Rs Aggarwal 2019 for Class 10 Math Chapter 2 - Polynomials

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Rs Aggarwal 2019 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 2019 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 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 48:

Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$x^{2} + 7 x + 12$

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 48:

Question 2:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$x^{2} - 2 x - 8$

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 48:

Question 3:

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 48:

Question 4:

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 48:

Question 5:

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 48:

Question 6:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$2 \sqrt{3} x^{2} - 5 x + \sqrt{3}$

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 48:

Question 7:

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 48:

Question 8:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$4 x^{2} - 4 x + 1$

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 49:

Question 9:

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 49:

Question 10:

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 49:

Question 11:

Find the zeros of the quadratic polynomial (5y² + 10y) and verify the relation between the zeros and the coefficients.

Answer:

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

Hence, the relation has been verified.

Page No 49:

Question 12:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$3 x^{2} - x - 4$

Answer:

$3 x^{2} - x - 4 = 0 \Rightarrow 3 x^{2} - 4 x + 3 x - 4 = 0 \Rightarrow x (3 x - 4) + 1 (3 x - 4) = 0$
$\Rightarrow (3 x - 4) (x + 1) = 0 \Rightarrow (3 x - 4) or (x + 1) = 0 \Rightarrow x = \frac{4}{3} or x = - 1$

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

Page No 49:

Question 13:

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 49:

Question 14:

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 49:

Question 15:

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 49:

Question 16:

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 49:

Question 17:

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 49:

Question 18:

Find the quadratic polynomial, the sum of whose root is $\sqrt{2}$ and their product is $\frac{1}{3}$

Answer:

We can find the quadratic equation if we know the sum of the roots and product of the roots by using the formula
x² − (Sum of the roots)x + Product of roots = 0
$\Rightarrow x^{2} - \sqrt{2} x + \frac{1}{3} = 0 \Rightarrow 3 x^{2} - 3 \sqrt{2} x + 1 = 0$

Page No 49:

Question 19:

If $x = \frac{2}{3}$ and $x = - 3$ are the roots of the quadratic equation $a x^{2} + 7 x + b = 0$ then find the values of a and b.

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 49:

Question 20:

If $(x + a)$ is a factor of the polynomial $2 x^{2} + 2 a x + 5 x + 10$ , find the value of a.

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 49:

Question 21:

One zero of the polynomial $3 x^{3} + 16 x^{2} + 15 x - 18$ is $\frac{2}{3}$ . Find the other zeroes of the polynomial.

Answer:

Given: $x = \frac{2}{3}$ is one of the zero of $3 x^{3} + 16 x^{2} + 15 x - 18$
Now, we have
$x = \frac{2}{3} \Rightarrow x - \frac{2}{3} = 0$
Now, we divide $3 x^{3} + 16 x^{2} + 15 x - 18$ by $x - \frac{2}{3}$ to find the quotient

So, the quotient is $3 x^{2} + 18 x + 27$
Now,
$3 x^{2} + 18 x + 27 = 0 \Rightarrow 3 x^{2} + 9 x + 9 x + 27 = 0 \Rightarrow 3 x (x + 3) + 9 (x + 3) = 0$

$\Rightarrow (x + 3) (3 x + 9) = 0 \Rightarrow (x + 3) = 0 or (3 x + 9) = 0 \Rightarrow x = - 3 or x = - 3$

Page No 58:

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 58:

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:

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

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Question 3:

Find a cubic polynomial whose zeroes are 2, −3 and 4

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 58:

Question 4:

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

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 58:

Question 5:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes are 5, −2 and −24 respectively.

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 58:

Question 6:

Find the quotient and the remainder when:
$f (x) = x^{3} - 3 x^{2} + 5 x - 3$ is divided by $g (x) = x^{2} - 2$

Answer:

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

Page No 58:

Question 7:

Find the quotient and remainder when:
$f (x) = x^{4} - 3 x^{2} + 4 x + 5$ is divided by $g (x) = x^{2} + 1 - x$

Answer:

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

Page No 58:

Question 8:

Find the quotient and remainder when
$f (x) = x^{4} - 5 x + 6$ is divided by $g (x) = 2 - x^{2}$

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 58:

Question 9:

By actual division, show that $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$

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 58:

Question 10:

On dividing, $3 x^{3} + x^{2} + 2 x + 5$ by a polynomial g(x), the quotient and remainder are $3 x - 5$ and $9 x + 10$ respectively. Find g(x)

Answer:

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

$∴ g (x) = x^{2} + 2 x + 1$

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Question 11:

Verify division algorithm for the polynomials $f (x) = 8 + 20 x + x^{2} - 6 x^{3}$ and $g (x) = 2 + 5 x - 3 x^{2}$

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

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

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Question 12:

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:

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

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

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Question 13:

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

Answer:

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

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

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Question 14:

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

Answer:

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

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

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Question 15:

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

Answer:

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

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

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Question 16:

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:

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

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

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Question 17:

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:

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

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

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Question 18:

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

Answer:

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

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

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Question 19:

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:

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

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

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Question 1:

If one zero of the polynomial $x^{2} - 4 x + 1$ is $2 + \sqrt{3}$ . Write the other zero. [CBSE 2010]

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 $2 - \sqrt{3}$ .

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Question 2:

Find the zeros of the polynomial x² + x − p(p + 1). [CBSE 2011]

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.

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Question 3:

Find the zeros of the polynomial x² − 3x − m(m + 3). [CBSE 2011]

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.

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Question 4:

If α, β are the zeros of a polynomial such that α + β = 6 and αβ = 4 the write the polynomial. [CBSE 2010]

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

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Question 5:

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

Answer:

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

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Question 6:

If 3 is a zero of the polynomial 2x² + x + k, find the value of k. [CBSE 2010]

Answer:

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

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Question 7:

If −4 is a zero of the quadratic polynomial x² − x − (2k + 2) then find the value of k.

Answer:

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

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Question 8:

If 1 is a zero of the polynomial ax² − 3(a − 1) x − 1, then find the value of a.

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
$a {(1)}^{2} - 3 (a - 1) 1 - 1 = 0 \Rightarrow a - 3 a + 3 - 1 = 0 \Rightarrow - 2 a = - 2 \Rightarrow a = 1$

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Question 9:

If −2 is a zero of the polynomial 3x² + 4x + 2k then find the value of k.

Answer:

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

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Question 10:

Write the zeros of the polynomial x² − x − 6

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.

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Question 11:

If the sum of the zeros of the quadratic polynomial kx² − 3x + 5 is 1, write the value of k.

Answer:

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

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Question 12:

If the product of the zeros of the quadratic polynomial x² − 4x + k is 3 then write the value of k.

Answer:

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

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Question 13:

If (x + a) is a factor of (2x² + 2ax + 5x + 10), find the value of a. [CBSE 2010]

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
$∴ 2 {(- a)}^{2} + 2 a (- a) + 5 (- a) + 10 = 0 \Rightarrow - 5 a + 10 = 0 \Rightarrow a = 2$

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Question 14:

If (a − b), a and (a + b) are zeros of the polynomial 2x³ − 6x² + 5x − 7, write the value of a.

Answer:

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

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Question 15:

If x³ + x² − ax + b is divisible by (x² − x), write the values of a and b.

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
${(1)}^{3} + 1^{2} - a (1) + 0 = 0 [∵ b = 0] \Rightarrow a = 2$

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Question 16:

If α and β are the zeroes of a polynomial 2x² + 7x + 5, write the value of α + β + αβ. [CBSE 2010]

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 = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- 7}{2} and α β = \frac{5}{2} Now, α + β + α β = \frac{- 7}{2} + \frac{5}{2} = - 1$

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Question 17:

State division algorithm for polynomials.

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).

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Question 18:

The sum of the zero and the product of zero of a quadratic polynomial are $\frac{- 1}{2}$ and −3 respectively, write the polynomial.

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

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Question 19:

Write the zeros of the quadratic polynomial f(x) = 6x² − 3

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 $\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}$ .

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Question 20:

Find the zeros of the quadratic polynomial $f (x) = 4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3}$

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 $- \frac{2}{\sqrt{3}} or \frac{\sqrt{3}}{4}$ .

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Question 21:

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

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 = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- (- 5)}{1} and α β = \frac{k}{1} \Rightarrow α + β = 5 and α β = \frac{k}{1}$
Solving α − β = 1 and α + β = 5, we will get
α = 3 and β = 2
Substituting these values in $α β = \frac{k}{1}$ , we will get
k = 6

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Question 22:

If α and β are the zeroes of a polynomial f(x) = 6x² + x − 2, find the value of $(\frac{α}{β} + \frac{β}{α})$

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 = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- 1}{6} and α β = - \frac{1}{3} Now, \frac{α}{β} + \frac{β}{α} = \frac{α^{2} + β^{2}}{α β} = \frac{α^{2} + β^{2} + 2 α β - 2 α β}{α β}$
$= \frac{{(α + β)}^{2} - 2 α β}{α β} = \frac{{(\frac{- 1}{6})}^{2} - 2 (- \frac{1}{3})}{- \frac{1}{3}} = \frac{\frac{1}{36} + \frac{2}{3}}{- \frac{1}{3}} = - \frac{25}{12}$

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Question 23:

If α and β are the zeroes of a polynomial f(x) = 5x² − 7x +1, find the value of $(\frac{1}{α} + \frac{1}{β})$

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 = $\frac{constant term}{coefficent of x^{2}}$
$∴ α + β = \frac{- (- 7)}{5} and α β = \frac{1}{5} \Rightarrow α + β = \frac{7}{5} and α β = \frac{1}{5} Now, \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$

$= \frac{\frac{7}{5}}{\frac{1}{5}} = 7$

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Question 24:

If α and β are the zeroes of a polynomial f(x) = x² + x − 2, find the value of $(\frac{1}{α} - \frac{1}{β})$

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 = $\frac{constant term}{coefficent of x^{2}}$

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

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

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

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Question 25:

If the zeros of the polynomial f(x) = x³ − 3x² + x + 1 are (a − b), a and (a + b), Find a and b.

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

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Question 1:

Which of the following is a polynomial?

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

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.

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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.

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Question 3:

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

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

Answer:

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

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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:

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

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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 64:

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:

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

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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:

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

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Question 8:

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:

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

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Question 9:

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:

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

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Question 10:

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) 10x² – x – 3

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
$10 x^{2} - x - 3$

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Question 11:

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:

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

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Question 12:

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:

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

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Question 13:

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:

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

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Question 14:

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:

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

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Question 15:

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:

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

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Question 16:

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:

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

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Question 17:

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:

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

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Question 18:

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:

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

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Question 19:

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:

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

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Question 20:

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:

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

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Question 21:

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:

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

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Question 22:

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:

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

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Question 23:

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:

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

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Question 24:

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:

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

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Question 25:

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:

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

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Question 26:

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:

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

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Question 27:

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:

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

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Question 28:

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

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

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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:
$(αβ + βγ + γα) = - 1$

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

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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 $= - \frac{b}{a}$
$= > α + α + 4 = \frac{- (- 8 k)}{4} = > 2 α + 4 = 2 k = > α + 2 = k = > α = (k - 2) . . . (i) Also, product of the roots, αβ = \frac{c}{a} = > α (α + 4) = \frac{9}{4} = > (k - 2) (k - 2 + 4) = \frac{9}{4} = > (k - 2) (k + 2) = \frac{9}{4} = > k^{2} - 4 = \frac{9}{4} = > 4 k^{2} - 16 = 9 = > 4 k^{2} = 25 = > k^{2} = \frac{25}{4} = > k = \frac{5}{2} (∵ k > 0)$

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Question 5:

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

Answer:

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

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

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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:

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

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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 = $x^{2} - (α + β) x + αβ$
$= x^{2} — (- 3) x + (- 10) = x^{2} + 3 x - 10$

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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 $(a - b), a and (a + b)$ .

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

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Question 9:

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

Answer:

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

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

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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:

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

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Question 12:

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

Answer:

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

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Question 13:

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

Answer:

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

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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:
$α β + β γ + γ α = - 5 and α β γ = - 1 ∴ (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}) = (\frac{β γ + α γ + α β}{α β γ}) = (\frac{- 5}{- 1}) = 5$

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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:

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

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Question 16:

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

Answer:

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

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Question 17:

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

Answer:

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

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

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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:

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

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

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Question 19:

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

Answer:

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

$∴ The quotient is 3 x^{2} - 4 x + 2$

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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:

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

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