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# NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.5

## NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5 (Ex 2.5)

Last updated date: 25th Jan 2023
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Vedantu provides the free PDF of NCERT Maths Class 9 Chapter 2.5 Polynomials. This exercise contains all the solutions to the questions given at the back of the CBSE textbook. The solutions and the reference notes are developed by the subject matter expert from Vedantu as per the NCERT (CBSE) latest guidelines. The NCERT Solutions are in pdf version which you can download anytime and on any device. These solutions will help you to revise the chapter thoroughly and score good marks in exams. The NCERT Solutions for Exercise 2.5 of chapter 2 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.

Chapter 2 discusses polynomials, their definition in one variable with many illustrated examples and counterexamples. This chapter also includes types of polynomial, coefficients of a polynomial, terms of a polynomial, and zero polynomial. Factors and Multiples which you have learned in the previous class are discussed again in this chapter. The solutions in Exercise 2.5 are based upon the degree of a polynomial, factorization, and algebraic identities. Below is a summary of what is covered in the chapter that will help you solve the NCERT Book Solutions Exercise 2.5. Students can also avail of NCERT Class 9 Science from our website.

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## Access NCERT solutions for Maths Chapter 2 – Polynomials

### Exercise 2.5

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

### Summary of Polynomial Chapter

A combination of constants and variables, linked by four fundamental arithmetical operations { +, -, x, ÷) is called an algebraic expression.

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.

Example: 3x3 + 2x2 - 7x + 5

The part of a polynomial that is separated from each other by arithmetic operations like + or - is called a term. Each term of a polynomial has a coefficient.

Example: x + 2, 3x2 + √5x - √6 etc.

Based on the number of terms:

(i) A polynomial that contains only one non-zero term is called a monomial.

(ii) A binomial is a polynomial that contains two non-zero terms.

(iii) A polynomial that contains three non-zero terms, is called a trinomial.

### Degree of a Polynomial

The exponent of the highest degree term in a polynomial is called its degree.

Example:

(i) f(x) = 4x -1 is a polynomial in x of degree 1.

(ii) q(y) = 2y2 - 3y + 5 is a polynomial in y of degree 2.

### Types of Polynomial

Let us see the different types of polynomials.

1. Constant Polynomial: It is a polynomial of degree zero.

Ex: f(x) = 5, g(x) = 2, etc.

The constant polynomial is called the zero polynomial. The degree of the zero polynomial is not defined as f(x) = 0, g(x) = 0x, h(x) = 0x2, p(x) = 0x3, etc. are all equal to the zero polynomial.

1. Linear Polynomial: It is a polynomial of degrees.

Ex: f(x) = 4x -1.

A polynomial q(y) = 3y2 + 4 is not linear as it has degree 2.

In general, any linear polynomial in x with real coefficients is of the form f(x) = ax + b where a & b are real numbers and a ≠ 0.

1. Quadratic Polynomial: A polynomial that is of a degree two is called a quadratic polynomial.

Ex: f(x) = 5x2 - 3x + 4.

In general, a quadratic polynomial is of the form f(x) = ax2 + bx + c, where a, b & c are real numbers and a ≠ 0.

1. Cubic Polynomial: A polynomial that is of a degree 3 is called a cubic polynomial.

Ex: f(x) = 2x3 – 5x2 + 7x – 9.

The general form of a cubic polynomial is f(x) = ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a ≠ 0.

1. Biquadratic Polynomial: It is a polynomial of degree 4.

### Value of a Polynomial

The value derived by putting a specific value of the variable in a polynomial is called the value of the polynomial at that value of the variable.

If f(x) is a polynomial and ‘a’ is a real number then the real number obtained by replacing x by ‘a’ , is called the value f(x) at x = a and is denoted by f(a). For example: if f(x) = 2x2 - 3x -2, then its value:

(i) At x = 1 is given by f(1) = 2(1)2 - 3 1 -2 = 2-3-2 = -3.

(ii) At x = -2, is given by f(-2) = 2(-2)2 -3(-2)-2 = 8 + 6 -2 = 12.

### Zero of a Polynomial

Zero of a polynomial f(x) is a number , if() = 0. It is also called the root of polynomial equation p(x) = 0.

Important Points  on Zeroes of a Polynomial are Given Below:

• Zero can be a zero of a polynomial.

• Every linear polynomial has only one zero.

• A non-zero constant polynomial has no zero.

• Every real number is a 0 of the zero polynomial.

• The number of 0 of a zero polynomial is infinite.

• The maximum number of 0 of a polynomial is equal to its degree.

### Remainder Theorem

To understand the remainder theorem, we must have knowledge about factors and multiples,  long division algorithm.

If f(x) is any polynomial of a degree ≥ 1 and f(x) is divided by the linear polynomial x – a, then the remainder is f(a).

### Factor Theorem

x – a is a factor of the polynomial f(x), if f(a) = 0. Also, if x – a is a factor of f(x), then f(a) = 0.

Sometimes a polynomial having values that are unknown and one of its factors is provided and we have to calculate the value of that unknown value. Sometimes, a linear polynomial is given and we have to verify whether it is a factor of given polynomial f(x)of degree greater than 1 or not. For solving such types of problems factor theorem is required.

### Algebraic Identities

Algebraic identities are expressions that are helpful to solve many problems such as factorization of the algebraic expression, finding a product without multiplying directly, and to evaluate the value of a number having exponent.

### Benefits of NCERT Solutions Provided by Vedantu

The NCERT Solutions Chapter 2 Exercise 2.5 for Class 9 is designed by Vedantu in a very unique style. The faculties of  Vedantu have given explanations to all the questions in the NCERT Solutions in a simple and self-explanatory way. There are NCERT Solutions for all other subjects for other classes too. NCERT Solutions are tested and verified by experienced teachers from reputed institutions across the country. By attempting these NCERT Solutions, students can practice for the examination thoroughly and master the topic. Countless students have shown remarkable improvement in their scores by solving the NCERT Solutions.

Vedantu is a well known online education website in the country that gives students a platform to master their skills and shape up their future. Vedantu also provides live sessions with our experienced teachers. To attend such sessions, you need to enroll with Vedantu and grab the opportunity to soar high in the sky.

## FAQs on NCERT Solutions for class 9 Maths Chapter 2- Polynomials Exercise 2.5

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.

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