Rs Aggarwal 2021 2022 for Class 10 Maths Chapter 2 - Polynomials

Share with your friends

Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 2 Polynomials are provided here with simple step-by-step explanations. These solutions for Polynomials are extremely popular among Class 10 students for Maths Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 52:

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 52:

Answer:

$x^{2} - 2 x - 8 = 0 \Rightarrow x^{2} - 4 x + 2 x - 8 = 0 \Rightarrow x (x - 4) + 2 (x - 4) = 0$
$\Rightarrow (x - 4) (x + 2) = 0 \Rightarrow (x - 4) = 0 or (x + 2) = 0 \Rightarrow x = 4 or x = - 2$

Sum of zeroes = $4 + (- 2) = 2 = \frac{2}{1} = \frac{- (coefficient of x)}{coefficient of x^{2}}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $4 (- 2) = - 8 = \frac{- 8}{1} = \frac{constant term}{coefficient of x^{2}}$ (−4)(−3)=121=constant termcoefficient of x2

Page No 52:

Answer:

$x^{2} + 7 x + 12 = 0 \Rightarrow x^{2} + 4 x + 3 x + 12 = 0 \Rightarrow x (x + 4) + 3 (x + 4) = 0$

$\Rightarrow (x + 4) (x + 3) = 0 \Rightarrow (x + 4) = 0 or (x + 3) = 0 \Rightarrow x = - 4 or x = - 3$

Sum of zeroes = $- 4 + (- 3) = \frac{- 7}{1} = \frac{- (coefficient of x)}{(coefficient of x^{2})}$
Product of zeroes = $(- 4) (- 3) = \frac{12}{1} = \frac{constant term}{coefficient of x^{2}}$

Page No 52:

Answer:

Let f(x) = 2x²– x – 6

$To find the zeroes, we put f (x) = 0 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0 \Rightarrow (x - 2) = 0 or (2 x + 3) = 0 \Rightarrow x = 2, - \frac{3}{2}$

Hence, all the zeroes of the polynomial f(x) are $2 and - \frac{3}{2}$ .

Now,
$f (2) = 2 {(2)}^{2} - 2 - 6 = 2 (4) - 8 = 8 - 8 = 0 f (- \frac{3}{2}) = 2 {(- \frac{3}{2})}^{2} - (- \frac{3}{2}) - 6 = 2 (\frac{9}{4}) + \frac{3}{2} - 6 = \frac{9}{2} + \frac{3}{2} - 6 = \frac{12}{2} - 6 = 6 - 6 = 0$

Hence, the relationship between the zeros and the coefficients is verified.

Page No 52:

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 52:

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 52:

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 52:

Answer:

$2 \sqrt{3} x^{2} - 5 x + \sqrt{3} \Rightarrow 2 \sqrt{3} x^{2} - 2 x - 3 x + \sqrt{3} \Rightarrow 2 x (\sqrt{3} x - 1) - \sqrt{3} (\sqrt{3} x - 1) = 0$

$\Rightarrow (\sqrt{3} x - 1) or (2 x - \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x - 1) = 0 or (2 x - \sqrt{3}) = 0 \Rightarrow x = \frac{1}{\sqrt{3}} or x = \frac{\sqrt{3}}{2} \Rightarrow x = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} or x = \frac{\sqrt{3}}{2}$

Sum of zeroes = $\frac{\sqrt{3}}{3} + \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{6} = \frac{- (coefficient of x)}{coefficient of x^{2}}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $\frac{\sqrt{3}}{3} \times \frac{\sqrt{3}}{2} = \frac{1}{2} = \frac{constant term}{coefficient of x^{2}}$ (−4)(−3)=121=constant termcoefficient of x2

Page No 52:

Answer:

$4 x^{2} - 4 x + 1 = 0 \Rightarrow {(2 x)}^{2} - 2 (2 x) (1) + {(1)}^{2} = 0 \Rightarrow {(2 x - 1)}^{2} = 0 [∵ a^{2} - 2 a b + b^{2} = {(a - b)}^{2}]$

$\Rightarrow {(2 x - 1)}^{2} = 0 \Rightarrow x = \frac{1}{2} or x = \frac{1}{2}$

Sum of zeroes = $\frac{1}{2} + \frac{1}{2} = 1 = \frac{1}{1} = \frac{- (coefficient of x)}{coefficient of x^{2}}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{constant term}{coefficient of x^{2}}$ (−4)(−3)=121=constant termcoefficient of x2

Page No 52:

Answer:

Let f(x) = 3x²– x – 4

$To find the zeroes, we put f (x) = 0 \Rightarrow 3 x^{2} - x - 4 = 0 \Rightarrow 3 x^{2} - 4 x + 3 x - 4 = 0 \Rightarrow x (3 x - 4) + 1 (3 x - 4) = 0 \Rightarrow (x + 1) (3 x - 4) = 0 \Rightarrow (x + 1) = 0 or (3 x - 4) = 0 \Rightarrow x = - 1, \frac{4}{3}$

Hence, all the zeroes of the polynomial f(x) are $- 1 and \frac{4}{3}$ .

Now,
$f (- 1) = 3 {(- 1)}^{2} - (- 1) - 4 = 3 (1) + 1 - 4 = 3 - 3 = 0 f (\frac{4}{3}) = 3 {(\frac{4}{3})}^{2} - \frac{4}{3} - 4 = 3 (\frac{16}{9}) - \frac{4}{3} - 4 = \frac{16}{3} - \frac{4}{3} - 4 = \frac{12}{3} - 4 = 4 - 4 = 0$

Hence, the relationship between the zeros and the coefficients is verified.

Page No 52:

Answer:

Let f(x) = 5x²+ 10x

$To find the zeroes, we put f (x) = 0 \Rightarrow 5 x^{2} + 10 x = 0 \Rightarrow 5 x (x + 2) = 0 \Rightarrow (x) (x + 2) = 0 \Rightarrow (x) = 0 or (x + 2) = 0 \Rightarrow x = 0, - 2$

Hence, all the zeroes of the polynomial f(x) are $0 and - 2$ .

Now,
$f (0) = 5 {(0)}^{2} + 10 (0) = 0 + 0 = 0 f (- 2) = 5 {(- 2)}^{2} + 10 (- 2) = 5 (4) - 20 = 20 - 20 = 0$

Hence, the relationship between the zeros and the coefficients is verified.

Page No 52:

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 52:

Answer:

Let α and β be the zeroes of the polynomial $p (x) = 2 x^{2} + 5 x + k$ .

$Sum of zeroes = - \frac{b}{a} \Rightarrow α + β = - \frac{5}{2} . . . (1) and Product of zeroes = \frac{c}{a} \Rightarrow α β = \frac{k}{2} . . . (2)$

Now, using (1)
$α + β = - \frac{5}{2} Squaring both sides, we get \Rightarrow {(α + β)}^{2} = {(- \frac{5}{2})}^{2} \Rightarrow α^{2} + β^{2} + 2 α β = \frac{25}{4} \Rightarrow α^{2} + β^{2} + α β + α β = \frac{25}{4} \Rightarrow \frac{21}{4} + α β = \frac{25}{4} \Rightarrow α β = \frac{25}{4} - \frac{21}{4} \Rightarrow α β = 1 \Rightarrow \frac{k}{2} = 1 (from (2)) \Rightarrow k = 2$

Hence, the value of k is 2.

Page No 52:

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 52:

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 52:

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 52:

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 52:

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 52:

Answer:

Given: $(x + a)$ is a factor of $2 x^{2} + 2 a x + 5 x + 10$
So, we have
$x + a = 0 \Rightarrow x = - a$
Now, It will satisfy the above polynomial.
Therefore, we will get
$2 {(- a)}^{2} + 2 a (- a) + 5 (- a) + 10 = 0 \Rightarrow 2 a^{2} - 2 a^{2} - 5 a + 10 = 0 \Rightarrow - 5 a = - 10 \Rightarrow a = 2$

Page No 53:

Answer:

Given: $a x^{2} + 7 x + b = 0$
Since, $x = \frac{2}{3}$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
$a {(\frac{2}{3})}^{2} + 7 (\frac{2}{3}) + b = 0 \Rightarrow \frac{4}{9} a + \frac{14}{3} + b = 0 \Rightarrow 4 a + 42 + 9 b = 0 \Rightarrow 4 a + 9 b = - 42 . . . (1)$

Since, $x = - 3$ is the root of the above quadratic equation
Hence, It will satisfy the above equation.
Therefore, we will get
$a {(- 3)}^{2} + 7 (- 3) + b = 0 \Rightarrow 9 a - 21 + b = 0 \Rightarrow 9 a + b = 21 . . . (2)$
From (1) and (2), we get
$a = 3, b = - 6$

Page No 59:

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 60:

Answer:

Page No 61:

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = $\frac{- (coefficient of x)}{coefficent of x^{2}}$ and Product of zeroes = $\frac{constant term}{coefficent of x^{2}}$

$\Rightarrow α + β = \frac{- 1}{1} and α β = \frac{- 2}{1} \Rightarrow α + β = - 1 and α β = - 2 Now, {(\frac{1}{α} - \frac{1}{β})}^{2} = {(\frac{β - α}{α β})}^{2}$

$= \frac{{(α + β)}^{2} - 4 α β}{{(α β)}^{2}} [∵ {(β - α)}^{2} = {(α + β)}^{2} - 4 α β] = \frac{{(- 1)}^{2} - 4 (- 2)}{{(- 2)}^{2}} [∵ α + β = - 1 and α β = - 2] = \frac{{(- 1)}^{2} - 4 (- 2)}{4} = \frac{9}{4}$

$∵ {(\frac{1}{α} - \frac{1}{β})}^{2} = \frac{9}{4} \Rightarrow \frac{1}{α} - \frac{1}{β} = \pm \frac{3}{2}$

Page No 61:

Answer:

By using the relationship between the zeroes of the cubic ploynomial.
We have, Sum of zeroes = $\frac{- (coefficient of x^{2})}{coefficent of x^{3}}$
$∴ a - b + a + a + b = \frac{- (- 3)}{1} \Rightarrow 3 a = 3 \Rightarrow a = 1$

Now, Product of zeros = $\frac{- (constant term)}{coefficent of x^{3}}$
$∴ (a - b) (a) (a + b) = \frac{- 1}{1} \Rightarrow (1 - b) (1) (1 + b) = - 1 [∵ a = 1] \Rightarrow 1 - b^{2} = - 1 \Rightarrow b^{2} = 2 \Rightarrow b = \pm \sqrt{2}$

Page No 63:

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 63:

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 63:

Answer:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
$x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x - a b c$ ...(1)
Let $a = 2, b = - 3 and c = 4$
Substituting the values in (1), we get
$x^{3} - (2 - 3 + 4) x^{2} + (- 6 - 12 + 8) x - (- 24) \Rightarrow x^{3} - 3 x^{2} - 10 x + 24$

Page No 63:

Answer:

If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
$x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x - a b c$ ...(1)
Let $a = \frac{1}{2}, b = 1 and c = - 3$
Substituting the values in (1), we get
$x^{3} - (\frac{1}{2} + 1 - 3) x^{2} + (\frac{1}{2} - 3 - \frac{3}{2}) x - (\frac{- 3}{2}) \Rightarrow x^{3} - (\frac{- 3}{2}) x^{2} - 4 x + \frac{3}{2} \Rightarrow 2 x^{3} + 3 x^{2} - 8 x + 3$

Page No 63:

Answer:

We know the sum, sum of the product of the zeroes taken two at a time and the product of the zeroes of a cubic polynomial then the cubic polynomial can be found as
x³ −(Sum of the zeroes)x² + (sum of the product of the zeroes taking two at a time)x − Product of zeroes
Therefore, the required polynomial is
$x^{3} - 5 x^{2} - 2 x + 24$

Page No 63:

Answer:

Quotient $q (x) = x - 3$
Remainder $r (x) = 7 x - 9$

Page No 63:

Answer:

Quotient $q (x) = x^{2} + x - 3$
Remainder $r (x) = 8$

Page No 63:

Answer:

We can write

$f (x) as x^{4} + 0 x^{3} + 0 x^{2} - 5 x + 6$ and $g (x) as - x^{2} + 2$

Quotient $q (x) = - x^{2} - 2$
Remainder $r (x) = - 5 x + 10$

Page No 63:

Answer:

Let $f (x) = 2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$ and $g (x) = x^{2} - 3$

Quotient $q (x) = 2 x^{2} + 3 x + 4$
Remainder $r (x) = 0$
Since, the remainder is 0.
Hence, $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$

Page No 63:

Answer:

Let f(x) = x⁴ + 2x³+ 8x²+ 12x + 18

It is given that when f(x) is divisible by x² + 5, the remainder comes out to be px + q.

On division, we get the quotient x²+ 2x + 3 and the remainder 2x + 3.

Since, the remainder comes out to be px + q.

Therefore, p = 2 and q = 3.

Hence, the values of p and q are 2 and 3 respectively.

Page No 63:

Answer:

By using division rule, we have
Divided = Quotient × Divisor + Remainder
$∴ 3 x^{3} + x^{2} + 2 x + 5 = (3 x - 5) g (x) + 9 x + 10 \Rightarrow 3 x^{3} + x^{2} + 2 x + 5 - 9 x - 10 = (3 x - 5) g (x) \Rightarrow 3 x^{3} + x^{2} - 7 x - 5 = (3 x - 5) g (x) \Rightarrow g (x) = \frac{3 x^{3} + x^{2} - 7 x - 5}{3 x - 5}$

$∴ g (x) = x^{2} + 2 x + 1$

Page No 63:

Answer:

We can write $f (x) as - 6 x^{3} + x^{2} + 20 x + 8$ and $g (x) as - 3 x^{2} + 5 x + 2$

Quotient = $2 x + 3$
Remainder = $x + 2$

By using division rule, we have

Divided = Quotient × Divisor + Remainder

$∴ - 6 x^{3} + x^{2} + 20 x + 8 = (- 3 x^{2} + 5 x + 2) (2 x + 3) + x + 2 \Rightarrow - 6 x^{3} + x^{2} + 20 x + 8 = - 6 x^{3} + 10 x^{2} + 4 x - 9 x^{2} + 15 x + 6 + x + 2 \Rightarrow - 6 x^{3} + x^{2} + 20 x + 8 = - 6 x^{3} + x^{2} + 20 x + 8$

Page No 63:

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 63:

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 63:

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 63:

Answer:

Let f(x) = 2x⁴ – 5x³ – 11x² + 20x + 12

It is given that 2 and –2 are two zeroes of f(x)

Thus, f(x) is completely divisible by (x + 2) and (x – 2).

Therefore, one factor of f(x) is (x² – 4).

We get another factor of f(x) by dividing it with (x² – 4).

On division, we get the quotient 2x² – 5x – 3.

$\Rightarrow f (x) = (x^{2} - 4) (2 x^{2} - 5 x - 3) = (x^{2} - 4) (2 x^{2} - 6 x + x - 3) = (x^{2} - 4) (2 x (x - 3) + 1 (x - 3)) = (x^{2} - 4) (2 x + 1) (x - 3) To find the zeroes, we put f (x) = 0 \Rightarrow (x^{2} - 4) (2 x + 1) (x - 3) = 0 \Rightarrow (x^{2} - 4) = 0 or (2 x + 1) = 0 or (x - 3) = 0 \Rightarrow x = \pm 2, - \frac{1}{2}, 3$

Hence, all the zeroes of the polynomial f(x) are $2, - 2, - \frac{1}{2} and 3 .$

Page No 64:

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 64:

Answer:

Let f(x) = x⁴+ x³– 14x²– 2x + 24

It is given that $\sqrt{2} and - \sqrt{2}$ are two zeroes of f(x)

Thus, f(x) is completely divisible by (x + $\sqrt{2}$ ) and (x – $\sqrt{2}$ ).

Therefore, one factor of f(x) is (x² – 2).

We get another factor of f(x) by dividing it with (x² – 2).

On division, we get the quotient x² + x – 12.

$\Rightarrow f (x) = (x^{2} - 2) (x^{2} + x - 12) = (x^{2} - 2) (x^{2} + 4 x - 3 x - 12) = (x^{2} - 2) (x (x + 4) - 3 (x + 4)) = (x^{2} - 2) (x + 4) (x - 3) To find the zeroes, we put f (x) = 0 \Rightarrow (x^{2} - 2) (x + 4) (x - 3) = 0 \Rightarrow (x^{2} - 2) = 0 or (x + 4) = 0 or (x - 3) = 0 \Rightarrow x = \pm \sqrt{2}, - 4, 3$

Hence, all the zeroes of the polynomial f(x) are $\sqrt{2}, - \sqrt{2}, - 4 and 3 .$

Page No 64:

Answer:

Let f(x) = 2x⁴– 13x³+ 19x²+ 7x – 3

It is given that $(2 + \sqrt{3}) and (2 - \sqrt{3})$ are two zeroes of f(x)

Thus, f(x) is completely divisible by (x $- 2 - \sqrt{3}$ ) and (x – $2 + \sqrt{3}$ ).

Therefore, one factor of f(x) is $({(x - 2)}^{2} - 3)$
$\Rightarrow$ one factor of f(x) is (x² – 4x + 1)

We get another factor of f(x) by dividing it with (x² – 4x + 1).

On division, we get the quotient 2x² – 5x – 3.

$\Rightarrow f (x) = (x^{2} - 4 x + 1) (2 x^{2} - 5 x - 3) = (x^{2} - 4 x + 1) (2 x^{2} - 6 x + x - 3) = (x^{2} - 4 x + 1) (2 x (x - 3) + 1 (x - 3)) = (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3}) (2 x + 1) (x - 3) To find the zeroes, we put f (x) = 0 \Rightarrow (x - 2 - \sqrt{3}) (x - 2 + \sqrt{3}) (2 x + 1) (x - 3) = 0 \Rightarrow (x - 2 - \sqrt{3}) = 0 or (x - 2 + \sqrt{3}) = 0 or (2 x + 1) = 0 or (x - 3) = 0 \Rightarrow x = 2 + \sqrt{3}, 2 - \sqrt{3,} - \frac{1}{2}, 3$

Hence, all the zeroes of the polynomial f(x) are $2 + \sqrt{3}, 2 - \sqrt{3}, - \frac{1}{2} and 3 .$

Page No 64:

Answer:

Let f(x) = 3x³ + 16x² + 15x – 18

It is given that one of its zeroes is $\frac{2}{3}$ .

Therefore, one factor of f(x) is (x – $\frac{2}{3}$ ).

We get another factor of f(x) by dividing it with (x – $\frac{2}{3}$ ).

On division, we get the quotient 3x²+ 18x + 27.

$\Rightarrow f (x) = (x - \frac{2}{3}) (3 x^{2} + 18 x + 27) = (x - \frac{2}{3}) (3 x^{2} + 9 x + 9 x + 27) = (x - \frac{2}{3}) (3 x (x + 3) + 9 (x + 3)) = (x - \frac{2}{3}) (3 x + 9) (x + 3) = 3 (x - \frac{2}{3}) (x + 3) (x + 3) = 3 (x - \frac{2}{3}) {(x + 3)}^{2} To find the zeroes, we put f (x) = 0 \Rightarrow 3 (x - \frac{2}{3}) {(x + 3)}^{2} = 0 \Rightarrow (x - \frac{2}{3}) = 0 or {(x + 3)}^{2} = 0 \Rightarrow x = \frac{2}{3}, - 3$

Hence, other zero of the polynomial f(x) is –3.

Page No 64:

Answer:

Let f(x) = 2x⁴– 3x³– 3x²+ 6x – 2

It is given that 1 and $\frac{1}{2}$ are two zeroes of f(x).

Thus, f(x) is completely divisible by (x – 1) and (x – $\frac{1}{2}$ ).

Therefore, one factor of f(x) is $(x - 1) (x - \frac{1}{2})$
$\Rightarrow$ one factor of f(x) is $(x^{2} - \frac{3}{2} x + \frac{1}{2})$

We get another factor of f(x) by dividing it with $(x^{2} - \frac{3}{2} x + \frac{1}{2})$ .

On division, we get the quotient 2x² – 4.

$\Rightarrow f (x) = (x^{2} - \frac{3}{2} x + \frac{1}{2}) (2 x^{2} - 4) = (x - 1) (x - \frac{1}{2}) (2 x^{2} - 4) To find the zeroes, we put f (x) = 0 \Rightarrow (x - 1) (x - \frac{1}{2}) (2 x^{2} - 4) = 0 \Rightarrow (x - 1) = 0 or (x - \frac{1}{2}) = 0 or (2 x^{2} - 4) = 0 \Rightarrow x = 1, \frac{1}{2}, \pm \sqrt{2}$

Hence, all the zeroes of the polynomial f(x) are $1, \frac{1}{2}, \sqrt{2} and - \sqrt{2} .$

Page No 64:

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 65:

Answer:

Let the other zeroes of $x^{2} - 4 x + 1$ be a.
By using the relationship between the zeroes of the quadratic ploynomial.
We have, Sum of zeroes = $\frac{- (coefficient of x)}{coefficent of x^{2}}$
$∴ 2 + \sqrt{3} + a = \frac{- (- 4)}{1} \Rightarrow a = 2 - \sqrt{3}$
Hence, the other zeroes of $x^{2} - 4 x + 1$ is $2 - \sqrt{3}$ .

Page No 65:

Answer:

$f (x) = x^{2} + x - p (p + 1)$
By adding and subtracting px, we get
$f (x) = x^{2} + p x + x - p x - p (p + 1) = x^{2} + (p + 1) x - p x - p (p + 1) = x [x + (p + 1)] - p [x + (p + 1)]$

$= [x + (p + 1)] (x - p) f (x) = 0 \Rightarrow [x + (p + 1)] (x - p) = 0 \Rightarrow [x + (p + 1)] = 0 or (x - p) = 0 \Rightarrow x = - (p + 1) or x = p$
So, the zeros of f(x) are −(p + 1) and p.

Page No 65:

Answer:

$f (x) = x^{2} - 3 x - m (m + 3)$
By adding and subtracting mx, we get
$f (x) = x^{2} - m x - 3 x + m x - m (m + 3) = x [x - (m + 3)] + m [x - (m + 3)] = [x - (m + 3)] (x + m)$

$f (x) = 0 \Rightarrow [x - (m + 3)] (x + m) = 0 \Rightarrow [x - (m + 3)] = 0 or (x + m) = 0 \Rightarrow x = m + 3 or x = - m$
So, the zeros of f(x) are −m and m + 3.

Page No 66:

Answer:

If the zeroes of the quadratic polynomial are α and β then the quadratic polynomial can be found as
x² − (α + β)x + αβ .....(1)
Substituting the values in (1), we get
x² − 6x + 4

Page No 66:

Answer:

Given: x = 2 is one zero of the quadratic polynomial kx² + 3x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$k {(2)}^{2} + 3 (2) + k = 0 \Rightarrow 4 k + 6 + k = 0 \Rightarrow 5 k + 6 = 0 \Rightarrow k = - \frac{6}{5}$

Page No 66:

Answer:

Given: x = 3 is one zero of the polynomial 2x² + x + k
Therefore, It will satisfy the above polynomial.
Now, we have
$2 {(3)}^{2} + 3 + k = 0 \Rightarrow 21 + k = 0 \Rightarrow k = - 21$

Page No 66:

Answer:

Given: x = −4 is one zero of the polynomial x² − x −(2k + 2)
Therefore, It will satisfy the above polynomial.
Now, we have
${(- 4)}^{2} - (- 4) - (2 k + 2) = 0 \Rightarrow 16 + 4 - 2 k - 2 = 0 \Rightarrow - 2 k = - 18 \Rightarrow k = 9$

Page No 66:

Answer:

Given: x = 1 is one zero of the polynomial ax² − 3(a − 1) x − 1
Therefore, It will satisfy the above polynomial.
Now, we have
$a {(1)}^{2} - 3 (a - 1) 1 - 1 = 0 \Rightarrow a - 3 a + 3 - 1 = 0 \Rightarrow - 2 a = - 2 \Rightarrow a = 1$

Page No 66:

Answer:

Given: x = −2 is one zero of the polynomial 3x² + 4x + 2k
Therefore, It will satisfy the above polynomial.
Now, we have
$3 {(- 2)}^{2} + 4 (- 2) + 2 k = 0 \Rightarrow 12 - 8 + 2 k = 0 \Rightarrow k = - 2$

Page No 66:

Answer:

$f (x) = x^{2} - x - 6 = x^{2} - 3 x + 2 x - 6 = x (x - 3) + 2 (x - 3)$

$= (x - 3) (x + 2) f (x) = 0 \Rightarrow (x - 3) (x + 2) = 0 \Rightarrow (x - 3) = 0 or (x + 2) = 0 \Rightarrow x = 3 or x = - 2$
So, the zeros of f(x) are 3 and −2.

Page No 66:

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Sum of zeroes = $\frac{- (coefficient of x)}{coefficent of x^{2}}$
$\Rightarrow 1 = \frac{- (- 3)}{k} \Rightarrow k = 3$

Page No 66:

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have
Product of zeroes = $\frac{constant term}{coefficent of x^{2}}$
$\Rightarrow 3 = \frac{k}{1} \Rightarrow k = 3$

Page No 66:

Answer:

Given: (x + a) is a factor of 2x² + 2ax + 5x + 10
We have
$x + a = 0 \Rightarrow x = - a$
Since, (x + a) is a factor of 2x² + 2ax + 5x + 10
Hence, It will satisfy the above polynomial
$∴ 2 {(- a)}^{2} + 2 a (- a) + 5 (- a) + 10 = 0 \Rightarrow - 5 a + 10 = 0 \Rightarrow a = 2$

Page No 66:

Answer:

By using the relationship between the zeroes of the cubic ploynomial.
We have
Sum of zeroes = $\frac{- (coefficient of x^{2})}{coefficent of x^{3}}$
$\Rightarrow a - b + a + a + b = \frac{- (- 6)}{2} \Rightarrow 3 a = 3 \Rightarrow a = 1$

Page No 66:

Answer:

Equating x² − x to 0 to find the zeros, we will get
$x (x - 1) = 0 \Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$

Since, x³ + x² − ax + b is divisible by x² − x.
Hence, the zeros of x² − x will satisfy x³ + x² − ax + b
$∴ {(0)}^{3} + 0^{2} - a (0) + b = 0 \Rightarrow b = 0$
and
${(1)}^{3} + 1^{2} - a (1) + 0 = 0 [∵ b = 0] \Rightarrow a = 2$

Page No 66:

Answer:

By using the relationship between the zeros of the quadratic ploynomial.
We have,
Sum of zeroes = $\frac{- (coefficient of x)}{coefficent of x^{2}}$ and Product of zeroes = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- 7}{2} and α β = \frac{5}{2} Now, α + β + α β = \frac{- 7}{2} + \frac{5}{2} = - 1$

Page No 66:

Answer:

“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, then there exists unique polynomials q(x) and r(x) such that

f(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x).

Page No 66:

Answer:

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula
x² − (Sum of the zeros)x + Product of zeros
$\Rightarrow x^{2} - (- \frac{1}{2}) x + (- 3) \Rightarrow x^{2} + \frac{1}{2} x - 3$
Hence, the required polynomial is $x^{2} + \frac{1}{2} x - 3$ .

⇒x2−2√x+13=0⇒3x2−32√x+1=0

Page No 66:

Answer:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
$∴ f (x) = 0 \Rightarrow 6 x^{2} - 3 = 0 \Rightarrow 3 (2 x^{2} - 1) = 0 \Rightarrow 2 x^{2} - 1 = 0$
$\Rightarrow 2 x^{2} = 1 \Rightarrow x^{2} = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}$
Hence, the zeros of the quadratic polynomial f(x) = 6x² − 3 are $\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}$ .

Page No 66:

Answer:

To find the zeros of the quadratic polynomial we will equate f(x) to 0
$∴ f (x) = 0 \Rightarrow 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0 \Rightarrow 4 \sqrt{3} x^{2} + 8 x - 3 x - 2 \sqrt{3} = 0 \Rightarrow 4 x (\sqrt{3} x + 2) - \sqrt{3} (\sqrt{3} x + 2) = 0$
$\Rightarrow (\sqrt{3} x + 2) (4 x - \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x + 2) = 0 or (4 x - \sqrt{3}) = 0 \Rightarrow x = - \frac{2}{\sqrt{3}} or x = \frac{\sqrt{3}}{4}$
Hence, the zeros of the quadratic polynomial $f (x) = 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3}$ are $- \frac{2}{\sqrt{3}} or \frac{\sqrt{3}}{4}$ .

Page No 66:

Answer:

Page No 66:

Answer:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = $\frac{- (coefficient of x)}{coefficent of x^{2}}$ and Product of zeroes = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- 1}{6} and α β = - \frac{1}{3} Now, \frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β} = \frac{α^{2} + β^{2} + 2 α β - 2 α β}{α β}$
$= \frac{{(α + β)}^{2} - 2 α β}{α β} = \frac{{(\frac{- 1}{6})}^{2} - 2 (- \frac{1}{3})}{- \frac{1}{3}} = \frac{\frac{1}{36} + \frac{2}{3}}{- \frac{1}{3}} = - \frac{25}{12}$

Page No 69:

Answer:

(d) is the correct option.
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 69:

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 69:

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 69:

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 69:

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 69:

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 70:

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 70:

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 70:

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 70:

Answer:

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

Page No 70:

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 70:

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 70:

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 70:

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 70:

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 70:

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 70:

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 70:

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 71:

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 71:

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 71:

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 71:

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 71:

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 71:

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 71:

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 71:

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 71:

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 71:

Answer:

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

Page No 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 75:

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 76:

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 76:

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 76:

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 76:

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