NCERT Solutions for Exercise 2.4 Class 9 Maths Chapter 2 - Polynomials

# NCERT Solutions for Exercise 2.4 Class 9 Maths Chapter 2 - Polynomials

Edited By Vishal kumar | Updated on Oct 04, 2023 10:40 AM IST

## NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.4- Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 2: Polynomials Exercise 2.4- Download Free PDF- NCERT Solutions for exercise 2.4 Class 9 Maths Chapter 2 Polynomials Exercise 2.4 is the part of NCERT solutions for Class 9 Maths. A polynomial expression is an equation made up of variables (or indeterminate variables), terms, exponents, and constants. If we talk about exercise 2.4 Class 9 Maths is an exercise of the chapter introduced and followed by exercise 2.3 that includes some numerical problems. Here in this exercise 2.4, we will be studying the factorisation of Polynomials.

The Class 9 Maths chapter 2 exercise 2.4 lists some basic level practice problems on the polynomials chapter that consist of factorization of higher degree Polynomials. The Class 9 Maths chapter 2 exercise 2.4 covers the topics like factorization theorem enclosed with examples. NCERT solutions for Class 9 Maths chapter 2 exercise 2.4 gives an end-to-end idea of the whole chapter.

The 9th class maths exercise 2.4 answers have been expertly crafted by subject experts, and presented in detail and easy-to-understand language. This exercise comprises a total of five questions, each with multiple parts. Students can conveniently access these class 9 maths chapter 2 exercise 2.4 solutions in PDF format, allowing offline usage without the need for an internet connection, and they are provided free of charge. Along with Class 9 Maths chapter 2 exercise 2.4, the following exercises are also present. Along with Class 9 Maths chapter 1 exercise 2.4 the following exercises are also present.

## Access Polynomials Class 9 Chapter 2 Exercise: 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 the 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 $(y-1)(2y+1)(y+1)$

## More About NCERT Solutions For Class 9 Maths Chapter 2 Exercise 2.4

The problems from the concepts of factorization of polynomials are covered in exercise 2.4 Class 9 Maths. The Initial questions of NCERT solutions for Class 9 Maths chapter 2 exercise 2.4 is to determine the factor of given polynomial expression. And later on questions of Class 9 Maths chapter 2 exercise, 2.4 is to factorize the given polynomial expression using splitting the middle term, the concept of division of polynomials will also be discussed in Class 9 Maths chapter 2 exercise 2.4.

## Key Features of Exercise 2.4 Class 9 Maths

• Comprehensive Exercise: 9th class maths exercise 2.4 answers is a comprehensive exercise that covers various topics related to polynomials.
• Conceptual Clarity: The primary objective of this class 9 maths chapter 2 exercise 2.4 is to help students develop a clear understanding of polynomial factorization and related theorems.
• Diverse Problem Set: exercise 2.4 class 9 maths offers a variety of problems with different levels of complexity, allowing students to enhance their skills in polynomial factorization.
• Expert-Created Solutions: class 9 maths ex 2.4 Solutions to the problems are typically provided in the exercise. These solutions are crafted by subject matter experts to ensure accuracy and clarity.
• PDF Availability: Students can often download a PDF version of the ex 2.4 class 9 solutions, allowing them to access and use them offline. This resource is provided at no cost.

Also Read| Polynomials Class 9 Notes

Also see-

## Subject Wise NCERT Exemplar Solutions

1. What is the middle term theorem?

Middle term theorem is used to find roots/zeros of equation ax² + bx + c= 0 that means we have to find that roots whose sum = -(b/a) and the product is equal to c/a

2. What is the degree of Constant Polynomial expression?

The degree of the constant polynomial is 0. Taking example is, 2 is a constant polynomial that is equal to 2x^0, so its degree is 0.

3. How many types of questions are there in chapter 2 exercise 2.4?

There are a total of 5 types of questions in the exercise 2.4

4. What is the sum of roots of polynomial 3x² +6x + 9=0?

Comparing the above  polynomial with  ax² + bx + c= 0, a= 3, b = 6 and c= 9, hence sum of  roots = -(b/a) = -(6/3) = -2

5. What is the product of roots of the above polynomial?

Product of roots = c/a = 9/3 = 3

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