NCERT Solutions For Maths Chapter 2 Exercise 2.4 Class 9 - 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,
\[ \left( x+4 \right)\left( x+10 \right)={{x}^{2}}+\left( 4+10 \right)x+\left( 4 \right)\left( 10 \right) \]

\[ ={{x}^{2}}+14x+40 \]

2. (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 $

3. ( 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 $

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

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} $

5. (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 $

2. \[\mathbf{95\times 96}\]

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 $

3. \[\mathbf{104\times 96}\]

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) $

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

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) $

3. \[\mathbf{{{x}^{2}}-\dfrac{{{y}^{2}}}{100}}\]

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 $

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

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 $

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

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 $

4. \[\mathbf{{{\left( 3a-7b-c \right)}^{2}}}\]

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 $

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

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 $

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

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) $

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

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 $

2. \[\mathbf{{{\left( 2a-3b \right)}^{3}}}\]

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}} $

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

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 $

4. \[\mathbf{{{\left( x-\dfrac{2}{3}y \right)}^{3}}}\]

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 $

2. \[\mathbf{{{\left( 102 \right)}^{3}}}\]

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 $

3. \[\mathbf{{{\left( 998 \right)}^{3}}}\]

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) $

2. \[\mathbf{8{{a}^{3}}-{{b}^{3}}-12{{a}^{2}}b+6a{{b}^{2}}}\]

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) $

3. \[\mathbf{27-125{{a}^{3}}-135a+225{{a}^{2}}}\]

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) $

4. \[\mathbf{64{{a}^{3}}-27{{b}^{3}}-144{{a}^{2}}b+108a{{b}^{2}}}\]

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) $

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

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.

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

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) $

2. \[\mathbf{64{{m}^{3}}-343{{n}^{3}}}\]

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 $

2. \[\mathbf{{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}}\]

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:

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 \[=\left( 5a-3 \right)\]

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 \[=\left( x-4 \right)\]

2. 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

Other Study Materials for CBSE Class 9 Maths Chapter 2

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.