# NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials - A polynomial is an algebraic expression which consists of variables and coefficient with operations such as additions, subtraction, multiplication, and non-negative exponents. In this particular chapter, you will learn the operations of two or more polynomials. NCERT solutions for class 9 maths chapter 2 Polynomials are there to help you while solving the problems related to this particular chapter. NCERT class 9 Polynomials, introduces a lot of important concepts that will be helpful for those students who are targeting exams like JEE, CAT, SSC, etc.

It is an important topic in maths that comes under the algebra unit which holds 20 marks in the CBSE class 9 maths final exams. In this particular chapter, you will study the definition of a polynomial, zeroes, coefficient, degrees, and terms of a polynomial, type of a polynomial. You will also study the remainder and factor theorems and the factorization of polynomials. In Polynomials, there are a total of 5 exercises that comprise of a total of 33 questions. Solutions of NCERT class 9 maths chapter 2 Polynomials will cover the detailed solution to each and every question present in the practice exercises including optional exercises. CBSE NCERT solutions for class 9 maths chapter 2 Polynomials can also be used while doing homework. It can be a good tool for the class 9 students as it is designed in such a manner so that a student can fetch the maximum marks available for the particular question. NCERT solutions are also available for other classes and subjects which can be downloaded by clicking on the link given.

## NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.1

### Q1 (i) Is the following expression polynomial in one variable? State reasons for your answer. $4x^2 - 3x + 7$

YES
Given polynomial $4x^2 - 3x + 7$ has only one variable which is x

YES
Given polynomial has only one variable which is y

NO
Because we can observe that the exponent of variable t in term $3\sqrt t$ is  $\frac{1}{2}$  which is not a whole number.
Therefore this expression is not a polynomial.

NO
Because we can observe that the exponent of variable y in term $\dpi{100} \frac{2}{y}$ is $-1$ which is not a whole number. Therefore this expression is not a polynomial.

NO
Because in the given polynomial $x^{10} + y^3 + t^{50}$  there are 3 variables which are x, y, t. That's why this is polynomial in three variable not in one variable.

Coefficient of $x^2$ in polynomial  $2 + x^2 +x$  is  1.

Coefficient of $x^2$ in  polynomial  $2 - x^2 + x^3$  is  -1.

Coefficient of  $x^2$  in polynomial  $\frac{\pi}{2}x^2 + x$  is  $\frac{\pi}{2}$

Coefficient of  $x^2$  in polynomial  $\sqrt2 x - 1$   is  0

Degree of a polynomial is the highest power of the variable in the polynomial.
In binomial, there are two terms
Therefore,  binomial of degree 35 is
Eg:-  $x^{35}+1$
In monomial, there is only one term in it.
Therefore, monomial of degree 100 can be written as $y^{100}$

Degree of a polynomial is the highest power of the variable in the polynomial.
Therefore, the degree of polynomial  $5x^3 + 4x^2 + 7x$  is  3.

Degree of a polynomial is the highest power of the variable in the polynomial.

Therefore, the degree of polynomial  $4 - y^2$  is  2.

Degree of a polynomial is the highest power of the variable in the polynomial.

Therefore, the degree of polynomial  $5t - \sqrt7$  is  1

Degree of a polynomial is the highest power of the variable in the polynomial.

In this case, only a constant value 3 is there and the degree of a constant polynomial is always 0.

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $x^2+x$  with degree 2

Therefore, it is a quadratic polynomial.

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $x - x^3$  with degree 3

Therefore, it is a cubic polynomial

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $y + y^2 + 4$  with degree 2

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $1 +x$  with degree 1

Therefore, it is linear polynomial

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $3t$  with degree 1

Therefore, it is linear polynomial

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $r^2$  with degree 2

Linear polynomial, quadratic polynomial, and cubic polynomial has its degrees as 1, 2, and 3 respectively

Given polynomial is  $7x^3$  with degree 3

Therefore, it is a cubic polynomial

## NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.2

### Q1 (i) Find the value of the polynomial $5x - 4x^2 +3$ at $x = 0$

Given polynomial is  $5x - 4x^2 +3$

Now, at $x = 0$  value is

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

Therefore, value of polynomial $5x - 4x^2 +3$ at x = 0 is  3

Given polynomial is  $5x - 4x^2 +3$

Now, at $x = -1$  value is

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

Therefore, value of polynomial $5x - 4x^2 +3$ at x = -1 is  -6

Given polynomial is  $5x - 4x^2 +3$

Now, at $x = 2$  value is

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

Therefore, value of polynomial $5x - 4x^2 +3$ at x = 2 is  -3

Given polynomial is

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

Now,

$p(0)= (0)^2 - 0 + 1= 1$

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

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

Therefore, values of p(0), p(1) and p(2) are 1 , 1 and 3 respectively.

Given polynomial is

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

Now,

$p(0) = 2 + 0 + 2(0)^2 - (0)^3 = 2$

$p(1) = 2 + 1 + 2(1)^2 - (1)^3 = 4$

$p(2) = 2 + 2 + 2(2)^2 - (2)^3 = 4$

Therefore, values of p(0), p(1) and p(2) are 2 , 4 and 4 respectively

Given polynomial is

$p(x) = x^3$

Now,

$p(0) = (0)^3 =0$

$p(1) = (1)^3=1$

$p(2) = (2)^3=8$

Therefore, values of p(0), p(1) and p(2) are 0 , 1 and 8 respectively

Given polynomial is

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

Now,

$p(0) = (0)^2-1 = -1$

$p(1) = (1)^2-1 = 0$

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

Therefore, values of p(0), p(1) and p(2) are -1 , 0 and 3 respectively

Given polynomial is   $p(x) = 3x + 1$

Now, at  $x = -\frac{1}{3}$   it's value is

$p\left ( -\frac{1}{3} \right )=3\times \left ( -\frac{1}{3} \right )+1 = -1+1=0$

Therefore, yes $x = -\frac{1}{3}$  is a zero of  polynomial  $p(x) = 3x + 1$

Given polynomial is   $p(x) = 5x - \pi$

Now, at  $x =\frac{4}{5}$   it's value is

$p\left ( \frac{4}{5} \right )=5\times \left ( \frac{4}{5} \right ) -\pi = 4-\pi \neq 0$
Therefore, no  $x =\frac{4}{5}$   is not a zero of  polynomial  $p(x) = 5x - \pi$

Given polynomial is   $p(x) = x^2-1$

Now, at  x = 1   it's value is

$p(1) = (1)^2-1 = 1 -1 =0$

And at x = -1

$p(-1) = (-1)^2-1 = 1 -1 =0$
Therefore, yes  x = 1 , -1   are zeros of  polynomial  $p(x) = x^2-1$

Given polynomial is   $p(x) = (x+1)(x-2)$

Now, at  x = 2   it's value is

$p(2) = (2+1)(2-2) = 0$

And at x = -1

$p(-1) = (-1+1)(-1-2) = 0$

Therefore, yes  x = 2 , -1   are zeros of  polynomial  $p(x) = (x+1)(x-2)$

Given polynomial is   $p(x) = x^2$

Now, at  x = 0   it's value is

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

Therefore, yes  x = 0  is a zeros of  polynomial  $p(x) = (x+1)(x-2)$

Given polynomial is   $p(x) = lx+m$

Now, at  $x = -\frac{m}{l}$   it's value is

$p\left ( -\frac{m}{l} \right )= l \times \left ( -\frac{m}{l} \right )+m = -m+m =0$

Therefore, yes  $x = -\frac{m}{l}$   is a zeros of  polynomial  $p(x) = lx+m$

Given polynomial is   $p(x) = 3x^2-1$

Now, at  $x = -\frac{1}{\sqrt3}$   it's value is

$p\left ( -\frac{1}{\sqrt3} \right )= 3 \times \left ( -\frac{1}{\sqrt3} \right )^2-1 = 1-1 =0$

And at  $x = \frac{2}{\sqrt3}$

$p\left ( \frac{2}{\sqrt3} \right )= 3 \times \left ( \frac{2}{\sqrt3} \right )^2-1 = 4-1 =3\neq 0$

Therefore,   $x = -\frac{1}{\sqrt3}$  is a zeros of polynomial$p(x) = 3x^2-1$

whereas $x = \frac{2}{\sqrt3}$  is not a zeros of  polynomial $p(x) = 3x^2-1$

Given polynomial is   $p(x) = 2x+1$

Now, at  $x = \frac{1}{2}$   it's value is

$p\left ( \frac{1}{2} \right )= 2 \times \left ( \frac{1}{2} \right )+1 = 1+1=2 \neq 0$

Therefore,   $x = \frac{1}{2}$  is not a zeros of  polynomial  $p(x) = 2x+1$

Given polynomial is  $p(x)= x + 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow x+5 = 0$

$\Rightarrow x=-5$

Therefore, x = -5 is the zero of polynomial  $p(x)= x + 5$

Given polynomial is  $p(x)= x - 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow x-5 = 0$

$\Rightarrow x=5$

Therefore, x = 5 is a zero of polynomial  $p(x)= x - 5$

Given polynomial is  $p(x)= 2x + 5$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 2x+5 = 0$

$\Rightarrow x=-\frac{5}{2}$

Therefore, $x=-\frac{5}{2}$ is a zero of polynomial  $p(x)= 2x + 5$

Given polynomial is  $p(x) = 3x - 2$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 3x-2 = 0$

$\Rightarrow x=\frac{2}{3}$
Therefore, $x=\frac{2}{3}$  is a zero of polynomial  $p(x) = 3x - 2$

Given polynomial is  $p(x) = 3x$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow 3x = 0$

$\Rightarrow x=0$

Therefore, $x=0$  is a zero of polynomial  $p(x) = 3x$

Given polynomial is  $p(x) = ax$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow ax = 0$

$\Rightarrow x=0$

Therefore, $x=0$  is a zero of polynomial  $p(x) = ax$

Given polynomial is  $p(x) = cx+d$

Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.

Now,

$p(x)=0$

$\Rightarrow cx+d = 0$

$\Rightarrow x=-\frac{d}{c}$

Therefore,  $x=-\frac{d}{c}$  is a zero of polynomial  $p(x) = cx+d$

## NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.3

### Q1 (i) Find the remainder when $x^3 + 3x^2 +3x + 1$ is divided by $x + 1$

When we divide  $x^3 + 3x^2 +3x + 1$  by  $x + 1$

By long division method, we will get

Therefore, remainder is  $0$.

## Q1 (ii) Find the remainder when $x^3 + 3x^2 +3x + 1$ is divided by $x - \frac{1}{2}$

When we divide  $x^3 + 3x^2 +3x + 1$  by  $x - \frac{1}{2}$

By long division method, we will get

Therefore, the remainder is  $\frac{27}{8}$

When we divide  $x^3 + 3x^2 +3x + 1$  by  $x$

By long division method, we will get

Therefore, remainder is  $1$.

When we divide  $x^3 + 3x^2 +3x + 1$  by  $x + \pi$

By long division method, we will get

Therefore, the remainder is  $1-3\pi + 3\pi^2-\pi^3$

When we divide  $x^3 + 3x^2 +3x + 1$  by  $5+2x$

By long division method, we will get

Therefore, the remainder is  $-\frac{27}{8}$

When we divide  $x^3 - ax^2 + 6x - a$  by  $x - a$

By long division method, we will get

Therefore, remainder is  $5a$

When we divide  $3x^3 + 7x$  by  $7 + 3x$

We can also write  $3x^3 + 7x$  as  $3x^3 +0x^2+ 7x$

By long division method, we will get

Since, remainder is not equal to  0

Therefore,  $7 + 3x$  is not a factor of  $3x^3 + 7x$

## NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.4

Zero of polynomial   $(x + 1)$  is -1.

If  $(x + 1)$  is a factor of polynomial    $p(x)=x^3 + x^2 +x + 1$

Then, $p(-1)$  must be equal to zero

Now,

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

$\Rightarrow p(-1)=-1+1-1+1 = 0$

Therefore,  $(x + 1)$  is a factor of polynomial    $p(x)=x^3 + x^2 +x + 1$

Zero of polynomial   $(x + 1)$  is -1.

If  $(x + 1)$  is a factor of polynomial    $p(x)=x^4 + x^3 + x^2 +x + 1$

Then, $p(-1)$  must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^4+(-1)^3+(-1)^2-1+1$

$\Rightarrow p(-1)=1-1+1-1+1 = 1\neq 0$

Therefore,  $(x + 1)$  is not a factor of polynomial   $p(x)=x^4 + x^3 + x^2 +x + 1$

Zero of polynomial   $(x + 1)$  is -1.

If  $(x + 1)$  is a factor of polynomial    $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$

Then, $p(-1)$  must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^4+3(-1)^3+3(-1)^2-1+1$

$\Rightarrow p(-1)=1-3+3-1+1 = 1\neq 0$

Therefore,  $(x + 1)$  is not a factor of polynomial   $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$

Zero of polynomial   $(x + 1)$  is -1.

If  $(x + 1)$  is a factor of polynomial    $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Then, $p(-1)$  must be equal to zero

Now,

$\Rightarrow p(-1)=(-1)^3-(-1)^2-(2+\sqrt2)(-1)+\sqrt2$

$\Rightarrow p(-1)=-1-1+2+\sqrt2+\sqrt2 = 2\sqrt2\neq 0$

Therefore,  $(x + 1)$  is not a factor of polynomial $p(x)=x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Zero of polynomial  $g(x)=x+1$  is  $-1$

If  $g(x)=x+1$  is factor of polynomial  $p(x) = 2x^3 + x^2 - 2x - 1$

Then,  $p(-1)$  must be equal to zero

Now,

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

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

Therefore,  $g(x)=x+1$   is factor of polynomial  $p(x) = 2x^3 + x^2 - 2x - 1$

Zero of polynomial  $g(x)=x+2$  is  $-2$

If  $g(x)=x+2$  is factor of polynomial  $p(x) = x^3 + 3x^2 + 3x + 1$

Then,  $p(-2)$  must be equal to zero

Now,

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

$\Rightarrow p(-2)= -8+12-6+1 = -1\neq 0$

Therefore,  $g(x)=x+2$   is not a factor of polynomial  $p(x) = x^3 + 3x^2 + 3x + 1$

Zero of polynomial  $g(x)=x-3$  is  $3$

If  $g(x)=x-3$  is factor of polynomial  $p(x) = x^3 - 4x^2 + x + 6$

Then,  $p(3)$  must be equal to zero

Now,

$\Rightarrow p(3) = (3)^3 - 4(3)^2 + 3 + 6$

$\Rightarrow p(3) = 27-36+3+6=0$

Therefore,  $g(x)=x-3$   is a factor of polynomial  $p(x) = x^3 - 4x^2 + x + 6$

Zero of polynomial   $x - 1$ is  $1$

If  $x - 1$  is factor of polynomial   $p(x) = x^2 + x + k$

Then,  $p(1)$  must be equal to zero

Now,

$\Rightarrow p(1) = (1)^2 + 1 + k$

$\Rightarrow p(1) =0$

$\Rightarrow 2+k = 0$

$\Rightarrow k = -2$

Therefore,  value of k is $-2$

Zero of polynomial   $x - 1$ is  $1$

If  $x - 1$  is factor of polynomial   $p(x) = 2x^2 + kx + \sqrt2$

Then,  $p(1)$  must be equal to zero

Now,

$\Rightarrow p(1) = 2(1)^2 + k(1) + \sqrt2$

$\Rightarrow p(1) =0$

$\Rightarrow 2+k +\sqrt2= 0$

$\Rightarrow k = -(2+\sqrt2)$

Therefore,  value of k is $-(2+\sqrt2)$

Zero of polynomial   $x - 1$ is  $1$

If  $x - 1$  is factor of polynomial   $p(x) = kx^2-\sqrt2 x + 1$

Then,  $p(1)$  must be equal to zero

Now,

$\Rightarrow p(1) = k(1)^2 -\sqrt2(1) + 1$

$\Rightarrow p(1) =0$

$\Rightarrow k -\sqrt2 +1= 0$

$\Rightarrow k = -1+\sqrt2$

Therefore,  value of k is $-1+\sqrt2$

Zero of polynomial   $x - 1$ is  $1$

If  $x - 1$  is factor of polynomial   $p(x) = kx^2 -3 x + k$

Then,  $p(1)$  must be equal to zero

Now,

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

$\Rightarrow p(1) =0$

$\Rightarrow k -3+k= 0$

$\Rightarrow k = \frac{3}{2}$
Therefore,  value of k is  $\frac{3}{2}$

Given polynomial is   $12x^2 - 7x + 1$

We need to factorise the middle term into two terms such that their product is equal to $12 \times 1 = 12$ and their sum is equal to $-7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 12x^2-3x-4x+1$                                                       $(\because -3 \times -4 = 12 \ \ and \ \ -3+(-4) = -7)$

$\Rightarrow 3x(4x-1)-1(4x-1)$

$\Rightarrow (3x-1)(4x-1)$

Given polynomial is   $2x^2 + 7x + 3$

We need to factorise the middle term into two terms such that their product is equal to $2 \times 3 = 6$ and their sum is equal to $7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 2x^2+6x+x+3$                                                       $(\because 6 \times 1 = 6 \ \ and \ \ 6+1 = 7)$

$\Rightarrow 2x(x+3)+1(x+3)$

$\Rightarrow (2x+1)(x+3)$

Given polynomial is   $6x^2 +5x - 6$

We need to factorise the middle term into two terms such that their product is equal to $6 \times -6 =-3 6$ and their sum is equal to $5$

We can solve it as

$\Rightarrow 6x^2 +5x - 6$

$\Rightarrow 6x^2 +9x -4x - 6$                                       $(\because 9 \times -4 = -36 \ \ and \ \ 9+(-4) = 5)$

$\Rightarrow 3x(2x+3)-2(2x+3)$

$\Rightarrow (2x+3)(3x-2)$

Given polynomial is   $3x^2 - x - 4$

We need to factorise the middle term into two terms such that their product is equal to $3 \times -4 =-12$ and their sum is equal to $-1$

We can solve it as

$\Rightarrow 3x^2 - x - 4$

$\Rightarrow 3x^2 -4 x+3x - 4$                                      $(\because 3 \times -4 = -12 \ \ and \ \ 3+(-4) = -1)$

$\Rightarrow x(3x-4)+1(3x-4)$

$\Rightarrow (x+1)(3x-4)$

Given polynomial is  $x^3 - 2x^2 - x +2$

Now, by hit and trial method we observed that  $(x+1)$  is one of the factors of the given polynomial.

By long division method, we will get

We know that  Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 2x^2 - x +2 = (x+1)(x^2-3x+2)+0$

$= (x+1)(x^2-2x-x+2)$

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

Therefore, on factorization of  $x^3 - 2x^2 - x +2$ we will get  $(x+1)(x-2)(x-1)$

Given polynomial is  $x^3 - 3x^2 -9x -5$

Now, by hit and trial method we observed that  $(x+1)$  is one of the factors of the given polynomial.

By long division method, we will get

We know that  Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)$

$= (x+1)(x^2-5x+x-5)$

$= (x+1)(x-5)(x+1)$

Therefore, on factorization of  $x^3 - 3x^2 -9x -5$ we will get  $(x+1)(x-5)(x+1)$

Given polynomial is  $x^3 + 13x^2 + 32x + 20$

Now, by hit and trial method we observed that  $(x+1)$  is one of the factore of given polynomial.

By long division method, we will get

We know that  Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 + 13x^2 + 32x + 20=(x+1)(x^2+12x+20)$

$= (x+1)(x^2+10x+2x+20)$

$= (x+1)(x+10)(x+2)$

Therefore, on factorization of  $x^3 + 13x^2 + 32x + 20$ we will get  $(x+1)(x+10)(x+2)$

Given polynomial is  $2y^3 + y^2 - 2y - 1$

Now, by hit and trial method we observed that  $(y-1)$  is one of the factors of the given polynomial.

By long division method, we will get

We know that  Dividend = (Divisor $\times$ Quotient) + Remainder

$2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)$

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

$= (y-1)(2y+1)(y+1)$

Therefore, on factorization of  $2y^3 + y^2 - 2y - 1$ we will get  $(y-1)(2y+1)(y+1)$

## NCERT solutions for class 9 maths chapter 2 Polynomials Excercise: 2.5

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put   $a = 4 \ \ and \ \ b = 10$

$(x+4)(x+10)= x^2+(10+4)x+10\times 4$

$= x^2+14x+40$

Therefore, $(x + 4) ( x + 10)$  is equal to  $x^2+14x+40$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put   $a = 8 \ \ and \ \ b = -10$

$(x+8)(x-10)= x^2+(-10+8)x+8\times (-10)$

$= x^2-2x-80$

Therefore, $(x+8)(x-10)$  is equal to  $x^2-2x-80$

We can write  $(3x+4)(3x - 5)$  as

$(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put   $a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}$

$\dpi{100} 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )$

$=9x^2-3x-20$

Therefore, $(3x+4)(3x - 5)$  is equal to  $9x^2-3x-20$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put   $x=y^2 \ \ and \ \ a = \frac{3}{2}$

$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2$

$= y^4-\frac{9}{4}$

Therefore, $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$  is equal to  $y^4-\frac{9}{4}$

We can write $(3 - 2x) (3 + 2x)$  as

$(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put   $a = \frac{3}{2}$

$-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )$

$=9-4x^2$

Therefore, $(3 - 2x) (3 + 2x)$  is equal to  $9-4x^2$

We can rewrite  $103 \times 107$  as

$\Rightarrow 103 \times 107= (100+3)\times (100+7)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put  $x =100 , a=3 \ \ and \ \ b = 7$

$(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7$

$=10000+1000+21= 11021$

Therefore, value of $103 \times 107$  is  $11021$

We can rewrite  $95 \times 96$  as

$\Rightarrow 95 \times 96= (100-5)\times (100-4)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put  $x =100 , a=-5 \ \ and \ \ b = -4$

$(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)$

$=10000-900+20= 9120$

Therefore, value of  $95 \times 96$  is  $9120$

We can rewrite  $104 \times 96$  as

$\Rightarrow 104 \times 96= (100+4)\times (100-4)$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put  $x =100 \ \ and \ \ a=4$

$(100+4)\times (100-4)= (100)^2-(4)^2$

$=10000-16= 9984$

Therefore, value of  $104 \times 96$  is  $9984$

We can rewrite $9x^2 + 6xy + y^2$ as

$\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2$

Using identity  $\Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2$

Here, $a= 3x \ \ and \ \ b = y$

Therefore,

$9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)$

We can rewrite $4y^2 - 4y + 1$ as

$\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2$

Using identity  $\Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2$

Here, $a= 2y \ \ and \ \ b = 1$

Therefore,

$4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)$

We can rewrite   $x^2 - \frac{y^2}{100}$  as

$\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2$

Using identity  $\Rightarrow a^2-b^2 = (a-b)(a+b)$

Here, $a= x \ \ and \ \ b = \frac{y}{10}$

Therefore,

$x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )$

Given is  $(x + 2y+4z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = x , b = 2y \ \ and \ \ c = 4z$

Therefore,

$(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x$

$= x^2+4y^2+16z^2+4xy+16yz+8zx$

Given is  $(2x - y + z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = -y \ \ and \ \ c = z$

Therefore,

$(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x$

$= 4x^2+y^2+z^2-4xy-2yz+4zx$

Given is  $(-2x + 3y + 2z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 3y \ \ and \ \ c = 2z$

Therefore,

$(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)$

$= 4x^2+9y^2+4z^2-12xy+12yz-8zx$

Given is  $(3a - 7b - c)^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =3a , y = -7b \ \ and \ \ z = -c$

Therefore,

$(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c)$$.3a$

$= 9a^2+49b^2+c^2-42ab+14bc-6ca$

Given is  $(-2x + 5y -3z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 5y \ \ and \ \ c = -3z$

Therefore,

$(-2x +5y-3z)^2$$= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)$

$= 4x^2+25y^2+9z^2-20xy-30yz+12zx$

Given is  $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1$

Therefore,

$\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$$=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )$

$= \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$

We can rewrite  $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$  as

$\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ $= (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = 3y \ \ and \ \ c = -4z$

Therefore,

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

$= (2x+3y-4z)(2x+3y-4z)$

We can rewrite  $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$  as

$\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ $= (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z$

Therefore,

$2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2$

$=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)$

Given is  $(2x + 1)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here,  $a= 2x \ \ and \ \ b= 1$

Therefore,

$(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2$

$= 8x^3+1+12x^2+6x$

Given is  $(2a-3b)^3$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x= 2a \ \ and \ \ y= 3b$

Therefore,

$(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2$

$= 8a^3-9b^3-36a^2b+54ab^2$

Given is  $\left[\frac{3}{2}x + 1\right ]^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here,  $a= \frac{3x}{2} \ \ and \ \ b= 1$

Therefore,

$\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2$

$= \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}$

Given is  $\left[x - \frac{2}{3} y\right ]^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here,  $a=x \ and \ \ b= \frac{2y}{3}$

Therefore,

$\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2$

$= x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}$

We can rewrite   $(99)^3$  as

$\Rightarrow (99)^3=(100-1)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here,  $a=100 \ and \ \ b= 1$

Therefore,

$(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2$

$= 1000000-1-30000+300= 970299$

We can rewrite   $(102)^3$  as

$\Rightarrow (102)^3=(100+2)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here,  $a=100 \ and \ \ b= 2$

Therefore,

$(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2$

$= 1000000+8+60000+1200= 1061208$

We can rewrite   $(998)^3$  as

$\Rightarrow (998)^3=(1000-2)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here,  $a=1000 \ and \ \ b= 2$

Therefore,

$(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2$

$= 1000000000-8-6000000+12000= 994011992$

We can rewrite   $8a^3 + b^3 + 12a^2 b + 6ab^2$  as

$\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2$ $= (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x+y)^3=x^3+y^3+3x^2y+3xy^2$

Here,  $x=2a \ \ and \ \ y= b$

Therefore,

$8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3$

$=(2a+b)(2a+b)(2a+b)$

We can rewrite   $8a ^3 - b^3 - 12a^2 b + 6ab^2$  as

$\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2$ $= (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x=2a \ \ and \ \ y= b$

Therefore,

$8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3$

$=(2a-b)(2a-b)(2a-b)$

We can rewrite   $27 - 125a^ 3 - 135a + 225a^2$  as

$\Rightarrow 27 - 125a^ 3 - 135a + 225a^2^{}$ $= (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x=3 \ \ and \ \ y= 5a$

Therefore,

$27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3$

$=(3-5a)(3-5a)(3-5a)$

We can rewrite   $64a^3 - 27b^3 - 144a^2 b + 108ab^2$  as

$\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2$ $= (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x=4a \ \ and \ \ y= 3b$

Therefore,

$64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2$

$=(4a-3b)(4a-3b)(4a-3b)$

We can rewrite   $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$  as

$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here,  $x=3p \ \ and \ \ y= \frac{1}{6}$

Therefore,

$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$

$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$

We know that

$(x+y)^3=x^3+y^3+3xy(x+y)$

Now,

$\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)$

$\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )$                                                $(\because (a+b)^2=a^2+b^2+2ab)$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )$

Hence proved.

We know that

$(x-y)^3=x^3-y^3-3xy(x-y)$

Now,

$\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)$

$\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )$                                                $(\because (a-b)^2=a^2+b^2-2ab)$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )$

Hence proved.

We know that

$a^3+b^3=(a+b)(a^2+b^2-ab)$

Now, we can write  $27y^3 + 125z^3$  as

$\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3$

Here,  $a = 3y \ \ and \ \ b = 5z$

Therefore,

$27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )$

$27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )$

We know that

$a^3-b^3=(a-b)(a^2+b^2+ab)$

Now, we can write  $64m^3 - 343n^3$  as

$\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3$

Here,  $a = 4m \ \ and \ \ b = 7n$

Therefore,

$64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )$

$64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )$

Q11 Factorise:     $27x^3 + y^3 + z^3 - 9xyz$

Given is  $27x^3 + y^3 + z^3 - 9xyz$

Now, we know that

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Now, we can write  $27x^3 + y^3 + z^3 - 9xyz$  as

$\Rightarrow 27x^3 + y^3 + z^3 - 9xyz$ $=(3x)^3+(y)^3+(z)^3-3.3x.y.z$

Here, $a= 3x , b = y \ \ and \ \ c = z$

Therefore,

$27x^3 + y^3 + z^3 - 9xyz$ $=(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )$

$=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )$

We know that

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

Now, multiply and divide the R.H.S. by 2

$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)$

$= \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)$

$= \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )$                                        $\left(\because a^2+b^2-2ab=(a-b)^2 \right )$

Hence proved.

We know that

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

Now, It is given that  $x + y + z = 0$

Therefore,

$x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz =0$

$x^3+y^3+z^3=3xyz$

Hence proved.

Given is   $(-12)^3 + (7)^3 + (5)^3$

We know that

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

Here, $x = -12 , y = 7 \ \ an d \ \ z = 5$

$\Rightarrow x+y+z = -12+7+5 = 0$

Therefore,

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

Therefore, value of  $(-12)^3 + (7)^3 + (5)^3$  is  $-1260$

Given is   $(28)^3 + (-15)^3 + (-13)^3$

We know that

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

Here, $x = 28 , y = -15 \ \ an d \ \ z = -13$

$\Rightarrow x+y+z =28-15-13 = 0$

Therefore,

$(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380$

Therefore, value of  $(28)^3 + (-15)^3 + (-13)^3$  is  $16380$

 $25a^2 - 35a + 12$

We know that

Area of rectangle is =  $length \times breadth$

It is given that area  =  $25a^2-35a+12$

Now, by splitting middle term method

$\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12$

$= 5a(5a-4)-3(5a-4)$

$= (5a-3)(5a-4)$

case (i) :-   Length = $(5a-4)$  and  Breadth = $(5a-3)$

case (ii) :-   Length = $(5a-3)$  and  Breadth = $(5a-4)$

 $35y^2 + 13y- 12$

We know that

Area of rectangle is =  $length \times breadth$

It is given that area  =  $35y^2 + 13y- 12$

Now, by splitting the middle term method

$\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12$

$= 7y(5y+4)-3(5y+4)$

$= (7y-3)(5y+4)$

case (i) :-   Length = $(5y+4)$  and  Breadth = $(7y-3)$

case (ii) :-   Length = $(7y-3)$  and  Breadth = $(5y+4)$

 Volume :    $3x^2 - 12x$

We know that

Volume of cuboid is =  $length \times breadth \times height$

It is given that volume  =  $3x^2-12x$

Now,

$\Rightarrow 3x^2-12x=3\times x\times (x-4)$

Therefore,one of the possible answer is possible

Length = $3$  and  Breadth = $x$  and  Height =  $(x-4)$

 Volume :    $12ky^2 + 8ky - 20k$

We know that

Volume of cuboid is =  $length \times breadth \times height$

It is given that volume  =  $12ky^2+8ky-20k$

Now,

$\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)$

$= k(12y^2+20y-12y-20)$

$= k\left(4y(3y+5)-4(3y+5) \right )$

$= k(3y+5)(4y-4)$

$= 4k(3y+5)(y-1)$

Therefore,one of the possible answer is possible

Length = $4k$  and  Breadth = $(3y+5)$  and  Height =  $(y-1)$

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 2 Polynomials?

• First of all, learn some basics and concepts regarding chapter polynomials.
• While reading the basics, go through the examples so that you can understand the applications of the concepts.
• Once you have done the above two points, then you can directly move to the practice exercises.
• While practising the exercises, if you stuck anywhere then you can take the help of the NCERT solutions for class 9 maths chapter 2 Polynomials.
• After the completion of practice exercises, you can go through some previous year question papers.

Keep Working Hard and Happy Learning!