# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials  Written by Team Trustudies
Updated at 2021-05-07

## NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.1

Q1 ) Which of the following expression are polynomials in one variable and which are not?
i)$$4x^2 - 3x + 7$$
ii)$$y^2 + \sqrt{2}$$
iii)$$3\sqrt{t} + t\sqrt{2}$$
iv)$$y + \frac{2}{y}$$
v)$$x^{10} + y^3 + t^{50}$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) $$4x^2 - 3x + 7$$ is a polynomail in one variable. As it is having only non-negative integral powers of x.

ii)$$y^2 + \sqrt{2}$$ is also a polynomial in one variable. As it is having only non-negative integral power of y.

iii)$$3\sqrt{t} + t\sqrt{2}$$ is not a polynomial. As the integral power of the variable is not a whole number.

iv)$$y + \frac{2}{y}$$ is not a polynomial. As it having a negative integral power of y.

v) $$x^{10} + y^3 + t^{50}$$ is a polynomial with three variables.

Q2 ) Write the coefficients of $$x^2$$ in each of the following :
i)$$2 + x^2 + x$$
ii)$$2 - x^2 + x^3$$
iii)$$\frac{\pi{x^2}}{2} + x$$
iv)$$\sqrt{2}x - 1$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) The coefficient of $$x^2$$ in $$2 + x^2 + x$$ is $$1$$.

ii) The coefficient of $$x^2$$ in $$2 - x^2 + x^3$$ is $$-1$$.

iii) The coefficient of $$x^2$$ in $$\frac{\pi{x^2}}{2} + x$$ is $$\frac{\pi }{2}$$.

iv) The coefficient of $$x^2$$ in $$\sqrt{2}x - 1$$ is $$0$$.

Q3 ) Give one example each of a binomial of degree 35, and of a monomial of degree 100.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

A binomial of degree 35 = $$4y^{35} + 3\sqrt{2}$$

and a monomial of degree 100 = $$9y^{100}$$

Q4 ) Write the degree of each of the following polynomials :
i)$$5x^3 + 4x^2 + 7x$$
ii)$$4 - y^2$$
iii)$$5t - \sqrt{7}$$
iv)$$3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$5x^3 + 4x^2 + 7x$$ is a polynomial with highest power of variable 3.
Hence, the degree of given polynomial is 3.

ii)$$4 - y^2$$ is a polynomial with highest power of variable 2.
Hence, the degree of given polynomial is 2.

iii)$$5t - \sqrt{7}$$ is a polynomial with highest power of variable 1.
Hence, the degree of given polynomial is 1.

iv)$$3$$ is a polynomial with highest power of variable 0.
Hence, the degree of given polynomial is 0.

Q5 ) Classify the following as linear, quadratic and cubic polynomials :
i)$$x^2 + x$$
ii)$$x - x^3$$
iii)$$y + y^2 + 4$$
iv)$$1 + x$$
v)$$3t$$
vi)$$r^2$$
vii)$$7x^3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) The degree of polynomial $$x^2 + x$$ is 2. Hence, it is a quadratic polynomial.

ii) The degree of polynomial $$x - x^3$$ is 3. Hence, it is a cubic polynomial.

iii) The degree of polynomial $$y + y^2 + 4$$ is 2. Hence, it is a quadratic polynomial.

iv) The degree of polynomial $$1 + x$$ is 1. Hence, it is a linear polynomial.

v) The degree of polynomial $$3t$$ is 1. Hence, it is a linear polynomial.

vi) The degree of polynomial $$r^2$$ is 2. Hence, it is a quadratic polynomial.

vii) The degree of polynomial $$7x^3$$ is 3. Hence, it is a cubic polynomial.

## NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.2

Q1 ) Find the value of the polynomial $$5x - 4x^2 + 3$$ at :
i) x = 0
ii) x = -1
iii) x = 2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Let f(x) = $$5x - 4x^2 + 3$$

i)The value of f(x) at x = 0 is :

$$= 5(0) - 4(0)^2 + 3$$
$$= 3$$

$$\therefore$$ f(0) = 3.

ii)The value of f(x) at x = -1 is :

$$= 5(-1) - 4(-1)^2 + 3$$
$$= -5 - 4 + 3$$
$$= -6$$

$$\therefore$$ f(-1) = -6.

iii)The value of f(x) at x = 2 is :

$$= 5(2) - 4(2)^2 + 3$$
$$= 10 - 16 + 3$$
$$= -3$$

$$\therefore$$ f(2) = -3.

Q2 ) Find p(0), p(1)and p(2) for each of the following polynomials :
i)$$p(y) = y^2 -y + 1$$
ii)$$p(t) = 2 + t + 2t^2 - t^3$$
iii)$$p(x) = x^3$$
iv)$$p(x) = (x - 1)(x + 1)$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$p(y) = y^2 - y + 1$$

Firstly, let us put x = 0,
$$\therefore p(0) = 0^2 - 0 +1$$
$$\Rightarrow p(0) = 1$$

Now, let us put x = 1,
$$\therefore p(1) = 1^2 - 1 +1$$
$$\Rightarrow p(1) = 1$$

Now, let us put x = 2,
$$\therefore p(2) = 2^2 - 2 +1$$
$$\Rightarrow p(2) = 4 - 2 +1$$
$$\Rightarrow p(2) = 3$$

ii)$$p(t) = 2 + t + 2t^2 - t^3$$

Firstly, let us put x = 0,
$$\therefore p(0) = 2 + 0 + 2{0}^2 - {0}^3$$
$$\Rightarrow p(0) = 2$$

Now, let us put x = 1,
$$\therefore p(1) = 2 + 1 + 2{1}^2 - {1}^3$$
$$\Rightarrow p(1) = 3 + 2 - 1$$
$$\Rightarrow p(1) = 4$$

Now, let us put x = 2,
$$\therefore p(2) = 2 + 2 + 2{2}^2 - {2}^3$$
$$\Rightarrow p(2) = 4 + 8 - 8$$
$$\Rightarrow p(2) = 4$$

iii)$$p(x) = x^3$$

Firstly, let us put x = 0,
$$\therefore, p(0) = {0}^3$$
$$\Rightarrow p(0) = 0$$

Now, let us put x = 1,
$$\therefore p(1) = {1}^3$$
$$\Rightarrow p(1) = 1$$

Now, let us put x = 2,
$$\therefore p(2) = {2}^3$$
$$\Rightarrow p(2) = 8$$

iv)$$p(x) = (x - 1)(x + 1)$$

Firstly, let us put x = 0,
$$\therefore p(0) = (0 - 1)(0 + 1)$$
$$\Rightarrow p(0) = -1$$

Now, let us put x = 1,
$$\therefore p(1) = (1 - 1)(1 + 1)$$
$$\Rightarrow p(1) = 0$$

Now, let us put x = 2,
$$\therefore, p(2) = (2 - 1)(2 + 1)$$
$$\Rightarrow p(2) = 1 × 3$$
$$\Rightarrow p(2) = 3$$

Q3 ) Verify whether the following are zeroes of the polynomial, indicated against them.
i)$$p(x) = 3x + 1, x = \frac{-1}{3}$$
ii)$$p(x) = 5x - \pi , x = \frac{4}{5}$$
iii)$$p(x) = x^2 - 1$$, x = 1, -1
iv)$$p(x) = (x + 1)(x - 2)$$, x = -1, 2
v)$$p(x) = x^2$$, x = 0
vi)$$p(x) = lx + m, x = \frac{-m}{l}$$
vii)$$p(x) = 3x^2 - 1$$, $$x = \frac{-1}{\sqrt{3}}$$, $$\frac{2}{\sqrt{3}}$$
viii)$$p(x) = 2x + 1, x = \frac{1}{2}$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) $$p(x) = 3x + 1$$

To check, if $$x = \frac{-1}{3}$$ is zero of given polynomial then $$p(\frac{-1}{3})$$ should be equal to 0.

$$\therefore p(\frac{-1}{3})$$
$$\Rightarrow p(\frac{-1}{3} ) = 3(\frac{-1}{3} ) + 1$$
$$\Rightarrow p(\frac{-1}{3})= -1 +1 = 0$$

$$\therefore$$ $$x = \frac{-1}{3}$$ is zero of given polynomial.

ii) $$p(x) = 5x - \pi$$

To check, if $$\frac{4}{5}$$is zero of given polynomial then $$p(\frac{4}{5} )$$ should be equal to 0.

$$\therefore p(\frac{4}{5} ) = 5(\frac{4}{5} ) - \pi$$
$$\Rightarrow p(\frac{4}{5} ) = 4 - \pi$$

$$\therefore x = \frac{4}{5}$$ is not a zero of given polynomial.

iii) $$p(x) = x^2 - 1$$

To check, if x = 1 is zero of given polynomial then p(1) should be equal to 0.

$$\therefore p(1) = 1^2 - 1$$
$$\Rightarrow p(1) = 1 - 1 = 0$$

$$\therefore x = 1$$ is a zero of given polynomial.

To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.

$$\therefore p(-1) = {-1}^2 - 1$$
$$\Rightarrow p(-1) = 1 - 1 = 0$$

$$\therefore, x = 1$$ is too, a zero of given polynomial.

iv) $$p(x) = (x + 1)(x - 2)$$

To check, if x = -1 is zero of given polynomial then p(-1) should be equal to 0.

$$\therefore p(-1) = (-1 + 1)(-1 - 2)$$
$$\Rightarrow p(-1) = 0 × -3 = 0$$

$$\therefore x = -1$$ is zero of given polynomial.

To check, if x = 2 is zero of given polynomial then p(2) should be equal to 0.

$$\therefore p(2) = (2 + 1)(2 - 2)$$
$$\Rightarrow p(2) = 3 × 0 = 0$$

Therefore, x = 2 is too, a zero of given polynomial.

v) $$p(x) = x^2$$

To check, if x = 0 is zero of given polynomial then p(0) should be equal to 0.

$$\therefore p(0) = (0^2) = 0$$

Therefore, x = 0 is zero of given polynomial.

vi) $$p(x) = lx + m$$

To check, if $$x = \frac{-m}{l}$$ is zero of given polynomial then $$p(\frac{-m}{l} )$$ should be equal to 0.

$$\therefore p(\frac{-m}{l} ) = l(\frac{-m}{l} ) + m$$
$$\Rightarrow p(\frac{-m}{l} )= -m + m = 0$$
$$\therefore, x = \frac{-m}{l}$$ is zero of given polynomial.

vii)$$p(x) = 3x^2 - 1$$

To check, if$$x = \frac{-1}{\sqrt{3}}$$ is zero of given polynomial then p($$\frac{-1}{\sqrt{3}}$$) should be equal to 0.

$$\therefore p(\frac{-1}{\sqrt{3}} ) = 3(\frac{-1}{\sqrt{3}} )^2 - 1$$
$$\Rightarrow p(\frac{-1}{\sqrt{3}} ) = 1 - 1 = 0$$

$$\therefore x = \frac{-1}{\sqrt{3}}$$ is zero of given polynomial.

To check, if $$x = \frac{2}{\sqrt{3}}$$ is zero of given polynomial then $$p(\frac{2}{\sqrt{3}}$$) should be equal to 0.

$$\therefore p(\frac{2}{\sqrt{3}} ) = 3(\frac{2}{\sqrt{3}} )^2 - 1$$
$$\Rightarrow p(\frac{2}{\sqrt{3}} ) = 3$$

$$\therefore, x = \frac{2}{\sqrt{3}}$$ is not a zero of given polynomial.

viii)$$p(x) = 2x + 1$$

To check, if $$x = \frac{1}{2}$$ is zero of given polynomial then$$p(\frac{1}{2} )$$ should be equal to 0.

$$\therefore p(\frac{1}{2} ) = 2(\frac{1}{2} ) + 1$$
$$\Rightarrow p(\frac{1}{2} ) = 1 + 1 = 2$$

$$\therefore x = \frac{1}{2}$$ is not a zero of given polynomial.

Q4 ) Find the Zero of the polynomial in each of the following cases.
i)$$p(x) = x + 5$$
ii)$$p(x) = x - 5$$
iii)$$p(x) = 2x + 5$$
iv)$$p(x) = 3x - 2$$
v)$$p(x) = 3x$$
vi)$$p(x) = ax, a \ne 0$$
vii)$$p(x) = cx + d ,c \ne 0$$, c,d, are real numbers.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) We have, $$p(x) = x + 5$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow x + 5 = 0$$
$$\Rightarrow x = -5$$

Hence, -5 is the zero of given ploynomial.

ii)We have, $$p(x) = x - 5$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow x - 5 = 0$$
$$\Rightarrow x = 5$$

Hence, 5 is the zero of given ploynomial.

iii)We have, $$p(x) = 2x + 5$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore, p(x) = 0$$
$$\Rightarrow 2x + 5 = 0$$
$$\Rightarrow x = \frac{-5}{2}$$

Hence,$$\frac{-5}{2}$$ is the zero of given ploynomial.

iv) We have, $$p(x) = 3x - 2$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow 3x - 2 = 0$$
$$\Rightarrow x = \frac{2}{3}$$

Hence, $$\frac{2}{3}$$ is the zero of given ploynomial.

v) We have, $$p(x) = 3x$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow 3x = 0$$
$$\Rightarrow x = 0$$

Hence, 0 is the zero of given ploynomial.

vi) We have, $$p(x) = ax$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow ax = 0$$
$$\Rightarrow x = 0, a \ne 0$$

Hence, 0 is the zero of given ploynomial.

vii) We have, $$p(x) = cx + d$$

We also know that, if p(x) = 0 then x is the Zero of the polynomial.

$$\therefore p(x) = 0$$
$$\Rightarrow cx + d = 0$$
$$\Rightarrow x = \frac{-d}{c} , c \ne 0$$

Hence, $$\frac{-d}{c}$$ is the zero of given ploynomial.

## NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.3

Q1 ) Find the remainder when $$x^3 + 3x^2 + 3x + 1$$ is divided by
i)$$x + 1$$
ii)$$x - \frac{1}{2}$$
iii)$$x$$
iv)$$x + \pi$$
v)$$5 + 2x$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Let $$p(x) = x^3 + 3x^2 + 3x + 1$$

i)The zero of $$x + 1$$ is -1
[$$\because x+1= 0 \Rightarrow x = -1$$ ]

$$\therefore p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1$$
$$\Rightarrow p(-1) = -1 + 3 - 3 +1$$
$$\Rightarrow p(-1) = 0$$

Hence, By remainder theorem, required remainder = 0.

ii)The zero of $$x - \frac{1}{2}$$ is x = $$\frac{1}{2}$$
[$$\because (x - \frac{1}{2} = 0 )$$]

$$\therefore p(\frac{1}{2} ) = (\frac{1}{2} )^3 + 3(\frac{1}{2} )^2 + 3(\frac{1}{2} )+ 1$$
$$\Rightarrow p(\frac{1}{2}) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} +1$$
$$\Rightarrow p(\frac{1}{2} ) = \frac{27}{8}$$

Hence, By remainder theorem, required remainder = $$\frac{27}{8}$$ .

iii)The zero of $$x$$ is x = 0 [$$\because (x = 0)$$ ]

$$\therefore p(0) = (0)^3 + 3(0)^2 + 3(0) + 1$$
$$\Rightarrow p(-1) = 1$$

Hence, By remainder theorem, required remainder = 1.

iv) The zero of $$x + \pi$$ is $$x = -\pi$$ [$$\because (x + \pi = 0)$$]

$$\therefore p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1$$

$$\Rightarrow p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1$$

Hence, By remainder theorem, required remainder$$= -\pi^3 + 3\pi^2 - 3\pi + 1$$.

v) The zero of $$5 + 2x$$ is x = $$\frac{-5}{2}$$ [$$\because (5 + 2x = 0)$$]

$$\therefore p(\frac{-5}{2} ) = (\frac{-5}{2})^3 + 3(\frac{-5}{2})^2 + 3(\frac{-5}{2} )+ 1$$
$$\Rightarrow p(\frac{-5}{2} ) = \frac{-125}{8} + \frac{75}{4} - \frac{15}{2} + 1$$
$$\Rightarrow p(\frac{-5}{2} ) = \frac{-27}{8}$$

Hence, By remainder theorem, required remainder = $$\frac{-27}{8}$$.

Q2 ) Find the remainder when $$x^3 - ax^2 + 6x - a$$ is divided by x - a.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Let $$p(x) = x^3 - ax^2 + 6x - a$$
The zero of $$x - a$$ is x = a
[$$\because (x - a = 0)$$ ]

$$\therefore p(a) = (a)^3 - a(a)^2 + 6(a) - a$$
$$\Rightarrow p(a) = a^3 - a^3 + 6a - a$$
$$\Rightarrow p(a) = 5a$$

Hence, By remainder theorem, required remainder = 5a.

Q3 ) Check whether 7 + 3x is a factor of $$3x^3 + 7x$$.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Let $$p(x) = 3x^3 + 7x$$

To check, whether 7 + 3x is a factor of $$3x^3 + 7x$$.
By factor theorem, $$7 + 3x = 0$$
$$\therefore x = \frac{-7}{3}$$

$$\therefore p(\frac{-7}{3} ) = 3(\frac{-7}{3} )^3 + 7(\frac{-7}{3} )$$
$$\Rightarrow p(\frac{-7}{3} ) = - 3×(\frac{343}{27} ) - \frac{49}{3}$$
$$\Rightarrow p(\frac{-7}{3} ) = \frac{-1470}{27}$$
$$\Rightarrow p(\frac{-7}{3} ) = \frac{-490}{9}$$

Hence, 7 + 3x is not the factor of $$3x^3 + 7x$$ since the remainder is not zero.

## NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.4

Q1 ) Determine which of the following polynomials has a factor :
i)$$x^3 + x^2 + x + 1$$
ii)$$x^4 + x^3 + x^2 + x + 1$$
iii)$$x^4 + 3x^3 + 3x^2 + x + 1$$
iv)$$x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

The zero of x + 1 is –1.

i)Let $$p(x) = x^3 + x^2 + x + 1$$
$$\therefore$$ To check, whether x + 1 is a factor of $$x^3 + x^2 + x + 1$$.

By factor theorem,
$$\therefore$$ $$p(–1) = {–1}^3 + {–1}^2 + {-1} + 1$$
$$\Rightarrow$$ $$p(–1) = –1 + 1 -1 + 1$$
$$\Rightarrow$$ $$p(-1) = 0$$

Hence, x + 1 is the factor of $$x^3 + x^2 + x + 1$$

ii)Let $$p(x) = x^4 + x^3 + x^2 + x + 1$$
$$\therefore$$ To check, whether x + 1 is a factor of $$x^4 + x^3 + x^2 + x + 1$$.

By factor theorem,
$$\therefore p(–1) = (-1)^4 + (–1)^3 + (–1)^2 + (-1) + 1$$
$$\Rightarrow$$ $$p(–1) = 1 – 1 + 1 - 1 + 1$$
$$\Rightarrow$$ $$p(-1) = 1$$

Hence, x + 1 is not the factor of $$x^4 + x^3 + x^2 + x + 1$$

iii) Let $$p(x) = x^4 + 3x^3 + 3x^2 + x + 1$$
$$\therefore$$ To check, whether x + 1 is a factor of $$x^4 + 3x^3 + 3x^2 + x + 1$$.

By factor theorem,
$$\therefore p(–1) = {-1}^4 + 3{–1}^3 + 3{–1}^2 + {-1} + 1$$
$$\Rightarrow p(–1) = 1 – 3 + 3 - 1 + 1$$
$$\Rightarrow p(-1) = 1$$

Hence, x + 1 is not the factor of $$x^4 + 3x^3 + 3x^2 + x + 1$$

iv) Let $$p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$$
$$\therefore$$ To check, whether x + 1 is a factor of $$x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$$.

By factor theorem,
$$\therefore p(–1) = {–1}^3 - {–1}^2 - (2 + \sqrt{2}){-1} + \sqrt{2}$$
$$\Rightarrow p(–1) = – 1 - 1 + 2 + \sqrt{2} + \sqrt{2}$$
$$\Rightarrow p(-1) = 2\sqrt{2}$$

Hence, x + 1 is not the factor of $$x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$$

Q2 ) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
i) $$p(x) = 2x^3 + x^2 - 2x - 1$$, $$g(x) = x + 1$$
ii)$$p(x) = x^3 + 3x^2 + 3x + 1$$, $$g(x) = x + 2$$
iii)$$p(x) = x^3 - 4x^2 + x + 6$$, $$g(x) = x - 3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)The zero of g(x) = x + 1 is –1.

Let $$p(x) = 2x^3 + x^2 - 2x - 1$$

so, To check, whether x + 1 is a factor of $$2x^3 + x^2 - 2x - 1$$.

By factor theorem,
$$\therefore p(–1) = 2{–1}^3 + {–1}^2 - 2{-1} - 1$$
$$\Rightarrow p(–1) = –2 + 1 + 2 - 1$$
$$\Rightarrow p(-1) = 0$$

Hence, g(x) = x + 1 is the factor of $$2x^3 + x^2 - 2x - 1$$

ii)The zero of $$g(x) = x + 2$$ is –2.

Let $$p(x) = x^3 + 3x^2 + 3x + 1$$

So, To check, whether x + 2 is a factor of $$x^3 + 3x^2 + 3x + 1$$.

By factor theorem,
$$\therefore p(–2) = (–2)^3 + 3(–2)^2 + 3(–2) + 1$$
$$\Rightarrow p(–2) = –8 + 12 - 6 + 1$$
$$\Rightarrow p(-2) = -1$$

Hence, $$g(x) = x + 2$$ is not the factor of $$x^3 + 3x^2 + 3x + 1$$

iii)The zero of $$g(x) = x - 3$$ is 3.

Let $$p(x) = x^3 - 4x^2 + x + 6$$

So, To check, whether x - 3 is a factor of $$p(x) = x^3 - 4x^2 + x + 6$$.

By factor theorem,

$$\therefore p(3) = (3)^3 - 4(3)^2 + (3) + 6$$
$$\Rightarrow$$ $$p(3) = 27 - 36 + 3 + 6$$
$$\Rightarrow$$ $$p(3) = 0$$

Hence, $$g(x) = x - 3$$ is the factor of $$p(x) = x^3 - 4x^2 + x + 6$$

Q3 ) Find the value of k, if x – 1 is a factor of p(x) in each of the following cases :
i)$$p(x) = x^2 + x + k$$
ii)$$p(x) = 2x^2 + kx + \sqrt{2}$$
iii)$$p(x) = kx^2 - \sqrt{2}x + 1$$
iv)$$p(x) = kx^2 - 3x + k$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

The zero of x - 1 is 1.

i)So, as x - 1 is a factor of $$p(x) = x^2 + x + k$$,
$$\therefore p(1) = 0$$
$$\Rightarrow 0 = (1)^2 + 1 + k$$
$$\Rightarrow 0 = 1 + 1 + k$$
$$\Rightarrow 0 = 2 + k$$
$$\Rightarrow k = -2$$

ii)So, as x - 1 is a factor of $$p(x) = 2x^2 + kx + \sqrt{2}$$,
$$\therefore p(1) = 0$$
$$\Rightarrow 0 = 2(1)^2 + k(1) + \sqrt{2}$$
$$\Rightarrow 0 = 2 + k + \sqrt{2}$$
$$\Rightarrow k = -2 - \sqrt{2}$$

iii)So, as x - 1 is a factor of $$p(x) = kx^2 - \sqrt{2}x + 1$$,
$$\therefore p(1) = 0$$
$$\Rightarrow 0 = k(1)^2 - ( \sqrt{2})1 + 1$$
$$\Rightarrow 0 = k - \sqrt{2} + 1$$
$$\Rightarrow k = \sqrt{2} - 1$$

iv)So, as x - 1 is a factor of $$p(x) = kx^2 - 3x + k$$,
$$\therefore p(1) = 0$$
$$\Rightarrow 0 = k(1)^2 - 3(1) + k$$
$$\Rightarrow 0 = k - 3 + k$$
$$\Rightarrow 3 = 2k$$
$$\Rightarrow k = \frac{3}{2}$$

Q4 ) Factorize :
i)$$12x^2 - 7x + 1$$
ii)$$2x^2 + 7x + 3$$
iii)$$6x^2 + 5x - 6$$
iv)$$3x^2 - x - 4$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$12x^2 - 7x + 1$$
$$= 12x^2 - 4x -3x + 1$$
(by splitting middle term)

$$= 4x(3x - 1) - 1(3x - 1)$$
$$=(3x - 1)(4x - 1)$$

ii) $$2x^2 + 7x + 3$$
$$= 2x^2 + 6x + x + 3$$...
(by splitting middle term)

$$= 2x(x + 3) + 1(x + 3)$$
$$=(x + 3)(2x + 1)$$

iii) $$6x^2 + 5x - 6$$
$$= 6x^2 + 9x - 4x - 6$$
(by splitting middle term)

$$= 3x(2x + 3) - 2(2x + 3)$$
$$=(2x + 3)(3x - 2)$$

iv) $$3x^2 - x - 4$$
$$= 3x^2 - 4x + 3x - 4$$
(by splitting middle term)

$$= x(3x - 4) + 1(3x - 4)$$
$$=(3x - 4)(x + 1)$$

Q5 ) Factorize :
i)$$x^3 - 2x^2 - x + 2$$
ii)$$x^3 - 3x^2 -9x - 5$$
iii)$$x^3 + 13x^2 + 32x + 20$$
iv)$$2y^3 + y^2 - 2y - 1$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)Let $$p(x) = x^3 - 2x^2 - x + 2$$

Here, constant term is 2. So, finding the factors of +2, we get, ±1 and ±2.

Now, $$p(1) = (1)^3 - 2(1)^2 - (1) + 2$$
$$= 1 - 2 - 1 + 2$$
= 0

Therefore, by trial method, we find that,
p(1) = 0.

Hence, x - 1 is a factor of p(x).

$$\therefore x^3 - 2x^2 - x + 2$$
$$= x^3 - x^2 - x^2 + x - 2x + 2$$
$$= x^2(x - 1) -x(x - 1) - 2(x - 1)$$
$$= (x - 1)(x^2 - x - 2)$$
now, splitting the middle term,
$$= (x - 1)(x^2 - 2x + x - 2)$$
$$= (x - 1)[x(x - 2) + 1(x - 2)]$$
$$= (x - 1)(x - 2)(x + 1)$$

ii)Let $$p(x) = x^3 - 3x^2 -9x - 5$$
Here, constant term is -5. So, finding the factors of -5, we get, 1 and -5 or vice versa.

Now, $$p(5) = (5)^3 - 3(5)^2 - 9(5) - 5$$
$$= 125 - 75 - 45 - 5$$
= 0

Therefore, by trial method, we find that, p(5) = 0.

Hence, x - 5 is a factor of p(x).

$$\therefore x^3 - 3x^2 -9x - 5$$
$$= x^3 - 5x^2 + 2x^2 - 10x + x - 5$$
$$= x^2(x - 5) + 2x(x - 5) + 1(x - 5)$$
$$= (x - 5)(x^2 + 2x + 1)$$
now, splitting the middle term,
$$= (x - 5)(x^2 + x + x + 1)$$
$$= (x - 5)[x(x + 1) + 1(x + 1)]$$
$$= (x - 5)(x + 1)^2$$

iii)Let $$p(x) = x^3 + 13x^2 + 32x + 20$$

Here, constant term is +20. So, finding the factors of 20, we get, ±1, ±2, ±4 and ±5 or vice versa.

Now, $$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$$
$$= -1 + 13 - 32 + 20$$
= 0
Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

$$\therefore x^3 + 13x^2 + 32x + 20$$
$$= x^3 + 2x^2 + 11x^2 + 22x + 10x + 20$$
$$= x^2(x + 2) + 11x(x + 2) + 10(x + 2)$$
$$= (x + 2)(x^2 + 11x + 10)$$
now, splitting the middle term,
$$= (x + 2)(x^2 + 10x + x + 10)$$
$$= (x + 2)[x(x + 10) + 1(x + 10)]$$
$$= (x + 2)(x + 1)(x + 10)$$

iv)Let $$p(x) = 2y^3 + y^2 - 2y - 1$$

Here, constant term is -1. So, finding the factors of -1, we get, -1.

Now, $$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1$$
$$= -2 + 1 + 2 - 1$$
= 0

Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

$$\therefore 2y^3 + y^2 - 2y - 1$$
$$= 2y^3 + 2y^2 - y^2 - y - y -1$$
$$= 2y^2(y + 1) - y(x + 1) - 1(y + 1)$$
$$= (y + 1)(2y^2 - y - 1)$$
now, splitting the middle term,
$$= (y + 1)(2y^2 - 2y + y - 1)$$
$$= (y + 1)[2y(y - 1) + 1(y - 1)]$$
$$= (y - 1)(y + 1)(2y + 1)$$

## NCERT solutions for class 9 Maths Chapter 2 Polynomials Exercise 2.5

Q1 ) Use suitable identities to find the following products :
i)$$(x + 4)(x + 10)$$
ii)$$(x + 8)(x - 10)$$
iii)$$(3x - 4)(3x - 5)$$
iv)$$(y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )$$
v)$$(3 - 2x)(3 + 2x)$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$(x + 4)(x + 10)$$
$$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$$

We have,
$$=(x + 4)(x + 10)$$
$$= x^2 + (4 + 10)x + (4)(10)$$
$$= x^2 + 14x + 40$$

ii) $$(x + 8)(x - 10)$$
$$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$$
We have,
$$= (x + 8)(x - 10)$$
$$= x^2 + (8 - 10)x + (8)(-10)$$
$$= x^2 - 2x - 80$$

iii) $$(3x + 4)(3x - 5)$$
$$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$$
We have,
$$(3x + 4)(3x - 5)$$
$$= (3x)^2 + (4 - 5)x + (4)(5)$$
$$= 9x^2 - x + 20$$

iv) $$(y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )$$
$$[\because (a^2 - b^2) = (a + b)(a - b)]$$
We have,
$$(y^2 + \frac{3}{2} )(y^2 - \frac{3}{2} )$$
$$= (y^2)^2 - (\frac{3}{2} )^2$$
$$= y^4 - \frac{9}{4}$$

v) $$(3 - 2x)(3 + 2x)$$
$$[\because (a^2 - b^2) = (a + b)(a - b)]$$
We have,
$$(3 - 2x)(3 + 2x)$$
$$= (3)^2 - (2x)^2)$$
$$= 9 - 4x^2$$

Q2 ) Evaluate the following products without multiplying directly
i)103 × 107
ii)95 × 96
iii)104 × 96

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)103 × 107 = (100 + 3)(100 + 7)
$$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$$

We have,
$$(100 + 3)(100 + 7)$$
$$= {100}^2 + (3 + 7)100 + (3)(7)$$
$$= 10000 + 1000 + 21$$
$$= 11021$$

ii)95 × 96 = (100 - 5)(100 - 4)
$$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$$

We have,
$$(100 - 5)(100 - 4)$$
$$= {100}^2 - (5 + 4)100 + (4)(5)$$
$$= 10000 - 900 + 20$$
$$= 9120$$

iii)104 × 96 = (100 + 4)(100 - 4)
$$[\because (a^2 - b^2) = (a + b)(a - b)]$$

We have,
$$(100 + 4)(100 - 4)$$
$$= {100}^2 - {4}^2$$
$$= 10000 - 16$$
$$= 9984$$

Q3 ) Factorize the following using appropriate identities :
i)$$9x^2 + 6xy + y^2$$
ii)$$4y^2 - 4y + 1$$
iii)$$x^2 - \frac{y^2}{100}$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i) $$9x^2 + 6xy + y^2$$
$$= {3x}^2 + 2(3x)(y) + {y}^2$$
$$[\because (a + b)^2 = a^2 + 2ab + b^2$$
$$= (3x + y)^2 ]$$

Therefore, the factors are (3x + y)(3x + y).

ii) $$4y^2 - 4y + 1$$
$$= {2y}^2 - 2(2y)(1) + {1}^2$$
$$[\because (a + b)^2 = a^2 + 2ab + b^2$$
$$= (2y - 1)^2 ]$$

Therefore, the factors are (2y - 1)(2y - 1).

iii)$$x^2 - \frac{y^2}{100}$$
$$= (x +\frac{y}{10} )(x - \frac{y}{10} )$$
$$[\because (a^2 - b^2) = (a + b)(a - b))]$$
$$=(x + \frac{y}{10} )(x - \frac{y}{10} )$$

Therefore, the factors are $$(x + \frac{y}{10} )(x - \frac{y}{10} ).$$

Q4 ) Factorize the following using appropriate identities :
i)$$(x + 2y + 4z)^2$$
ii)$$(2x - y + z)^2$$
iii)$$(-2x + 3y + 2z)^2$$
iv)$$(3a - 7b - c)^2$$
v)$$(-2x + 5y - 3z)^2$$
vi)$$[\frac{1}{4} a - \frac{1}{2} b + 1]^2$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$(x + 2y + 4z)^2$$
$$= (x)^2 + (2y)^2 +(4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)$$
$$[\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz$$

ii)$$(2x - y + z)^2$$
$$= (2x)^2 + (-y)^2 + (z)^2 - 2(2x)(y) - 2(y)(z) + 2(2x)(z)$$
$$[ \because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$$

iii)$$(-2x + 3y + 2z)^2$$
$$= (-2x)^2 + (3y)^2 +(2z)^2 - 2(2x)(3y) + 2(3y)(2z) - 2(2x)(2z)$$
$$[\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz$$

iv)$$(3a - 7b - c)^2$$
$$= (3a)^2 + (-7b)^2 + (-c)^2 - 2(3a)(7b) + 2(-7b)(-c) - 2(3a)(c)$$
$$[\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac$$

v)$$(-2x + 5y - 3z)^2$$
$$= (-2x)^2 + (5y)^2 +(-3z)^2 - 2(2x)(5y) - 2(5y)(3z) + 2(2x)(3z)$$
$$[\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz$$

vi)$$[\frac{1}{4} a - \frac{1}{2} b + 1]^2$$
$$= (\frac{1}{4} a )^2 + (\frac{-1}{2} b )^2 + (1)^2 - 2(\frac{1}{4} a )(\frac{1}{2} b ) - 2(\frac{-1}{2} b )(1) + 2(\frac{1}{4} a )(1)$$
$$[\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]$$
$$= \frac{a^2}{16} + \frac{b^2}{4} + 1 - \frac{ab}{4} - b + \frac{a}{2}$$

Q5 ) Factorize :
i)$$4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$$
ii)$$2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz$$
$$= (2x)^2 + (3y)^2 - (4z)^2 + 2(2x)(3y) - 2(3y)(4z) - 2(2x)(4z)$$
$$=(2x + 3y - 4z)^2$$

ii)$$2x^2 + y^2 + 8z^2 - 2\sqrt{2}xy + 4\sqrt{2}yz - 8xz$$
= $$-\sqrt{2}x)^2 + (y)^2 + (2\sqrt{2}z)^2 - 2(\sqrt{2}x)(y) + 2(y)(2\sqrt{2}z) - 2(\sqrt{2}x)(2\sqrt{2}z)$$
$$=(-\sqrt{2}x + y + 2\sqrt{2}z)^2$$

Q6 ) Write the following cubes in expanded form.
i)$$(2x + 1)^3$$
ii)$$(2a - 3b)^3$$
iii)$$[\frac{3x}{2} + 1]^3$$
iv)$$(x - \frac{2y}{3} )^3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$(2x + 1)^3$$
$$= (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= 8x^3 + 1 + 6x(2x + 1))$$
$$= 8x^3 + 1 + 12x^2 + 6x$$
$$= 8x^3 + 12x^2 + 6x + 1$$

ii)$$(2a - 3b)^3$$
$$= (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= 8a^3 - 27b^3 - 18ab(2a - 3b))$$
$$= 8a^3 - 27b^3 - 36a^2b + 54ab^2)$$
$$= 8a^3 - 36a^2b + 54ab^2 - 27b^3$$

iii)$$[\frac{3x}{2} + 1]^3$$
$$= (\frac{3x}{2} )^3 + (1)^3 + 3(\frac{3x}{2} )(1)(\frac{3x}{2} + 1)$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= \frac{27x^3}{8} + 1 + \frac{9x}{2} (\frac{3x}{2} + 1))$$
$$= \frac{27x^3}{8} + 1 + \frac{27x^2}{4} + \frac{9x}{2}$$
$$= \frac{27x^3}{8} + \frac{27x^2}{4} + \frac{9x}{2} + 1$$

iv)$$(x - \frac{2y}{3} )^3 = (x)^3 - (\frac{2y}{3} )^3 - 3(x)(\frac{2y}{3} )(x - \frac{2y}{3} )$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= x^3 - \frac{8y^3}{27} - 2xy(x - \frac{2y}{3} )$$
$$= x^3 - \frac{8y^3}{27} - 2x^2y + \frac{4xy^2}{3}$$
$$= x^3 - 2x^2y + \frac{4xy^2}{3} - \frac{8y^3}{27}$$

Q7 ) Evaluate the following using suitable identities.
i)$${99}^3$$
ii)$${102}^3$$
iii)$${998}^3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$${99}^3$$
$$= (100 - 1)^3$$
$$= (100)^3 - (1)^3 - 3(100)(1)(100 - 1)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= 1000000 - 1 - 300(99)$$
$$= 1000000 - 1 - 2700)$$
$$= 970299$$

ii)$${102}^3$$
$$= (100 + 2)^3$$
$$= (100)^3 + (2)^3 + 3(100)(2)(100 + 2)$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= 1000000 + 8 + 600(102)$$
$$= 1000000 + 8 + 61200)$$
$$= 1061208$$

iii)$${998}^3$$
$$= (1000 - 2)^3$$
$$= (1000)^3 - (2)^3 - 3(1000)(2)(1000 - 2)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= 1000000000 - 8 - 6000(998)$$
$$= 1000000000 - 1 - 5988000)$$
$$= 994011992$$

Q8 ) Factorize each of the following :
i)$$8a^3 + b^3 + 12a^2b + 6ab^2$$
ii)$$8a^3 - b^3 - 12a^2b + 6ab^2$$
iii)$$27 - 125a^3 - 135a + 225a^2$$
iv)$$64a^3 - 27b^3 - 144a^2b + 108ab^2$$
v)$$27p^3 - \frac{1}{216} - \frac{9p^2}{2} + \frac{p}{4}$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$8a^3 + b^3 + 12a^2b + 6ab^2$$
$$= (2a)^3 + b^3 +3(2a)(b)(2a + b))$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= (2a)^3 + b^3$$

ii)$$8a^3 - b^3 - 12a^2b + 6ab^2$$
$$= (2a)^3 - b^3 -3(2a)(b)(2a - b)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= (2a)^3 - b^3$$

iii)$$27 - 125a^3 - 135a + 225a^2$$
$$= (3)^3 - (5a)^3 -3(3)(5a)(3 - 5a)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= (3)^3 - 5a^3$$

iv)$$64a^3 - 27b^3 - 144a^2b + 108ab^2$$
$$= (4a)^3 - (3b)^3 -3(4a)(4b)(4a - 3b))$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= (4a)^3 - (3b)^3$$

v)$$(27p^3 - \frac{1}{216} - \frac{9p^2}{2} + \frac{p}{4}$$
$$= (3p)^3 - (\frac{1}{6} )^3 -3(3p)(\frac{1}{6})(3p - 1/6))$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= (3p)^3 - (\frac{1}{6})^3$$

Q9 ) Verify :
i)$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$
ii)$$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)We know that,
$$(x + y)^3 = x^3 + y^3 + 3xy(x + y)$$
$$\Rightarrow x^3 + y^3 = (x + y)^3 - 3xy(x + y)$$
$$= (x + y)[(x + y)^2 - 3xy]$$
$$= (x + y)[(x^2 + y^2 +2xy - 3xy]$$
$$= (x + y)[(x^2 + y^2 - xy]$$
$$= R.H.S$$
Hence, Proved.

ii)We know that,
$$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$$
$$\Rightarrow x^3 - y^3 = (x - y)^3 + 3xy(x - y)$$
$$= (x - y)[(x - y)^2 + 3xy]$$
$$= (x - y)[(x^2 + y^2 - 2xy + 3xy]$$
$$= (x - y)[(x^2 + y^2 + xy]$$
$$= R.H.S$$
Hence, Proved.

Q10 ) Factorize each of the following :
i)$$27y^3 + 125z^3$$
ii)$$64m^3 + 343n^3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)$$27y^3 + 125z^3$$
$$= (3y)^3 + (5z)^3$$
$$= (3y + 5z)[((3y)^2 + (5y)^2 - (3y)(5z)]$$
$$= (3y + 5z)[(9y^2 - 15yz + 25z^2)]$$

ii)$$64m^3 + 343n^3$$
$$= (4m)^3 + (7n)^3$$
$$= (4m + 7n)[((4m)^2 + (7n)^2 - (4m)(7n)]$$
$$= (4m + 7n)[(16m^2 - 48mn + 49n^2)]$$

Q11 ) Factorize $$27x^3 + y^3 + z^3 - 9xyz$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

$$27x^3 + y^3 + z^3 - 9xyz$$
$$= (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$$
$$= (3x + y +z)((3x)^2 + y^2 + z^2 - 3xy - yz - z(3x)$$
$$[ \because a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) ]$$
We get,
$$= (3x + y +z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$$

Q12 ) Verify that $$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

We know that, $$x^3 + y^3 + z^3 - 3xyz$$
$$= (x + y + z)[x^2 + y^2 + z^2 - xy - yz - zx]$$
$$= \frac{1}{2} (x + y + z)[2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx]$$
$$= \frac{1}{2} (x + y + z)[x^2 + x^2 + y^2 + y^2 + z^2 + z^2 - 2xy - 2yz - 2zx]$$
$$= \frac{1}{2} (x + y + z)[x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + z^2 + x^2 - 2zx]$$
$$= \frac{1}{2} (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]$$

Q13 ) If show that x + y + z = 0, show that $$x^3 + y^3 + z^3 = 3xyz$$.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

We know that,
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$
by using identity $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
As it is given that (x + y + z = 0)
$$x^3 + y^3 + z^3 - 3xyz = (0)(x^2 + y^2 + z^2 - xy - yz - zx)$$
$$\Rightarrow x^3 + y^3 + z^3 = 3xyz$$
Hence, proved.

Q14 ) Without actually calculating the cubes, find the value of each of the following :
i)$$(-12)^3 + (5)^3 + (7))^3$$
ii)$$(28)^3 + (-15)^3 + (-13)^3$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

We know that,
$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$$

Also, if x + y + z = 0 then, $$x^3 + y^3 + z^3 = 3xyz$$

i) Here, -12 + 7 + 5 = 0

$$\therefore (-12)^3 + (5)^3 + (7))^3$$
$$= 3(-12)(7)(5)$$
$$= -1260$$

ii)Here, 28 - 15 - 13 = 0
$$\therefore (28)^3 + (-15)^3 + (-13))^3$$
$$= 3(-13)(28)(-15)$$
$$= 16380$$

Q15 ) Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
i)Area $$25a^2 - 35a + 12$$
ii)Area $$35y^2 + 13y - 12$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)We have area of rectangle $$= 25a^2 - 35a + 12$$
$$= 25a^2 - 20a - 15a + 12$$
$$= 5a(5a - 4) - 3(5a - 4)$$
$$= (5a - 4)(5a - 3)$$
One possible answers : Length = 5a - 3, Breadth = 5a - 4.

ii)We have area of rectangle $$= 35y^2 + 13y - 12$$
$$= 35y^2 - 15y + 28y - 12$$
$$= 5y(7y - 3) + 4(7y - 3)$$
$$= (7y - 3)(5y + 4)$$
One possible answers : Length = 7y - 3, Breadth = 5y + 4.

Q16 ) What are the possible expressions for the dimensions of the cuboids whose volumes are given below ?
i)Volume $$3x^2 - 12x$$
ii)Volume $$12ky^2 + 8ky - 20k$$

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

i)We have Volume of cuboid $$3x^2 - 12x = 3x(x - 4)$$
One possible expressions for the dimensions of the cuboid is 3, x and x-4.

ii)We have Volume of cuboid $$12ky^2 + 8ky - 20k$$
$$= 12ky^2 + 20ky - 12ky - 20k$$
$$= 4ky(3y - 5) - 4k(3y - 5)$$
$$= (3y - 5)(4ky - 4k)$$
$$= (3y - 5)4k(y - 1)$$
One possible expressions for the dimensions of the cuboid is 4k, 3y - 5 and y - 1.

##### FAQs Related to NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
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