . Then α + α = \frac{- b}{a} > 2 α = \frac{- b}{a} = > α = \frac{- b}{2 a} Also, α \times α = \frac{c}{a} = > α^{2} = \frac{c}{a} > 0 ∴ c and a must have the same sign .$

Page No 58:

Question 20:

The zeros of the quadratic polynomial x2 + kx + k, where k > 0

(a) are both positive
(b) are both negative
(c) are always equal
(d) are always unequal

Answer:

$(b) are both negative Let α and β be the zeroes of x^{2} + k x + k . Then α + β = - k and α \times β = k This is possible only when α and β are both negative .$

Page No 58:

Question 21:

If one zero of 3x² + 8x + k be the reciprocal of the other, then k = ?

(a) 3
(b) −3
(c) $\frac{1}{3}$
(d) $\frac{- 1}{3}$

Answer:

$(a) k = 3 Let α and \frac{1}{α} be the zeroes of 3 x^{2} - 8 x + k . Then product of zeroes= \frac{k}{3} = > α \times \frac{1}{α} = \frac{k}{3} = > 1 = \frac{k}{3} = > k = 3$

Page No 58:

Question 22:

If the sum of the zeros of the quadratic polynomial kx² + 2x + 3k is equal to the product of its zeros, then k = ?

(a) $\frac{1}{3}$
(b) $\frac{- 1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{- 2}{3}$

Answer:

$(d) \frac{- 2}{3} Let α and β be the zeroes of k x^{2} + 2 x + 3 k . Then α + β = \frac{- 2}{k} and αβ = \frac{3 k}{k} = 3 = > α + β = αβ = > \frac{- 2}{k} = 3 = > k = \frac{- 2}{3}$

Page No 59:

Question 23:

If α, β are the zeros of f (x) = 2x² + 6x − 6, then

(a) α + β = αβ
(b) α + β > αβ
(c) α + β < αβ
(d) α + β + αβ = 0

Answer:

$(a) α + β = αβ Given: α and β are zeroes of 2 x^{2} + 6 x - 6 . Then α + β = \frac{- 6}{2} = - 3 and αβ = \frac{- 6}{2} = - 3 ∴ α + β = αβ$

Page No 59:

Question 24:

If α, β are the zeros of the polynomial x² − 5x + c and α − β = 1, then c = ?

(a) 0
(b) 1
(c) 4
(d) 6

Answer:

$(d) 6 Given : α and β are the zeroes of x^{2} - 5 x + c . Also, α - β =1 ∴ α + β = 5 Solving α + β = 5 and α - β =1, we get α =3 and β =2. Since 3 is a zero of the polynomial x^{2} - 5 x + c, we have : 3^{2} - 5 \times 3 + c = 0 = > c = 6$

Page No 59:

Question 25:

If α, β are the zeros of the polynomial x² + 6x + 2, then $(\frac{1}{α} + \frac{1}{β}) = ?$
(a) 3
(b) −3
(c) 12
(d) −12

Answer:

$(b) -3 Since α and β are the zeroes of x^{2} + 6 x + 2,we have: α + β = - 6 and αβ = 2 ∴ (\frac{1}{α} + \frac{1}{β}) = (\frac{α + β}{α β}) = \frac{- 6}{2} = - 3$

Page No 59:

Question 26:

If α, β, γ are the zeros of the polynomial x³ − 6x² − x + 30, then (αβ + βγ + γα) = ?

(a) −1
(b) 1
(c) −5
(d) 30

Answer:

$(a) -1 It is given that α, β and γ are the zeroes of x^{3} - 6 x^{2} - x + 30 . ∴ (αβ + βγ + γα) = \frac{co-efficient of x}{co-efficient of x^{3}} = \frac{- 1}{1} = - 1$

Page No 59:

Question 27:

If α, β, γ are the zeros of the polynomial 2x³ + x² − 13x + 6, then αβγ = ?

(a) −3
(b) 3
(c) $\frac{- 1}{2}$
(d) $\frac{- 13}{2}$

Answer:

$(a) - 3 Since α, β and γ are the zeroes of 2 x^{3} + x^{2} - 13 x + 6, we have: αβγ = \frac{- (constant term)}{co-efficient of x^{3}} = \frac{- 6}{2} = - 3$

Page No 59:

Question 28:

If α, β, γ be the zeros of the polynomial p(x) such that (α + β + γ) = 3, (αβ + βγ + γα) = −
10 and αβγ = −24, then p(x) = ?

(a) x³ + 3x² − 10x + 24
(b) x³ + 3x² + 10x −24
(c) x³ − 3x² −10x + 24
(d) None of these

Answer:

$(c) x^{3} - 3 x^{2} - 10 x + 24 Given: α, β and γ are the zeroes of polynomial p (x) . Also, (α + β + γ) = 3, (αβ + βγ + γα) = - 10 and α β γ = - 24 ∴ p (x) = x^{3} - (α + β + γ) x^{2} + (αβ + βγ + γα) x - αβγ = x^{3} - 3 x^{2} - 10 x + 24$

Page No 59:

Question 29:

If two of the zeros of the cubic polynomial ax³ + bx² + cx + d is 0, then the third zeros is

(a) $\frac{- b}{a}$
(b) $\frac{b}{a}$
(c) $\frac{c}{a}$
(d) $\frac{- d}{a}$

Answer:

$(a) \frac{- b}{a} Let α, 0 and 0 be the zeroes of a x^{3} + b x^{2} + c x + d = 0. Then sum of the zeroes= \frac{- b}{a} = > α + 0 + 0 = \frac{- b}{a} = > α = \frac{- b}{a} Hence, the third zero is \frac{- b}{a} .$

Page No 59:

Question 30:

If one of the zeros of the cubic polynomial ax³ + bx2 + cx + d is 0, then the product of the other two zeros is

(a) $\frac{- c}{a}$
(b) $\frac{c}{a}$
(c) 0
(d) $\frac{- b}{a}$

Answer:

$(b) \frac{c}{a} Let α, β and 0 be the zeroes of a x^{3} + b x^{2} + c x + d . Then, sum of the products of zeroes taking two at at a time is given by (αβ + β \times 0 + α \times 0) = \frac{c}{a} = > αβ = \frac{c}{a} ∴ T he product of the other two zeroes is \frac{c}{a} .$

Page No 59:

Question 31:

If one of the zeros of the cubic polynomial x³+ ax² + bx + c is −1, then the product of the other two zeros is

(a) a − b − 1
(b) b − a − 1
(c) 1 − a + b
(d) 1 + a − b

Answer:

$(c) 1 - a + b Since - 1 is a zero of x^{3} + a x^{2} + b x + c, we have: {(- 1)}^{3} + a \times {(- 1)}^{2} + b \times (- 1) + c = 0 = > a - b + c - 1 = 0 = > c = 1 - a + b Also, product of all zeroes is given by α β \times (- 1) = - c = > α β = c = > α β = 1 - a + b$

Page No 59:

Question 32:

If the zeros of the polynomial x³ − 3x² + x + 1 are a −d, a and a + d, then a + d is

(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number

Answer:

$(d) an irrational number Since a - d, a and a + d are the zeroes of x^{3} - 3 x^{2} + x + 1, we have: Sum of zeroes=3 => a - d + a + a + d = 3 = > 3 a = 3 = > a = 1 So, the zeroes are (1 - d), 1 and (1 + d) . Now, product of the zeroes = - 1 = > (1 - d) \times 1 \times (1 + d) = - 1 = > 1 - d^{2} = - 1 = > d^{2} = 2 = > d = \sqrt{2} ∴ a + d = (1 \pm \sqrt{2}), which is irrational .$

Page No 59:

Question 33:

If α, β be the zeros of the polynomial x² − 8x + k such that α² + β² = 40, then k = ?

(a) 6
(b) 9
(c) 12
(d) −12

Answer:

$(c) 12 Since α and β are the zeroes of x^{2} - 8 x + k, we have: α + β = 8 and α β = k Also, it is given that α^{2} + β^{2} = 40 . Now, {(α + β)}^{2} = α^{2} + β^{2} + 2 αβ = > 8^{2} = 40 + 2 k = > 2 k = 24 = > k = 12$

Page No 60:

Question 34:

If α, β be the zero of the polynomial 2x² + 5x + k such that α² + β² + αβ = $\frac{21}{4}$ , then k = ?
(a) 3
(b) −3
(c) −2
(d) 2

Answer:

$(d) 2 {Since α and β are the zeroes of 2x}^{2} + 5 x + k, we have : α + β = \frac{- 5}{2} and αβ = \frac{k}{2} Also, it is given that α^{2} + β^{2} + αβ = \frac{21}{4} . = > {(α + β)}^{2} - αβ = \frac{21}{4} = > {(\frac{- 5}{2})}^{2} - \frac{k}{2} = \frac{21}{4} = > \frac{25}{4} - \frac{k}{2} = \frac{21}{4} = > \frac{k}{2} = \frac{25}{4} - \frac{21}{4} = \frac{4}{4} = 1 = > k = 2$

Page No 60:

Question 35:

On dividing a polynomial p(x) by a non-zero polynomial q(x), let g(x) be the quotient and r(x) be the remainder, than p(x) = q(x)⋅g(x) + r(x), where

(a) r(x) = 0 always
(b) deg r (x) <deg g(x) always
(c) either r(x) = 0 or deg r(x) <deg g(x)
(d) r(x) = g(x)

Answer:

$(c) either r (x) = 0 o r \deg r (x) < \deg g (x) By division algorithm on polynomials, either r (x) = 0 or deg r (x) < deg g (x) .$

Page No 60:

Question 36:

Which of the following is a true statement?

(a) x² + 5x − 3 is a linear polynomial.
(b) x²+ 4x − 1 is a binomial.
(c) x + 1 is a monomial.
(d) 5x³ is a monomial.

Answer:

$(d) 5 x^{2} is a monomial . 5 x^{2} consists of one term only . So, it is a monomial .$

Page No 60:

Question 37:

If α, β are the zeros of the polynomial ax² + bx + c, then (α² + β²) = ?

(a) $\frac{a^{2} - 2 b c}{b^{2}}$
(b) $\frac{b^{2} - 2 a c}{a^{2}}$
(c) $\frac{a^{2} + 2 b c}{b^{2}}$
(d) $\frac{b^{2} + 2 a c}{a^{2}}$

Answer:

$(b) \frac{b^{2} - 2 a c}{a^{2}} Here, α + β = \frac{- b}{a} and αβ = \frac{c}{a} ∴ (α^{2} + β^{2}) = {(α + β)}^{2} - 2 αβ = \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{b^{2} - 2 a c}{a^{2}}$

Page No 60:

Question 38:

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

(a)
(b)
(c)
(d)

Answer:

(c)

$The graph cuts the x - a x i s at three distinct points . So, it represents a cubic polynomial .$

Page No 60:

Question 39:

Read the statements given below:
I. If α, β are the zeros of the polynomial x² − p(x + 1) −c, then (a + 1)(β + 1) = 1 − c.

II. If α, β are the zeros of the polynomial x² + px + q, then the polynomial having $\frac{1}{α}, \frac{1}{β}$ as zeros is qx² + px + 1.
III. When x³+ 3x² − 5x + 4 is divided by (x + 1), then the remainder is 9.

Which of the above statements is false?

(a) I only
(b) II only
(c) III only
(d) I and III both

Answer:

$(c) III only The g iven polynomial is x^{2} - p x - (p - c) . ∴ α + β = p and αβ = - (p - c) = > (α + 1) (β + 1) = (α + β) + αβ + 1 = p - (p - c) + 1 = 1 + c ∴ I is true . \begin{array}{l} (II) α and β are the zeroes of x^{2} + p x + q . \end{array} α + β = - p and αβ = q \frac{1}{α} + \frac{1}{β} = \frac{α + β}{αβ} = \frac{- p}{q} and \frac{1}{α} \times \frac{1}{β} = \frac{1}{q} ∴ R equired polynomial= x^{2} - \frac{p}{q} x + \frac{1}{q}, i .e ., q x^{2} - p x + 1 ∴ II is true . (III) Let p (x) = x^{3} + 3 x^{2} - 5 x + 4 be divided by (x + 1) . Then remainder= p (- 1) = {(- 1)}^{3} + 3 \times {(- 1)}^{2} - 5 \times (- 1) + 4 = - 1 + 3 + 5 + 4 = 11 ∴ III is false .$

Page No 61:

Question 40:

Read the statements given below:

I. If the polynomial p(x) = 2x³ − kx² + 5x + 2 is exactly divisible by (x + 2), then k = −6.

II. If the polynomial q(x) = x³ − 7x + k when divided by (x − 1) leaves the remainder 2, then k = 6.

III. If two zeros of the polynomial f(x) = x³ − 5x² − 16x + 80 are equal in magnitude and opposite in sigh, then the third zero is 5.

Which of the above statements is not true?

(a) I only
(b) II only
(c) III only
(d) I as well as II

Answer:

$(b) II only (I) p (- 2) = 0 = > 2 \times {(- 2)}^{3} - k {(- 2)}^{2} + 5 \times (- 2) + 2 = 0 = > - 16 - 4 k - 10 + 2 = 0 = > 4 k = - 24 = > k = - 6 ∴ I is true . (II) p (1) = 2 = > 1^{3} - 7 \times 1 - k = 2 = > k = - 8 ∴ II is false . (III) Let the zeroes of f (x) be α, - α and β . Then sum of zeroes= \frac{- (co-efficient of x^{2})}{(co-efficient of x^{3})} = 5 ∴ α + (- α) + β = 5 = > β = 5 \begin{array}{l} So the third zero is 5 . \\ ∴ III is true . \end{array}$

Page No 61:

Question 41:

Assertion (A)
If one zero of the polynomial p(x) = (k² + 4) x² + 9x + 4k is the reciprocal of the other zero, then k = 2.

Reason (R)
If (x − α) is a factor of the polynomial p(x), then α is a zero of p(x).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

$(b) The Reason (R) is the factor theorem . Let α and \frac{1}{α} be the zeroes of p (x) . Then product of the zeroes= α \times \frac{1}{α} = \frac{constant term}{co-efficient of x^{2}} = \frac{4 k}{k^{2} + 4} = > \frac{4 k}{k^{2} + 4} = 1 = > k^{2} - 4 k + 4 = 0 = > {(k - 2)}^{2} = 0 = > k = 2 ∴ Assertion (A) is true, but reason (R) is not its correct explanation.$

Page No 61:

Question 42:

Assertion (A)
The polynomial p(x) = x³ + x has one real zero.

Reason (R)
A polynomial of nth degree has at most n zeros.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

$(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A) . p (x) = 0 = > (x^{3} + x) = 0 = > x (x^{2} + 1) = 0 = > x = 0 or x^{2} + 1 = 0 But x^{2} + 1 \neq 0, for any real value of x . [∵ x^{2} + 1 > 0] ∴ p (x) has one real zero, i .e ., 0 . ∴ Asseration (A) is true and reason (R) does not clearly explain Assertion (A) .$

Page No 62:

Question 43:

Assertion (A)
If on dividing the polynomial p(x) = x² − 3ax + 3a − 7 by (x + 1), we get 6 as remainder, then a= 3.

Reason (R)
When a polynomial p(x) is divided by (x − α), then the remainder is p(α).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

$(d) Assertion (A) is false . By Remainder Theorem, when p (x) is divided by (x + 1), then the remainder is p (- 1) . So, p (- 1) = 6 = > {(- 1)}^{2} - 3 a \times (- 1) + 3 a - 7 = 6 = > 6 a = 12 = > a = 2 ∴ Assertion (A) is false .$

Page No 62:

Question 44:

Assertion (A)
A monic quadratic polynomial having 4 and −2 as zeroes is x² − 2x − 8.

Reason (R)
The monic quadratic polynomial having α and β as zeroes is given by p(x) = x² − (α + β) x + αβ.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

$(a) Both the assertion (A) and reason (R) are true and reason (R) is a correct explanation of (A) . Let α =4 and β = - 2. Then α + β =2 and α β = - 8 ∴ Required polynomial = x^{2} - (α + β) x + αβ = x^{2} - 2 x - 8 i . e ., a ssertion (A) is true .$

Page No 62:

Question 45:

If the zeros of a quadratic polynomial ax² + bx + c are both negative, then a, b, c will have the same sign.

(a) True
(b) False

Answer:

$(a) True Let α and β be the negative zeroes of a x^{2} + b x + c . Then α + β = \frac{- b}{a} < 0 and αβ = \frac{c}{a} = > \frac{b}{a} > 0 and \frac{c}{a} > 0 = > a, b, c have the same sign . ∴ The g iven statement is true .$

Page No 62:

Question 46:

Match the following columns:

Column I	Column II
(a) If α and β be the zeros of the polynomial x² − 5x + k such that (α − β) = 1, then k = ......... .	(p) 10
(b) If one zero of 4x² + 17x + p is the reciprocal of the other, then p = ......... .	(q) −3
(c) If the zeros of x³ − 6x² + 3x + m are (a − d), a and (a + d), then m = ......... .	(r) 4
(d) If the zeros of x³ + 9x² + 23x + 15 are (a − d), a and (a + d), then a = ......... .	(s) 6

Answer:

$(a) α - β = 1 and α + β =5 = > 2 α = 6 = > α = 3 As α is a zero of the given polynomial, it will satisfy the given polynomial . ∴ 3^{2} - 5 \times 3 + k = 0 = > k = 6 (b) If one zero is α, then the other zero is \frac{1}{α} . Product of the roots = \frac{p}{4} \Rightarrow α \times \frac{1}{α} = \frac{p}{4} = > p = 4 (c) (a - d) + a + (a + d) = 6 = > 3 a = 6 = > a = 2 As a is a zero of the given polynomial, it will satisfy the given equation . ∴ 2^{3} - 6 \times 2^{2} + 3 \times 2 + m = 0 = > 8 - 24 + 6 + m = 0 = > m = 10 (d) (a - d) + a + (a + d) = - 9 = > 3 a = - 9 = > a = - 3 ∴ T he correct matches are (a) - (s), (b) - (r), (c) - (p) and (d) - (q) .$

Page No 63:

Question 47:

Match the following columns:

Column I	Column II
(a) The polynomial whose zeros are 2 and −3 is ......... .	(p) x²− 4x + 1
(b) The polynomial whose zeros are $(2 + \sqrt{3}) and (2 - \sqrt{3})$ is ......... .	(q) $x^{2} - 2 \sqrt{3} x + 2$
(c) The polynomial whose zeros are $\frac{3}{2} and - \frac{1}{2}$ is ......... .	(r) x² + x − 6
(d) The polynomial whose zeros are $(\sqrt{3} + 1) and (\sqrt{3} - 1)$	(s) 4x²− 4x − 3

Answer:

$(a) α = 2 and β = - 3 α + β = 2 + (- 3) = - 1 αβ = 2 \times (- 3) = - 6 ∴ The required polynomial will be x^{2} - (α + β) + αβ = x^{2} + x - 6 . (b) α = 2 + \sqrt{3} and β = 2 - \sqrt{3} α + β = 2 + \sqrt{3} + 2 - \sqrt{3} = 4 α β = (2 + \sqrt{3}) (2 - \sqrt{3}) = 4 - 3 = 1 ∴ The required polynomial will be x^{2} - (α + β) + αβ = x^{2} - 4 x + 1 . (c) α = \frac{3}{2} and β = \frac{- 1}{2} α + β = \frac{3}{2} - \frac{1}{2} = 1 αβ = \frac{- 3}{4} \begin{array}{l} ∴ R equired polynomial = x^{2} - (α + β) + αβ = x^{2} - x - \frac{3}{4}, i .e ., 4 x^{2} - 4 x - 3 \end{array} (d) α = \sqrt{3} + 1 and β = \sqrt{3} - 1 α + β = 2 \sqrt{3} α β = 3 - 1 = 2 ∴ T he required polynomial is x^{2} - (α + β) + α = x^{2} - 2 \sqrt{3} x + 2 . Thus, the correct matches are (a) - (r), (b) - (p), (c) - (s) and (d) - (q) .$

Page No 69:

Question 1:

Zeros of p(x) = x² − 2x − 3 are

(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3

Answer:

(b) 3,-1
Here, ${p (x) = x}^{2} - 2 x - 3$

$Let x^{2} - 2 x - 3 = 0 = > x^{2} - (3 - 1) x - 3 = 0 = > x^{2} - 3 x + x - 3 = 0 = > x (x - 3) + 1 (x - 3) = 0 = > (x - 3) (x + 1) = 0 = > x = 3, - 1$

Page No 69:

Question 2:

If α, β, γ are the zeros of the polynomial x³ − 6x² − x + 30, then the value of (αβ + βγ + γα) is

(a) −1
(b) 1
(c) −5
(d) 30

Answer:

(a) −1
Here, $p (x) = x^{3} - 6 x^{2} - x + 3$

Comparing the given polynomial with $x^{3} - (α + β + γ) x^{2} + (αβ + βγ + γα) x - αβγ$ , we get:
$(αβ + βγ + γα) = - 1$

Page No 69:

Question 3:

If α, β are the zeroes of kx² − 2x + 3k such that α + β = αβ, then k = ?

(a) $\frac{1}{3}$
(b) $\frac{- 1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{- 2}{3}$

Answer:

(c) $\frac{2}{3}$
Here, $p (x) = x^{2} - 2 x + 3 k$
Comparing the given polynomial with $a x^{2} + b x + c$ , we get:
$a = 1, b = - 2 and c = 3 k$
It is given that $α and β$ are the roots of the polynomial.
$∴ α + β = - \frac{b}{a} = > α + β = - (\frac{- 2}{1}) = > α + β = 2 . . . (i)$

Also, $α β$ = $\frac{c}{a}$
$= > αβ = \frac{3 k}{1} = > αβ = 3 k . . . (ii) Now, α + β = αβ = > 2 = 3 k [Using (i) and (ii)] = > k = \frac{2}{3}$

Page No 69:

Question 4:

It is given that the difference between the zeroes of 4x² − 8kx + 9 is 4 and k > 0. Then, k = ?

(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $\frac{5}{2}$
(d) $\frac{7}{2}$

Answer:

(c) $\frac{5}{2}$
Let the zeroes of the polynomial be $α and α + 4$ .
Here, p $(x) = 4 x^{2} - 8 k x + 9$
Comparing the given polynomial with $a x^{2} + b x + c$ , we get:
a = 4, b = −8k and c = 9
Now, sum of the roots $= - \frac{b}{a}$
$= > α + α + 4 = \frac{- (- 8 k)}{4} = > 2 α + 4 = 2 k = > α + 2 = k = > α = (k - 2) . . . (i) Also, product of the roots, αβ = \frac{c}{a} = > α (α + 4) = \frac{9}{4} = > (k - 2) (k - 2 + 4) = \frac{9}{4} = > (k - 2) (k + 2) = \frac{9}{4} = > k^{2} - 4 = \frac{9}{4} = > 4 k^{2} - 16 = 9 = > 4 k^{2} = 25 = > k^{2} = \frac{25}{4} = > k = \frac{5}{2} (∵ k > 0)$

Page No 69:

Question 5:

Find the zeros of the polynomial x² + 2x − 195.

Answer:

Here, p ${(x) = x}^{2} + 2 x - 195$

$L e t p (x) = 0 = > x^{2} + (15 - 13) x - 195 = 0 = > x^{2} + 15 x - 13 x - 195 = 0 = > x (x + 15) - 13 (x + 15) = 0 = > (x + 15) (x - 13) = 0 = > x = - 15,13 Hence, the zeroes are - 15 and 13 .$

Page No 69:

Question 6:

If one zero of the polynomial (a² + 9)x² + 13x + 6a is the reciprocal of the other, find the value of a.

Answer:

$(a + 9) x^{2} - 13 x + 6 a = 0 Here, A = (a^{2} + 9), B = 13 and C = 6 a Let α and \frac{1}{α} be the two zeroes . Then, product of the zeroes = \frac{C}{A} = > α . \frac{1}{α} = \frac{6 a}{a^{2} + 9} = > 1 = \frac{6 a}{a^{2} + 9} = > a^{2} + 9 = 6 a = > a^{2} - 6 a + 9 = 0 = > a^{2} - 2 \times a \times 3 + 3^{2} = 0 = > {(a - 3)}^{2} = 0 = > a - 3 = 0 = > a = 3$

Page No 69:

Question 7:

Find a quadratic polynomial whose zeros are 2 and −5.

Answer:

It is given that the two roots of the polynomial are 2 and −5.
Let $α = 2 and β = - 5$
Now, sum of the zeroes, $α + β$ = 2 + (−5) = −3
Product of the zeroes, $αβ$ = 2 $\times$ −5 = −10
∴ Required polynomial = $x^{2} - (α + β) x + αβ$
$= x^{2} — (- 3) x + (- 10) = x^{2} + 3 x - 10$

Page No 69:

Question 8:

If the zeroes of the polynomial x³− 3x² + x + 1 are (a − b), a and (a + b), find the values of a and b.

Answer:

The given polynomial $= x^{3} - 3 x^{2} + x + 1$ and its roots are $(a - b), a and (a + b)$ .

$C o m p a r i n g t h e g i v e n p o l y n o m i a l w i t h A x^{3} + B x^{2} + C x + D, we have : A = 1, B = - 3, C = 1 and D = 1 Now, (a - b) + a + (a + b) = \frac{- B}{A} = > 3 a = - \frac{- 3}{1} = > a = 1 Also, (a - b) \times a \times (a + b) = \frac{- D}{A} = > a (a^{2} - b^{2}) = \frac{- 1}{1} = > 1 (1^{2} - b^{2}) = - 1 = > 1 - b^{2} = - 1 = > b^{2} = 2 = > b = \pm \sqrt 2 ∴ a = 1 and b = \pm \sqrt 2$

Page No 69:

Question 9:

Verify that 2 is a zero of the polynomial x³ + 4x² − 3x − 18.

Answer:

Let p $(x) = x^{3} + 4 x^{2} - 3 x - 18$

$Now, p (2) = 2^{3} + 4 \times 2^{2} - 3 \times 2 - 18 = 0 ∴ 2 is a zero of p (x) .$

Page No 69:

Question 10:

Find the quadratic polynomial, the sum of whose zeroes is −5 and their product is 6.

Answer:

Given:
Sum of the zeroes = −5
Product of the zeroes = 6
∴ Required polynomial = $x^{2} - (sum of the zeroes) x + product of the zeroes$
$= x^{2} - (- 5) x + 6 = x^{2} + 5 x + 6$

Page No 70:

Question 11:

Find a cubic polynomial whose zeros are 3, 5 and −2.

Answer:

$Let α, β and γ be the zeroes of the required polynomial . Then we have : α + β + γ = 3 + 5 + (- 2) = 6 α β + β γ + γ α = 3 \times 5 + 5 \times (- 2) + (- 2) \times 3 = - 1 a n d α β γ = 3 \times 5 \times - 2 = - 30 Now, p (x) = x^{3} - x^{2} (α + β + γ) + x (α β + β γ + γ α) - α β γ = x^{3} - x^{2} \times 6 + x \times (- 1) - (- 30) = x^{3} - 6 x^{2} - x + 30 So, t h e r e q u i r e d p olyn o m i a l i s p (x) = x^{3} - 6 x^{2} - x + 30 .$

Page No 70:

Question 12:

Using remainder theorem, find the remainder when p(x) = x³ + 3x² − 5x + 4 is divided by (x − 2).

Answer:

$Given : p (x) = x^{3} + 3 x^{2} - 5 x + 4 Now, p (2) = 2^{3} + 3 {(2}^{2}) - 5 (2) + 4 = 8 + 12 - 10 + 4 = 14$

Page No 70:

Question 13:

Show that (x + 2) is a factor of f(x) = x³ + 4x² + x − 6.

Answer:

$Given : f (x) = x^{3} + 4 x^{2} + x - 6 Now, f (- 2) = {(- 2)}^{3} + 4 {(- 2)}^{2} + (- 2) - 6 = - 8 + 16 - 2 - 6 = 0 ∴ (x + 2) is a factor of f (x) = x^{3} + 4 x^{2} + x - 6 .$

Page No 70:

Question 14:

If α, β, γ are the zeroes of the polynomial p(x) = 6x³ + 3x²− 5x + 1, find the value of $(\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$

Answer:

$Given : p (x) = 6 x^{3} + 3 x^{2} - 5 x + 1 = 6 x^{2} - (- 3) x^{2} + (- 5) x - (- 1)$

Comparing the polynomial with $x^{3} - x^{2} (α + β + γ) + x (α β + β γ + γ α) - α β γ$ , we get:
$α β + β γ + γ α = - 5 and α β γ = - 1 ∴ (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) = (\frac{β γ + α γ + α β}{α β γ}) = (\frac{- 5}{- 1}) = 5$

Page No 70:

Question 15:

If α, β are the zeros of the polynomial f(x) = x² − 5x + k such that α − β = 1, find the value of k.

Answer:

$Given : f (x) = x^{2} - 5 x + k The co - efficients are a = 1, b = - 5 and c = k . ∴ α + β = \frac{- b}{a} = > α + β = - \frac{(- 5)}{1} = > α + β = 5 \dots (1) Also, α - β = 1 \dots (2) From (1) & (2), we get : 2 α = 6 = > α = 3 Pu t t i n g t h e v a l u e o f α in (1), w e g e t β = 2 . Now, α β = \frac{c}{a} = > 3 \times 2 = \frac{k}{1} ∴ k = 6$

Page No 70:

Question 16:

Show that the polynomial f(x) = x⁴ + 4x² + 6 has no zeroes.

Answer:

$Let t = x^{2} So, f (t) = t^{2} + 4 t + 6 Now, to find the zeroes, we will equate f (t) = 0 . \Rightarrow t^{2} + 4 t + 6 = 0 Now, t = \frac{- 4 \pm \sqrt{16 - 24}}{2} = \frac{- 4 \pm \sqrt{- 8}}{2} = - 2 \pm \sqrt{- 2} i . e ., x^{2} = - 2 \pm \sqrt{- 2} \Rightarrow x = \sqrt{- 2 \pm \sqrt{- 2}}, which is not a real number . The zeroes of a polynomial should be real number s . ∴ The given f (x) has no zeroes .$

Page No 70:

Question 17:

If one zero of the polynomial p(x) = x3 − 6x² + 11x − 6 is 3, find the other two zeroes.

Answer:

$Given : p (x) = x^{3} - 6 x^{2} + 11 x - 6 and its factor, x + 3 Let us divide p (x) by (x - 3) .$

$Here, x^{3} - 6 x^{2} + 11 x - 6 = (x - 3) (x^{2} - 3 x + 2) = (x - 3) [x^{2} - (2 + 1) x + 2] = (x - 3) (x^{2} - 2 x - x + 2) = (x - 3) [x (x - 2) - 1 (x - 2)] = (x - 3) (x - 1) (x - 2) ∴ The other two zeroes are 1 and 2 .$

Page No 70:

Question 18:

If two zeroes of the polynomial p(x) = 2x⁴ − 3x³ − 3x² + 6x − 2 are $\sqrt{2}$ and $- \sqrt{2}$ , find its other two zeroes.

Answer:

$Given : p (x) = 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 a n d t h e t w o z e r o e s, \sqrt{2} and - \sqrt{2} So, the polynomial is (x + \sqrt{2}) (x - \sqrt{2}) = x^{2} - 2 . Let us divide p (x) b y (x^{2} - 2) .$

$Here, 2 x^{4} - 3 x^{3} - 3 x^{2} + 6 x - 2 = (x^{2} - 2) (2 x^{2} - 3 x + 1) = (x^{2} - 2) [2 x^{2} - (2 + 1) x + 1] = (x^{2} - 2) (2 x^{2} - 2 x - x + 1) = (x^{2} - 2) [(2 x (x - 1) - 1 (x - 1)] = (x^{2} - 2) (2 x - 1) (x - 1) ∴ Th e o t h e r t w o z e r o e s a r e \frac{1}{2} a n d 1 .$

Page No 70:

Question 19:

Find the quotient when p(x) = 3x⁴ + 5x³ − 7x² + 2x + 2 is divided by (x² + 3x + 1).

Answer:

$Given : p (x) = 3 x^{4} + 5 x^{3} - 7 x^{2} + 2 x + 2 Dividing p (x) by (x^{2} + 3 x + 1), we have :$

$∴ The quotient is 3 x^{2} - 4 x + 2$

Page No 70:

Question 20:

Use remainder theorem to find the value of k, it being given that when x³ + 2x² + kx + 3 is divided by (x − 3), then the remainder is 21.

Answer:

$Let p (x) = x^{3} + 2 x^{2} + k x + 3 Now, p (3) = {(3)}^{3} + 2 {(3)}^{2} + 3 k + 3 = 27 + 18 + 3 k + 3 = 48 + 3 k It is given that the remainder is 21 ∴ 3 k + 48 = 21 \Rightarrow 3 k = - 27 \Rightarrow k = - 9$

View NCERT Solutions for all chapters of Class 10

Login or Create a free account