NCERT Solutions for Class 9 Maths Chapter 2: Polynomials - Exercise 2.5

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,
\[ \left( x+4 \right)\left( x+10 \right)={{x}^{2}}+\left( 4+10 \right)x+\left( 4 \right)\left( 10 \right) \]

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

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 \[=\left( y-1 \right)\]

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.

2. 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.

3. 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.

4. 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.

5. 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.