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

Last updated date: 09th Aug 2024
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## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4 - FREE PDF Download

Vedantu provides the free PDF of NCERT Class 9 Maths Ex 2.4 Chapter 2 Polynomials. This exercise contains all the solutions to the questions given at the back of the CBSE textbook. NCERT Solutions for Class 9 Maths and the reference notes of class 9 ex 2.4 are developed by the subject matter expert from Vedantu as per the NCERT (CBSE) latest guidelines. These solutions will help you to revise the chapter thoroughly and score good marks in exams. The NCERT class 9 maths chapter 2 exercise 2.4 solutions are 100% accurate, and you can verify your answers with them. If you have any doubts relating to the topic, then you can reach out to our experienced teachers for clarification. You can register with Vedantu for one to one interaction with our experts.

Table of Content
1. NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 2 Exercise 2.4 Class 9 | Vedantu
3. Access NCERT Solutions for Maths Class 9 Chapter 2 Polynomials Exercise 2.4
4. Class 9 Maths Chapter 2: Exercises Breakdown
5. Other Study Materials for CBSE Class 9 Maths Chapter 2
6. Chapter-Specific NCERT Solutions for Class 9 Maths
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## Glance on NCERT Solutions Maths Chapter 2 Exercise 2.4 Class 9 | Vedantu

• In class 9th maths chapter 2 exercise 2.4 you will learn skills for solving Polynomial equations

• This exercise deals with applying algebraic identities to solve problems.

• Algebraic identities are equations that hold true for all permissible values of the variables involved.

• Some important identities covered in this exercise include:

• (x + y)² = x² + 2xy + y² (identity for squaring a sum)

• (x - y)² = x² - 2xy + y² (identity for squaring a difference)

• x² - y² = (x + y) (x - y) (identity for difference of squares)

• (x + a) (x + b) = x² + (a + b)x + ab (identity for expanding a linear expression)

• Ex 2.4 Class 9 maths NCERT Solutions has an overall 16 Questions.

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

1. Use suitable identities to find the following products:

1. (x+4)(x+10)

Ans: Using the identity, $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$

Here we have, $a=4,b=10$

We get,
$\left( x+4 \right)\left( x+10 \right)={{x}^{2}}+\left( 4+10 \right)x+\left( 4 \right)\left( 10 \right)$

$={{x}^{2}}+14x+40$

1. (x+8)(x-10)

Ans: Using the identity, $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$

Here we have, $a=8,b=-10$

We get,
$\left( x+8 \right)\left( x+\left( -10 \right) \right)={{x}^{2}}+\left( 8+\left( -10 \right) \right)x+\left( 8 \right)\left( -10 \right)$

$\left( x+8 \right)\left( x-10 \right)={{x}^{2}}+\left( 8-10 \right)x-80$

$={{x}^{2}}-2x-80$

1. ( 3x+4)( 3x-5)

Ans: Using the identity, $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$

Here we have, $a=4,b=-5$

We get,
$\left( 3x+4 \right)\left( 3x+\left( -5 \right) \right)={{\left( 3x \right)}^{2}}+\left( 4+\left( -5 \right) \right)3x+\left( 4 \right)\left( -5 \right)$

$\left( 3x+4 \right)\left( 3x-5 \right)=9{{x}^{2}}+\left( 4-5 \right)3x-20$

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

1. $\mathbf{(y^{2}+\frac{3}{2})(y^{2}-\frac{3}{2})}$

Ans: Using the identity, $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$

Here we have, $x={{y}^{2}},y=\dfrac{3}{2}$

We get,
$\left( {{y}^{2}}+\dfrac{3}{2} \right)\left( {{y}^{2}}-\dfrac{3}{2} \right)={{\left( {{y}^{2}} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$

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

1. (3-2x)(3+2x)

Ans: Using the identity, $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$

Here we have, $x=3,y=2x$

We get,
$\left( 3+2x \right)\left( 3-2x \right)={{\left( 3 \right)}^{2}}-{{\left( 2x \right)}^{2}}$

$=9-4{{x}^{2}}$

2. Evaluate the following products without multiplying directly:

1. $\mathbf{103\times 107}$

Ans: $103\times 107=\left( 100+3 \right)\times \left( 100+7 \right)$

By using the identity, $\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab$

Here we have, $x=100,~~a=3,~~b=7$

We get,

$\left( 100+3 \right)\left( 100+7 \right)={{\left( 100 \right)}^{2}}+\left( 3+7 \right)100+\left( 3 \right)\left( 7 \right)$

$\left( 103 \right)\times \left( 107 \right)=10000+1000+21$

$=11021$

1. $\mathbf{95\times 96}$

Ans: $95\times 96=\left( 100-5 \right)\times \left( 100-4 \right)$

By using the identity, $\left( x-a \right)\left( x-b \right)={{x}^{2}}-\left( a+b \right)x+ab$

Here we have, $x=100,~~a=5,~~b=4$

We get,
$\left( 100-5 \right)\left( 100-4 \right)={{\left( 100 \right)}^{2}}-\left( 5+4 \right)100+\left( 5 \right)\left( 4 \right)$

$\left( 95 \right)\times \left( 96 \right)=10000-900+20$

$=9120$

1. $\mathbf{104\times 96}$

Ans: $104\times 96=\left( 100+4 \right)\times \left( 100-4 \right)$

By using the identity, $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$

Here we have, $x=100,~~y=4$

We get,
$\left( 100+4 \right)\left( 100-4 \right)={{\left( 100 \right)}^{2}}-{{\left( 4 \right)}^{2}}$

$\left( 104 \right)\times \left( 96 \right)=10000-16$

$=9984$

3. Factorize the following using appropriate identities:

1. $\mathbf{9{{x}^{2}}+6xy+{{y}^{2}}}$

Ans: $9{{x}^{2}}+6xy+{{y}^{2}}={{\left( 3x \right)}^{2}}+2\left( 3x \right)\left( y \right)+{{\left( y \right)}^{2}}$

By using the identity, ${{x}^{2}}+2xy+{{y}^{2}}={{\left( x+y \right)}^{2}}$

Here, $x=3x,~~y=y$

$9{{x}^{2}}+6xy+{{y}^{2}}={{\left( 3x \right)}^{2}}+2\left( 3x \right)\left( y \right)+{{\left( y \right)}^{2}}$

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

$=\left( 3x+y \right)\left( 3x+y \right)$

1. $\mathbf{4{{y}^{2}}-4y+1}$

Ans: $4{{y}^{2}}-4y+1={{\left( 2y \right)}^{2}}-2\left( 2y \right)\left( 1 \right)+{{\left( 1 \right)}^{2}}$

By using the identity, ${{x}^{2}}-2xy+{{y}^{2}}={{\left( x-y \right)}^{2}}$

Here, $x=2y,~~y=1$

$4{{y}^{2}}-4y+1={{\left( 2y \right)}^{2}}-2\left( 2y \right)\left( 1 \right)+{{\left( 1 \right)}^{2}}$

$={{\left( 2y-1 \right)}^{2}}$

$=\left( 2y-1 \right)\left( 2y-1 \right)$

1. $\mathbf{{{x}^{2}}-\dfrac{{{y}^{2}}}{100}}$

Ans: ${{x}^{2}}-\dfrac{{{y}^{2}}}{100}={{\left( x \right)}^{2}}-{{\left( \dfrac{y}{10} \right)}^{2}}$

By using the identity, ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$

Here, $x=x,~~y=\dfrac{y}{10}$

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

$=\left( x-\dfrac{y}{10} \right)\left( x+\dfrac{y}{10} \right)$

4. Expand each of the following, using suitable identities:

1. $\mathbf{{{\left( x+2y+4z \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

Here, $x=x,~~y=2y,~~z=4z$

${{\left( x+2y+4z \right)}^{2}}={{\left( x \right)}^{2}}+{{\left( 2y \right)}^{2}}+{{\left( 4z \right)}^{2}}+2\left( x \right)\left( 2y \right)+2\left( 2y \right)\left( 4z \right)+2\left( 4z \right)\left( x \right)$

$={{x}^{2}}+4{{y}^{2}}+16{{z}^{2}}+4xy+16yz+8xz$

1. $\mathbf{{{\left( 2x-y+z \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

Here, $x=2x,~~y=-y,~~z=z$

${{\left( 2x-y+z \right)}^{2}}={{\left( 2x \right)}^{2}}+{{\left( -y \right)}^{2}}+{{\left( z \right)}^{2}}+2\left( 2x \right)\left( -y \right)+2\left( -y \right)\left( z \right)+2\left( z \right)\left( 2x \right)$

$=4{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4xy-2yz+4xz$

1. $\mathbf{{{\left( -2x+3y+2z \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

Here, $x=-2x,~~y=3y,~~z=2z$

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

$=4{{x}^{2}}+9{{y}^{2}}+4{{z}^{2}}-12xy+12yz-8xz$

1. $\mathbf{{{\left( 3a-7b-c \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

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

${{\left( 3a-7b-c \right)}^{2}}={{\left( 3a \right)}^{2}}+{{\left( -7b \right)}^{2}}+{{\left( -c \right)}^{2}}+2\left( 3a \right)\left( -7b \right)+2\left( -7b \right)\left( -c \right)+2\left( -c \right)\left( 3a \right)$

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

1. $\mathbf{{{\left( -2x+5y-3z \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

Here, $x=-2x,~~y=5y,~~z=-3z$

${{\left( -2x+5y-3z \right)}^{2}}={{\left( -2x \right)}^{2}}+{{\left( 5y \right)}^{2}}+{{\left( -3z \right)}^{2}}+2\left( -2x \right)\left( 5y \right)+2\left( 5y \right)\left( -3z \right)+2\left( -3z \right)\left( -2x \right)$

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

1. $\mathbf{{{\left( \dfrac{1}{4}a-\dfrac{1}{2}b+1 \right)}^{2}}}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

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

${{\left( \dfrac{1}{4}a-\dfrac{1}{2}b+1 \right)}^{2}}={{\left( \dfrac{1}{4}a \right)}^{2}}+{{\left( -\dfrac{1}{2}b \right)}^{2}}+{{\left( 1 \right)}^{2}}+2\left( \dfrac{1}{4}a \right)\left( -\dfrac{1}{2}b \right)+2\left( -\dfrac{1}{2}b \right)\left( 1 \right)+2\left( 1 \right)\left( \dfrac{1}{4}a \right)$

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

5. Factorise:

1. $\mathbf{4{{x}^{2}}+9{{y}^{2}}+16{{z}^{2}}+12xy-24yz-16xz}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

We can see that, ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx={{\left( x+y+z \right)}^{2}}$

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

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

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

1. $\mathbf{2{{x}^{2}}+{{y}^{2}}+8{{z}^{2}}-2\sqrt{2}xy+4\sqrt{2}yz-8xz}$

Ans: By using the identity, ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$

We can see that, ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx={{\left( x+y+z \right)}^{2}}$

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

$={{\left( -\sqrt{2}x+y-2\sqrt{2}z \right)}^{2}}$

$=\left( -\sqrt{2}x+y-2\sqrt{2}z \right)\left( -\sqrt{2}x+y-2\sqrt{2}z \right)$

6. Write the following cubes in expanded form:

1. $\mathbf{{{\left( 2x+1 \right)}^{3}}}$

Ans: By using the identity, ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

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

$=8{{x}^{3}}+1+6x\left( 2x+1 \right)$

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

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

1. $\mathbf{{{\left( 2a-3b \right)}^{3}}}$

Ans: By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

${{\left( 2a-3b \right)}^{3}}={{\left( 2a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3\left( 2a \right)\left( 3b \right)\left( 2a-3b \right)$

$=8{{x}^{3}}-27{{b}^{3}}-18ab\left( 2a-3b \right)$

$=8{{x}^{3}}-27{{b}^{3}}-36{{a}^{2}}b+54a{{b}^{2}}$

1. $\mathbf{{{\left( \dfrac{3}{2}x+1 \right)}^{3}}}$

Ans: By using the identity, ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

${{\left( \dfrac{3}{2}x+1 \right)}^{3}}={{\left( \dfrac{3}{2}x \right)}^{3}}+{{\left( 1 \right)}^{3}}+3\left( \dfrac{3}{2}x \right)\left( 1 \right)\left( \dfrac{3}{2}x+1 \right)$

$=\dfrac{27}{8}{{x}^{3}}+1+\dfrac{9}{2}x\left( \dfrac{3}{2}x+1 \right)$

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

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

1. $\mathbf{{{\left( x-\dfrac{2}{3}y \right)}^{3}}}$

Ans: By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

${{\left( x-\dfrac{2}{3}y \right)}^{3}}={{\left( x \right)}^{3}}-{{\left( \dfrac{2}{3}y \right)}^{3}}-3\left( x \right)\left( \dfrac{2}{3}y \right)\left( x-\dfrac{2}{3}y \right)$

$={{x}^{3}}-\dfrac{8}{27}{{y}^{3}}-2xy\left( x-\dfrac{2}{3}y \right)$

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

7. Evaluate the following using suitable identities:

1. $\mathbf{{{\left( 99 \right)}^{3}}}$

Ans: Here we can write ${{\left( 99 \right)}^{3}}$ as ${{\left( 100-1 \right)}^{3}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

${{\left( 100-1 \right)}^{3}}={{\left( 100 \right)}^{3}}-{{\left( 1 \right)}^{3}}-3\left( 100 \right)\left( 1 \right)\left( 100-1 \right)$

$=1000000-1-300\left( 100-1 \right)$

$=1000000-1-30000+300$

$=970299$

1. $\mathbf{{{\left( 102 \right)}^{3}}}$

Ans: Here we can write ${{\left( 102 \right)}^{3}}$ as ${{\left( 100+2 \right)}^{3}}$

By using the identity, ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

${{\left( 100+2 \right)}^{3}}={{\left( 100 \right)}^{3}}+{{\left( 2 \right)}^{3}}+3\left( 100 \right)\left( 2 \right)\left( 100+2 \right)$

$=1000000+8+600\left( 100+2 \right)$

$=1000000+8+60000+1200$

$=1061208$

1. $\mathbf{{{\left( 998 \right)}^{3}}}$

Ans: Here we can write ${{\left( 998 \right)}^{3}}$ as ${{\left( 1000-2 \right)}^{3}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

${{\left( 1000-2 \right)}^{3}}={{\left( 1000 \right)}^{3}}-{{\left( 2 \right)}^{3}}-3\left( 1000 \right)\left( 2 \right)\left( 1000-2 \right)$

$=1000000000-8-6000\left( 1000-2 \right)$

$=1000000000-8-6000000+12000$

$=994011992$

8. Factorise each of the following:

1. $\mathbf{8{{a}^{3}}+{{b}^{3}}+12{{a}^{2}}b+6a{{b}^{2}}}$

Ans: Here we can write $8{{a}^{3}}+{{b}^{3}}+12{{a}^{2}}b+6a{{b}^{2}}$ as

${{\left( 2a \right)}^{3}}+{{\left( b \right)}^{3}}+3{{\left( 2a \right)}^{2}}\left( b \right)+3\left( 2a \right){{\left( b \right)}^{2}}$

By using the identity, ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

Here, $x=2a,~~y=b$

$8{{a}^{3}}+{{b}^{3}}+12{{a}^{2}}b+6a{{b}^{2}}={{\left( 2a \right)}^{3}}+{{\left( b \right)}^{3}}+3{{\left( 2a \right)}^{2}}\left( b \right)+3\left( 2a \right){{\left( b \right)}^{2}}$

$={{\left( 2a+b \right)}^{3}}$

$=\left( 2a+b \right)\left( 2a+b \right)\left( 2a+b \right)$

1. $\mathbf{8{{a}^{3}}-{{b}^{3}}-12{{a}^{2}}b+6a{{b}^{2}}}$

Ans: Here we can write $8{{a}^{3}}-{{b}^{3}}-12{{a}^{2}}b+6a{{b}^{2}}$ as

${{\left( 2a \right)}^{3}}-{{\left( b \right)}^{3}}-3{{\left( 2a \right)}^{2}}\left( b \right)+3\left( 2a \right){{\left( b \right)}^{2}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

Here, $x=2a,~~y=b$

$8{{a}^{3}}-{{b}^{3}}-12{{a}^{2}}b+6a{{b}^{2}}={{\left( 2a \right)}^{3}}-{{\left( b \right)}^{3}}-3{{\left( 2a \right)}^{2}}\left( b \right)+3\left( 2a \right){{\left( b \right)}^{2}}$

$={{\left( 2a-b \right)}^{3}}$

$=\left( 2a-b \right)\left( 2a-b \right)\left( 2a-b \right)$

1. $\mathbf{27-125{{a}^{3}}-135a+225{{a}^{2}}}$

Ans: Here we can write $27-125{{a}^{3}}-135a+225{{a}^{2}}$ as

${{\left( 3 \right)}^{3}}-{{\left( 5a \right)}^{3}}-3{{\left( 3 \right)}^{2}}\left( 5a \right)+3\left( 3 \right){{\left( 5a \right)}^{2}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

Here, $x=3,~~y=5a$

$27-125{{a}^{3}}-135a+225{{a}^{2}}={{\left( 3 \right)}^{3}}-{{\left( 5a \right)}^{3}}-3{{\left( 3 \right)}^{2}}\left( 5a \right)+3\left( 3 \right){{\left( 5a \right)}^{2}}$

$={{\left( 3-5a \right)}^{3}}$

$=\left( 3-5a \right)\left( 3-5a \right)\left( 3-5a \right)$

1. $\mathbf{64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}}$

Ans: Here we can write $64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}$ as

${{\left( 4a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3{{\left( 4a \right)}^{2}}\left( 3b \right)+3\left( 4a \right){{\left( 3b \right)}^{2}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

Here, $x=4a,~~y=3b$

$64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}={{\left( 4a \right)}^{3}}-{{\left( 3b \right)}^{3}}-3{{\left( 4a \right)}^{2}}\left( 3b \right)+3\left( 4a \right){{\left( 3b \right)}^{2}}$

$={{\left( 4a-3b \right)}^{3}}$

$=\left( 4a-3b \right)\left( 4a-3b \right)\left( 4a-3b \right)$

1. $\mathbf{27{{p}^{3}}-\dfrac{1}{216}-\dfrac{9}{2}{{p}^{2}}+\dfrac{1}{4}p}$

Ans: Here we can write $27{{p}^{3}}-\dfrac{1}{216}-\dfrac{9}{2}{{p}^{2}}+\dfrac{1}{4}p$ as

${{\left( 3p \right)}^{3}}-{{\left( \dfrac{1}{6} \right)}^{3}}-3{{\left( 3p \right)}^{2}}\left( \dfrac{1}{6} \right)+3\left( 3p \right){{\left( \dfrac{1}{6} \right)}^{2}}$

By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

Here, $x=3p,~~y=\dfrac{1}{6}$

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

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

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

9. Verify:

1. $\mathbf{{{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)}$

Ans: By using the identity, ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$

${{x}^{3}}+{{y}^{3}}={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)$

${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left[ {{\left( x+y \right)}^{2}}-3xy \right]$Taking $\left( x+y \right)$ common

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

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

Hence, verified.

1. $\mathbf{{{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)}$

Ans: By using the identity, ${{\left( x-y \right)}^{3}}={{x}^{3}}-{{y}^{3}}-3xy\left( x-y \right)$

${{x}^{3}}-{{y}^{3}}={{\left( x-y \right)}^{3}}+3xy\left( x+y \right)$

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

Taking $\left( x-y \right)$ common

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

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

Hence, verified.

10. Factorise each of the following:

1. $\mathbf{27{{y}^{3}}+125{{z}^{3}}}$

Ans: Here $27{{y}^{3}}+125{{z}^{3}}$ can be written as ${{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}$

$27{{y}^{3}}+125{{z}^{3}}={{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}$

As we know that, ${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$

$27{{y}^{3}}+125{{z}^{3}}={{\left( 3y \right)}^{3}}+{{\left( 5z \right)}^{3}}$

$27{{y}^{3}}+125{{z}^{3}}=\left( 3y+5z \right)\left[ {{\left( 3y \right)}^{2}}-\left( 3y \right)\left( 5z \right)+{{\left( 5z \right)}^{2}} \right]$

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

1. $\mathbf{64{{m}^{3}}-343{{n}^{3}}}$

Ans: Here $64{{m}^{3}}-343{{n}^{3}}$ can be written as ${{\left( 4y \right)}^{3}}-{{\left( 7z \right)}^{3}}$

$64{{m}^{3}}-343{{n}^{3}}={{\left( 4y \right)}^{3}}-{{\left( 7z \right)}^{3}}$

As we know that, ${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$

$64{{m}^{3}}-343{{n}^{3}}={{\left( 4y \right)}^{3}}-{{\left( 7z \right)}^{3}}$

$64{{m}^{3}}-343{{n}^{3}}=\left( 4m-7n \right)\left[ {{\left( 4m \right)}^{2}}+\left( 4m \right)\left( 7n \right)+{{\left( 7n \right)}^{2}} \right]$

$=\left( 4m-7n \right)\left( 16{{m}^{2}}+28mn+49{{n}^{2}} \right)$

11. Factorise: $\mathbf{27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz}$

Ans: Here $27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz$ can be written as ${{\left( 3x \right)}^{3}}+{{\left( y \right)}^{3}}+{{\left( z \right)}^{3}}-3\left( 3x \right)\left( y \right)\left( z \right)$

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

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

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

$=\left( 3x+y+z \right)\left[ {{\left( 3x \right)}^{2}}+{{y}^{2}}+{{z}^{2}}-3xy-yz-3xz \right]$

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

12. Verify that $\mathbf{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\dfrac{1}{2}\left( x+y+z \right)\left[ {{\left( x-y \right)}^{2}}+{{\left( y-z \right)}^{2}}+\left( z-{{x}^{2}} \right) \right]}$

Ans: As we know that, ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$,

Dividing the equation by $\dfrac{1}{2}$ and multiply by $2$

$\Rightarrow {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\dfrac{1}{2}\left( x+y+z \right)\left[ 2\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right) \right]$

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

$=\dfrac{1}{2}\left( x+y+z \right)\left[ \left( {{x}^{2}}+{{x}^{2}}+{{y}^{2}}+{{y}^{2}}+{{z}^{2}}+{{z}^{2}}-2xy-2yz-2zx \right) \right]$

$=\dfrac{1}{2}\left( x+y+z \right)\left[ \left( {{x}^{2}}+{{y}^{2}}-2xy \right)+\left( {{y}^{2}}+{{z}^{2}}-2yz \right)+\left( {{x}^{2}}+{{z}^{2}}-2zx \right) \right]$

$=\dfrac{1}{2}\left( x+y+z \right)\left[ {{\left( x-y \right)}^{2}}+\left( y-z \right){}^{2}+{{\left( z-x \right)}^{2}} \right]$

13. If $\mathbf{x+y+z=0}$, show that $\mathbf{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz}$

Ans: As we know that ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$

Given, $x+y+z=0$, then

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

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( 0 \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$

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

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

Hence, proved.

14. Without actually calculating the cubes, find the value of each of the following:

1. $\mathbf{{{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}}}$

Ans: Let ${{\left( -12 \right)}^{3}}+{{\left( 7 \right)}^{3}}+{{\left( 5 \right)}^{3}}$

$a=-12,~~~b=7,~~~c=5$

We know that if, $x+y+z=0$ then $x{}^{3}+{{y}^{3}}+z{}^{3}=3xyz$

Here, $-12+7+5=0$

$\left( -12 \right){}^{3}+{{\left( 7 \right)}^{3}}+\left( 5 \right){}^{3}=3xyz$

$=3\left( -12 \right)\left( 7 \right)\left( 5 \right)$

$=-1260$

1. $\mathbf{{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}}$

Ans: Let ${{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}$

$a=28,~~~b=-15,~~~c=-13$

We know that if, $x+y+z=0$ then $x{}^{3}+{{y}^{3}}+z{}^{3}=3xyz$

Here, $28-15-13=0$

$\left( 28 \right){}^{3}+{{\left( -15 \right)}^{3}}+\left( -13 \right){}^{3}=3xyz$

$=3\left( 28 \right)\left( -15 \right)\left( -13 \right)$

$=16380$

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

 Area:$\mathbf{25{{a}^{2}}-35a+12}$

Ans: Area: $25{{a}^{2}}-35a+12$

Using the splitting the middle term method,

We’ve to find a number whose sum $=-35$and product $25\times 12=300$

We’ll get $-15$ and $-20$ as the numbers $\left[ -15-20=-35 \right]$ and $\left[ -15\times \left( -20 \right)=300 \right]$

$25{{a}^{2}}-35a+12$

$25{{a}^{2}}-15a-20a+12$

$5a\left( 5a-3 \right)-4\left( 5a-3 \right)$

$\left( 5a-3 \right)\left( 5a-4 \right)$

Possible expression for length $=\left( 5a-4 \right)$

Possible expression for breadth $=\left( 5a-3 \right)$

 Area:$\mathbf{35{{y}^{2}}+13y-12}$

Ans: Area: $35{{y}^{2}}+13y-12$

Using the splitting the middle term method,

We’ve to find a number whose sum $=13$and product $35\times 12=420$

We’ll get $-15$ and $28$ as the numbers $\left[ -15+28=13 \right]$ and $\left[ 15\times 28=420 \right]$

$35{{y}^{2}}+13y-12$

$35{{y}^{2}}-15a+28y-12$

$5y\left( 7y-3 \right)+4\left( 7y-3 \right)$

$\left( 7y-3 \right)\left( 5y+4 \right)$

Possible expression for length $=\left( 5y+4 \right)$

Possible expression for breadth $=\left( 7y-3 \right)$

16. What are the possible expressions for the dimensions of the cuboids whose volume are given below?

1. Volume: $\mathbf{3{{x}^{2}}-12x}$

Ans: $3{{x}^{2}}-12x$ can be written as $3x\left( x-4 \right)$ by taking $3x$ common from both the terms.

Possible expression for length $=3$

Possible expression for length $=x$

Possible expression for length $=\left( x-4 \right)$

1. Volume: $\mathbf{12k{{y}^{2}}+8ky-20k}$

Ans: $12k{{y}^{2}}+8ky-20k$ can be written as $4k\left( 3{{y}^{2}}+2y-5 \right)$ by taking $4k$ common from both the terms.

$12k{{y}^{2}}+8ky-20k=4k\left( 3{{y}^{2}}+2y-5 \right)$

Here, we can write $4k\left( 3{{y}^{2}}+2y-5 \right)$ as $4k\left( 3{{y}^{2}}+5y-3y-5 \right)$ by using the splitting the middle term method

$4k\left( 3{{y}^{2}}+5y-3y-5 \right)$

$4k\left[ y\left( 3y+5 \right)-1\left( 3y+5 \right) \right]$

$4k\left( 3y+5 \right)\left( y-1 \right)$

Possible expression for length $=4k$

Possible expression for length $=\left( 3y+5 \right)$

Possible expression for length $=\left( y-1 \right)$

## Conclusion

Class 9 math exercise 2.4 in Chapter 2 is designed to solidify your understanding of factorizing polynomials through rigorous practice. By mastering these techniques, you build a strong foundation in algebra that will be beneficial in higher-level mathematics and various practical applications. The exercise helps to reinforce the concept that any polynomial can be expressed as a product of its factors, which is a critical skill for solving polynomial equations and simplifying complex expressions.

## Class 9 Maths Chapter 2: Exercises Breakdown

 Chapter 2 - Polynomials All Exercises in PDF Format Exercise 2.1 5 Question & Solutions Exercise 2.2 4 Questions & Solutions Exercise 2.3 5 Questions & Solutions

## Other Study Materials for CBSE Class 9 Maths Chapter 2

 S.No. Important Links for Chapter 2 Polynomials 1 Class 9 Polynomials Revision Notes 2 3 Class 9 Polynomials Formula 4 Class 9 Polynomials NCERT Exemplar Solution 5 Class 9 Polynomials RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions For Maths Chapter 2 Exercise 2.4 Class 9 - Polynomials Exercise 2.4

1. What do You Mean by Algebraic Expression and Polynomial?

An algebraic expression is a combination of constants and variables, linked by four fundamental arithmetical operations { +, -, x, ÷). An algebraic expression in which there is a single variable and has only one whole number with only positive integral power is called a polynomial.

2. What are the Different Kinds of Polynomials that you have Learned in this Chapter?

The different kinds of polynomials that are taught in this chapter are constant, linear, quadratic, cubic, biquadratic polynomials.

3. Where will I get the Best Reference for NCERT Solutions for Class 9 Chapter 2 - Polynomial Exercise 2.5?

You will get the best reference for NCERT Solutions for Class 9 Chapter 2 - Polynomial Exercise 2.5 on the official website of Vedantu, the leading education portal in India. The solutions and the reference notes for NCERT Solutions Chapter 2 - Polynomials are created by the in-house subject experts stepwise as per the latest guidelines of NCERT (CBSE). These notes and solutions will give you a better understanding of the topic and will help you with revision for your exams.

4. Can I Download the NCERT Solutions for Class 9 Chapter 2 - Polynomials Exercise 2.5 from the Vedantu App?

You can definitely download the free pdf of NCERT for Class 9 Chapter 2 - Polynomials Exercise 2.5 from the Vedantu app. You can download them on your mobile phone, laptops, tablets, and desktops. You can also register for Maths online tuitions on www. Vedantu.com.

5. How many questions are there in Chapter 2 of Class 9 Maths?

Chapter 2 in the Class 9 Maths textbook is Polynomials. The chapter has five exercises with a set of questions in each exercise. Each exercise covers a particular topic. There are examples that explain the topic before each exercise. The student should attempt all the exercise questions for maximum practice and assess their performance. They can also refer to the Class 9 Maths Solutions PDF available for each exercise.

6. What are the types of Polynomials?

Polynomials can be of five types- constant, linear, quadratic, cubic and biquadratic polynomials. Constant polynomials have zero degrees, linear has only one degree and a quadratic polynomial has two degrees. Similarly, a cubic polynomial has a degree of value three and biquadratic is the one in which the degree is of value four. For more explanations and examples, you can refer to the Solutions PDF for Class 9 Maths Chapter 2.

7. How many questions are there in Exercise- 2.5 of Chapter 2 Polynomials?

Exercise 2.5 from Chapter 2 Polynomials is the last exercise in the chapter. It focuses mostly on the degree and the factorization of the polynomials. Questions asked are related to finding values of the equations while using the theorems and formulas discussed in the chapter. The students can refer to the solutions PDF for easy step by step explanations for answers to each question. The solutions PDF also has extra questions for students to practice.

8. Why should I practice sample papers to prepare for the chapter?

Practicing sample papers is a great way to assess your knowledge of the chapter and how much of the concepts have you understood. It also helps you manage your time while practicing the paper. Attempting sample papers or papers from the previous years helps you understand the kind of questions to expect and how to answer them. The solutions PDF provided by Vedantu has enough sample papers for the students to practice from.

9. How should I prepare Chapter 2 Class 9 Maths for the exam?

Any chapter needs consistent practice for any student to score full marks. Students should practice all the NCERT questions and examples of this chapter to be thorough with the topics. Write down all the formulas so that you memorise them better and it's easy to revise. Attempt as many sample papers and papers of this chapter from previous years. You can also refer to the solutions PDF and Vedantus’ revision notes.