NCERT Solutions for Class 8 Maths Chapter 12: Factorisation - Exercise 12.2

Exercise 12.2

1. Factorise the following expression.

1. Factorise the expression  $ {a^2} + 8a + 16 $ .

Ans: To factorise the expression  $ {a^2} + 8a + 16 $ write  $ {a^2} $ ,  $ 8a $  and 16 as a product of different numbers.

 $ {a^2} $ can be written as  $ {\left( a \right)^2} $ .

 $ 8a $ can be written as  $ 2 \times 4 \times a $ .

16 can be written as  $ {\left( 4 \right)^2} $ .

Substitute  $ {\left( a \right)^2} $ for  $ {a^2} $ ,  $ 2 \times 4 \times a $  for  $ 8a $  and  $ {\left( 4 \right)^2} $  for 16 in expression  $ {a^2} + 8a + 16 $ .

 $ {a^2} + 8a + 16 = {\left( a \right)^2} + 2 \times 4 \times a + {\left( 4 \right)^2} $ 

Simplify  $ {a^2} + 8a + 16 = {\left( a \right)^2} + 2 \times 4 \times a + {\left( 4 \right)^2} $ using algebraic identity  $ {\left( x \right)^2} + 2xy + {\left( y \right)^2} = {\left( {x + y} \right)^2} $ .

 $ {\left( a \right)^2} + 2 \times 4 \times a + {\left( 4 \right)^2} = {\left( {a + 4} \right)^2} $ 

Therefore  $ {a^2} + 8a + 16 $  can be factorized as  $ {\left( {a + 4} \right)^2} $ .

2. Factorise the expression  $ {p^2} - 10p + 25 $ .

Ans: To factorise the expression  $ {p^2} - 10p + 25 $ write  $ {p^2} $ ,  $ 10p $  and 25 as a product of different numbers.

 $ {p^2} $ can be written as  $ {\left( p \right)^2} $ .

 $ 10p $ can be written as  $ 2 \times 5 \times p $ .

25 can be written as  $ {\left( 5 \right)^2} $ .

Substitute  $ {\left( p \right)^2} $ for  $ {p^2} $ ,  $ 2 \times 5 \times p $  for  $ 10p $  and  $ {\left( 5 \right)^2} $  for 25 in expression  $ {p^2} - 10p + 25 $ .

 $ {p^2} - 10p + 25 = {\left( p \right)^2} - 2 \times 5 \times p + {\left( 5 \right)^2} $ 

Simplify  $ {p^2} - 10p + 25 = {\left( p \right)^2} - 2 \times 5 \times p + {\left( 5 \right)^2} $ using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

 $ {\left( p \right)^2} - 2 \times 5 \times p + {\left( 5 \right)^2} = {\left( {p - 5} \right)^2} $ 

Therefore  $ {p^2} - 10p + 25 $  can be factorized as  $ {\left( {p - 5} \right)^2} $ .

3. Factorise the expression  $ 25{m^2} + 30m + 9 $ .

Ans: To factorise the expression  $ 25{m^2} + 30m + 9 $ write  $ 25{m^2} $ ,  $ 30m $  and 9 as a product of different numbers.

 $ 25{m^2} $ can be written as  $ {\left( {5m} \right)^2} $ .

 $ 30m $ can be written as  $ 2 \times 5m \times 3 $ .

9 can be written as  $ {\left( 3 \right)^2} $ .

Substitute  $ {\left( {5m} \right)^2} $ for  $ 25{m^2} $ ,  $ 2 \times 5m \times 3 $  for  $ 30m $  and  $ {\left( 3 \right)^2} $  for 9 in expression  $ 25{m^2} + 30m + 9 $ .

 $ 25{m^2} + 30m + 9 = {\left( {5m} \right)^2} + 2 \times 5m \times 3 + {\left( 3 \right)^2} $ 

Simplify  $ 25{m^2} + 30m + 9 = {\left( {5m} \right)^2} + 2 \times 5m \times 3 + {\left( 3 \right)^2} $ using algebraic identity  $ {\left( x \right)^2} + 2xy + {\left( y \right)^2} = {\left( {x + y} \right)^2} $ .

 $ {\left( {5m} \right)^2} + 2 \times 5m \times 3 + {\left( 3 \right)^2} = {\left( {5m + 3} \right)^2} $ 

Therefore  $ 25{m^2} + 30m + 9 $  can be factorized as  $ {\left( {5m + 3} \right)^2} $ .

4. Factorise the expression  $ 49{y^2} + 84yz + 36{z^2} $ .

Ans: To factorise the expression  $ 49{y^2} + 84yz + 36{z^2} $ write  $ 49{y^2} $ ,  $ 84yz $  and  $ 36{z^2} $  as a product of different numbers.

 $ 49{y^2} $ can be written as  $ {\left( {7y} \right)^2} $ .

 $ 84yz $ can be written as  $ 2 \times 7y \times 6z $ .

 $ 36{z^2} $ can be written as  $ {\left( {6z} \right)^2} $ .

Substitute  $ {\left( {7y} \right)^2} $ for  $ 49{y^2} $ ,  $ 2 \times 7y \times 6z $  for  $ 84yz $  and  $ {\left( {6z} \right)^2} $  for  $ 36{z^2} $  in expression  $ 49{y^2} + 84yz + 36{z^2} $ .

 $ 49{y^2} + 84yz + 36{z^2} = {\left( {7y} \right)^2} + 2 \times 7y \times 6z + {\left( {6z} \right)^2} $ 

Simplify  $ 49{y^2} + 84yz + 36{z^2} = {\left( {7y} \right)^2} + 2 \times 7y \times 6z + {\left( {6z} \right)^2} $ using algebraic identity  $ {\left( x \right)^2} + 2xy + {\left( y \right)^2} = {\left( {x + y} \right)^2} $ .

 $ {\left( {7y} \right)^2} + 2 \times 7y \times 6z + {\left( {6z} \right)^2} = {\left( {7y + 6z} \right)^2} $ 

Therefore  $ 49{y^2} + 84yz + 36{z^2} $  can be factored as  $ {\left( {7y + 6z} \right)^2} $ .

5. Factorise the expression  $ 4{x^2} - 8x + 4 $ .

Ans: To factorise the expression  $ 4{x^2} - 8x + 4 $ write  $ 4{x^2} $ ,  $ 8x $  and 4 as a product of different numbers.

 $ 4{x^2} $ can be written as  $ {\left( {2x} \right)^2} $ .

 $ 8x $ can be written as  $ 2 \times 2x \times 2 $ .

4 can be written as  $ {\left( 2 \right)^2} $ .

Substitute  $ {\left( {2x} \right)^2} $ for  $ 4{x^2} $ ,  $ 2 \times 2x \times 2 $  for  $ 8x $  and  $ {\left( 2 \right)^2} $  for 4 in expression  $ 4{x^2} - 8x + 4 $ .

 $ 4{x^2} - 8x + 4 = {\left( {2x} \right)^2} - 2 \times 2x \times 2 + {\left( 2 \right)^2} $ 

Simplify  $ 4{x^2} - 8x + 4 = {\left( {2x} \right)^2} - 2 \times 2x \times 2 + {\left( 2 \right)^2} $ using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

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

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

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

Therefore  $ 4{x^2} - 8x + 4 $  can be factorized as  $ 4{\left( {x - 1} \right)^2} $ .

6. Factorise the expression  $ 121{b^2} - 88bc + 16{c^2} $ .

Ans: To factorise the expression  $ 121{b^2} - 88bc + 16{c^2} $ write  $ 121{b^2} $ ,  $ 88bc $  and  $ 16{c^2} $  as a product of different numbers.

 $ 121{b^2} $ can be written as  $ {\left( {11b} \right)^2} $ .

 $ 88bc $ can be written as  $ 2 \times 11b \times 4c $ .

 $ 16{c^2} $ can be written as  $ {\left( {4c} \right)^2} $ .

Substitute  $ {\left( {11b} \right)^2} $ for  $ 121{b^2} $ ,  $ 2 \times 11b \times 4c $  for  $ 88bc $  and  $ {\left( {4c} \right)^2} $  for  $ 16{c^2} $  in expression  $ 121{b^2} - 88bc + 16{c^2} $ .

 $ 121{b^2} - 88bc + 16{c^2} = {\left( {11b} \right)^2} - 2 \times 11b \times 4c + {\left( {4c} \right)^2} $ 

Simplify  $ 121{b^2} - 88bc + 16{c^2} = {\left( {11b} \right)^2} - 2 \times 11b \times 4c + {\left( {4c} \right)^2} $ using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

 $ {\left( {11b} \right)^2} - 2 \times 11b \times 4c + {\left( {4c} \right)^2} = {\left( {11b - 4c} \right)^2} $ 

Therefore  $ 121{b^2} - 88bc + 16{c^2} $  can be factorized as  $ {\left( {11b - 4c} \right)^2} $ .

7. Factorise the expression  $ {\left( {l + m} \right)^2} - 4lm $ .

Ans: To factorise the expression  $ {\left( {l + m} \right)^2} - 4lm $ expand  $ {\left( {l + m} \right)^2} $  using algebraic identity  $ {\left( {x + y} \right)^2} = {\left( x \right)^2} + 2xy + {\left( y \right)^2} $  and simplify.

 $ {\left( {l + m} \right)^2} = {\left( l \right)^2} + 2lm + {\left( m \right)^2} $ 

 $ {\left( {l + m} \right)^2} = {l^2} + 2lm + {m^2} $ 

Substitute  $ {l^2} + 2lm + {m^2} $  for  $ {\left( {l + m} \right)^2} $  in expression  $ {\left( {l + m} \right)^2} - 4lm $ and simplify.

 $ {\left( {l + m} \right)^2} - 4lm = {l^2} + 2lm + {m^2} - 4lm $ 

 $ {\left( {l + m} \right)^2} - 4lm = {l^2} - 2lm + {m^2} $ 

Now factorise $ {l^2} - 2lm + {m^2} $ .

 $ {l^2} $ can be written as  $ {\left( l \right)^2} $ .

 $ 2lm $ can be written as  $ 2 \times l \times m $ .

 $ {m^2} $ can be written as  $ {\left( m \right)^2} $ .

Substitute  $ {\left( l \right)^2} $ for  $ {l^2} $ ,  $ 2 \times l \times m $  for  $ 2lm $  and  $ {\left( m \right)^2} $  for  $ {m^2} $  in expression  $ {l^2} - 2lm + {m^2} $ .

 $ {l^2} - 2lm + {m^2} = {\left( l \right)^2} - 2 \times l \times m + {\left( m \right)^2} $ 

Simplify  $ {l^2} - 2lm + {m^2} = {\left( l \right)^2} - 2 \times l \times m + {\left( m \right)^2} $ using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

 $ {\left( l \right)^2} - 2 \times l \times m + {\left( m \right)^2} = {\left( {l - m} \right)^2} $ 

Therefore  $ {\left( {l + m} \right)^2} - 4lm $  can be factorized as  $ {\left( {l - m} \right)^2} $ .

8. Factorise the expression  $ {a^4} + 2{a^2}{b^2} + {b^4} $ .

Ans: To factorise the expression  $ {a^4} + 2{a^2}{b^2} + {b^4} $ write  $ {a^4} $ ,  $ 2{a^2}{b^2} $  and  $ {b^4} $  as a product of different numbers.

 $ {a^4} $ can be written as  $ {\left( {{a^2}} \right)^2} $ .

 $ 2{a^2}{b^2} $ can be written as  $ 2 \times {a^2} \times {b^2} $ .

 $ {b^4} $ can be written as  $ {\left( {{b^2}} \right)^2} $ .

Substitute  $ {\left( {{a^2}} \right)^2} $ for  $ {a^4} $ ,  $ 2 \times {a^2} \times {b^2} $  for  $ 2{a^2}{b^2} $  and  $ {\left( {{b^2}} \right)^2} $  for  $ {b^4} $  in expression  $ {a^4} + 2{a^2}{b^2} + {b^4} $ .

 $ {a^4} + 2{a^2}{b^2} + {b^4} = {\left( {{a^2}} \right)^2} + 2 \times {a^2} \times {b^2} + {\left( {{b^2}} \right)^2} $ 

Simplify  $ {a^4} + 2{a^2}{b^2} + {b^4} = {\left( {{a^2}} \right)^2} + 2 \times {a^2} \times {b^2} + {\left( {{b^2}} \right)^2} $ using algebraic identity  $ {\left( x \right)^2} + 2xy + {\left( y \right)^2} = {\left( {x + y} \right)^2} $ .

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

Therefore  $ {a^4} + 2{a^2}{b^2} + {b^4} $  can be factorized as  $ {\left( {{a^2} + {b^2}} \right)^2} $ .

2. Factorise the following expression.

1. Factorise the expression  $ 4{p^2} - 9{q^2} $ .

Ans: To factorise $ 4{p^2} - 9{q^2} $ rewrite the expression as the square terms.

 $ 4{p^2} $ can be written as  $ {\left( {2p} \right)^2} $ .

 $ 9{q^2} $ can be written as  $ {\left( {3q} \right)^2} $ .

Substitute  $ {\left( {2p} \right)^2} $  for  $ 4{p^2} $  and  $ {\left( {3q} \right)^2} $  for  $ 9{q^2} $  in expression  $ 4{p^2} - 9{q^2} $ .

 $ 4{p^2} - 9{q^2} = {\left( {2p} \right)^2} - {\left( {3q} \right)^2} $ 

Simplify  $ 4{p^2} - 9{q^2} = {\left( {2p} \right)^2} - {\left( {3q} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ {\left( {2p} \right)^2} - {\left( {3q} \right)^2} = \left( {2p + 3q} \right)\left( {2p - 3q} \right) $ 

Therefore  $ 4{p^2} - 9{q^2} $  can be factorized as  $ \left( {2p + 3q} \right)\left( {2p - 3q} \right) $ .

2. Factorise the expression  $ 63{a^2} - 112{b^2} $ .

Ans: To factorise $ 63{a^2} - 112{b^2} $ rewrite the expression as the square terms.

Take 7 as a common factor from expression  $ 63{a^2} - 112{b^2} $ .

 $ 63{a^2} - 112{b^2} = 7\left( {9{a^2} - 16{b^2}} \right) $ 

 $ 9{a^2} $ can be written as  $ {\left( {3a} \right)^2} $ .

 $ 16{b^2} $ can be written as  $ {\left( {4b} \right)^2} $ .

Substitute  $ {\left( {3a} \right)^2} $  for  $ 9{a^2} $  and  $ {\left( {4b} \right)^2} $  for  $ 16{b^2} $  in expression  $ 7\left( {9{a^2} - 16{b^2}} \right) $ .

 $ 7\left( {9{a^2} - 16{b^2}} \right) = 7\left[ {{{\left( {3a} \right)}^2} - {{\left( {4b} \right)}^2}} \right] $ 

Simplify  $ 7\left( {9{a^2} - 16{b^2}} \right) = 7\left[ {{{\left( {3a} \right)}^2} - {{\left( {4b} \right)}^2}} \right] $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ 7\left[ {{{\left( {3a} \right)}^2} - {{\left( {4b} \right)}^2}} \right] = 7\left( {3a + 4b} \right)\left( {3a - 4b} \right) $ 

Therefore  $ 63{a^2} - 112{b^2} $  can be factorized as  $ 7\left( {3a + 4b} \right)\left( {3a - 4b} \right) $ .

3. Factorise the expression  $ 49{x^2} - 36 $ .

Ans: To factorise $ 49{x^2} - 36 $ rewrite the expression as the square terms.

 $ 49{x^2} $ can be written as  $ {\left( {7x} \right)^2} $ .

36 can be written as  $ {\left( 6 \right)^2} $ .

Substitute  $ {\left( {7x} \right)^2} $  for  $ 49{x^2} $  and  $ {\left( 6 \right)^2} $  for 36 in expression  $ 49{x^2} - 36 $ .

 $ 49{x^2} - 36 = {\left( {7x} \right)^2} - {\left( 6 \right)^2} $ 

Simplify  $ 49{x^2} - 36 = {\left( {7x} \right)^2} - {\left( 6 \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ {\left( {7x} \right)^2} - {\left( 6 \right)^2} = \left( {7x + 6} \right)\left( {7x - 6} \right) $ 

Therefore  $ 49{x^2} - 36 $  can be factorized as  $ \left( {7x + 6} \right)\left( {7x - 6} \right) $ .

4. Factorise the expression  $ 16{x^5} - 144{x^3} $ .

Ans: To factorise $ 16{x^5} - 144{x^3} $ rewrite the expression as the square terms.

Take  $ 16{x^3} $  as a common factor from the expression  $ 16{x^5} - 144{x^3} $ .

 $ 16{x^5} - 144{x^3} = 16{x^3}\left( {{x^2} - 9} \right) $ 

 $ {x^2} $ can be written as  $ {\left( x \right)^2} $ .

9 can be written as  $ {\left( 3 \right)^2} $ .

Substitute  $ {\left( x \right)^2} $  for  $ {x^2} $  and  $ {\left( 3 \right)^2} $  for 9 in expression  $ 16{x^3}\left( {{x^2} - 9} \right) $ .

 $ 16{x^3}\left( {{x^2} - 9} \right) = 16{x^3}\left[ {{{\left( x \right)}^2} - {{\left( 3 \right)}^2}} \right] $ 

Simplify  $ 16{x^3}\left( {{x^2} - 9} \right) = 16{x^3}\left[ {{{\left( x \right)}^2} - {{\left( 3 \right)}^2}} \right] $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ 16{x^3}\left[ {{{\left( x \right)}^2} - {{\left( 3 \right)}^2}} \right] = 16{x^3}\left( {x + 3} \right)\left( {x - 3} \right) $ 

Therefore  $ 16{x^5} - 144{x^3} $  can be factorized as  $ 16{x^3}\left( {x + 3} \right)\left( {x - 3} \right) $ .

5. Factorise the expression  $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} $ .

Ans: To factorise $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} $ rewrite the expression as the square terms.

Simplify  $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} = \left( {\left( {l + m} \right) + \left( {l - m} \right)} \right)\left( {\left( {l + m} \right) - \left( {l - m} \right)} \right) $ 

Simplify right hand side of expression  $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} = \left( {\left( {l + m} \right) + \left( {l - m} \right)} \right)\left( {\left( {l + m} \right) - \left( {l - m} \right)} \right) $ using addition.

 $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} = \left( {l + m + l - m} \right)\left( {l + m - l + m} \right) $ 

 $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} = 2l \cdot 2m $ 

 $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} = 4lm $ 

Therefore  $ {\left( {l + m} \right)^2} - {\left( {l - m} \right)^2} $  can be factorized as  $ 4lm $ .

6. Factorise the expression  $ 9{x^2}{y^2} - 16 $ .

Ans: To factorise $ 9{x^2}{y^2} - 16 $ rewrite the expression as the square terms.

 $ 9{x^2}{y^2} $ can be written as  $ {\left( {3xy} \right)^2} $ .

16 can be written as  $ {\left( 4 \right)^2} $ .

Substitute  $ {\left( {3xy} \right)^2} $  for  $ 9{x^2}{y^2} $  and  $ {\left( 4 \right)^2} $  for 16 in expression  $ 9{x^2}{y^2} - 16 $ .

 $ 9{x^2}{y^2} - 16 = {\left( {3xy} \right)^2} - {\left( 4 \right)^2} $ 

Simplify  $ 9{x^2}{y^2} - 16 = {\left( {3xy} \right)^2} - {\left( 4 \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ {\left( {3xy} \right)^2} - {\left( 4 \right)^2} = \left( {3xy + 4} \right)\left( {3xy - 4} \right) $ 

Therefore  $ 9{x^2}{y^2} - 16 $  can be factorized as  $ \left( {3xy + 4} \right)\left( {3xy - 4} \right) $ .

7. Factorise the expression  $ \left( {{x^2} - 2xy + {y^2}} \right) - {z^2} $

Ans: To factorise $ \left( {{x^2} - 2xy + {y^2}} \right) - {z^2} $ rewrite the expression as the square terms.

 $ \left( {{x^2} - 2xy + {y^2}} \right) $ can be written as  $ {\left( {x - y} \right)^2} $  using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

 $ {z^2} $ can be written as  $ {\left( z \right)^2} $ .

Substitute  $ {\left( {x - y} \right)^2} $  for  $ \left( {{x^2} - 2xy + {y^2}} \right) $  and  $ {\left( z \right)^2} $  for  $ {z^2} $  in expression  $ \left( {{x^2} - 2xy + {y^2}} \right) - {z^2} $ .

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

Simplify  $ \left( {{x^2} - 2xy + {y^2}} \right) - {z^2} = {\left( {x - y} \right)^2} - {\left( z \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Therefore  $ \left( {{x^2} - 2xy + {y^2}} \right) - {z^2} $  can be factorized as  $ \left( {x - y + z} \right)\left( {x - y - z} \right) $ .

8. Factorise the expression  $ 25{a^2} - 4{b^2} + 28bc - 49{c^2} $ .

Ans: To factorise $ 25{a^2} - 4{b^2} + 28bc - 49{c^2} $ rewrite the expression as the square terms.

Rewrite  $ 25{a^2} - 4{b^2} + 28bc - 49{c^2} $ as  $ 25{a^2} - \left( {4{b^2} - 28bc + 49{c^2}} \right) $ .

 $ \left( {4{b^2} - 28bc + 49{c^2}} \right) $ can be written as  $ {\left( {2b - 7c} \right)^2} $  using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

 $ 25{a^2} $ can be written as  $ {\left( {5a} \right)^2} $ .

Substitute  $ {\left( {2b - 7c} \right)^2} $  for  $ \left( {4{b^2} - 28bc + 49{c^2}} \right) $  and  $ {\left( {5a} \right)^2} $  for  $ 25{a^2} $  in expression  $ 25{a^2} - \left( {4{b^2} - 28bc + 49{c^2}} \right) $ .

 $ 25{a^2} - \left( {4{b^2} - 28bc + 49{c^2}} \right) = {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2} $ 

Simplify  $ 25{a^2} - \left( {4{b^2} - 28bc + 49{c^2}} \right) = {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

 $ {\left( {5a} \right)^2} - {\left( {2b - 7c} \right)^2} = \left( {5a + 2b - 7c} \right)\left( {5a - 2b + 7c} \right) $ 

Therefore  $ 25{a^2} - 4{b^2} + 28bc - 49{c^2} $  can be factorized as  $ \left( {5a + 2b - 7c} \right)\left( {5a - 2b + 7c} \right) $ .

3. Factorise the expressions.

1. Factorise the expression  $ a{x^2} + bx $ .

Ans: To factorise the expression  $ a{x^2} + bx $ write  $ a{x^2} $  and  $ bx $  as a product.

 $ a{x^2} $ can be written as  $ a \times x \times x $ .

 $ bx $ can be written as  $ b \times x $ .

Substitute  $ a \times x \times x $  for  $ a{x^2} $  and  $ b \times x $  for  $ bx $  in expression  $ a{x^2} + bx $ .

 $ a{x^2} + bx = \left( {a \times x \times x} \right) + \left( {b \times x} \right) $ 

Take  $ x $  as a common factor from right hand side of expression  $ a{x^2} + bx = \left( {a \times x \times x} \right) + \left( {b \times x} \right) $ .

 $ a{x^2} + bx = x\left[ {\left( {a \times x} \right) + \left( b \right)} \right] $ 

 $ a{x^2} + bx = x\left( {ax + b} \right) $ 

Therefore  $ a{x^2} + bx $  can be factorized as  $ x\left( {ax + b} \right) $ .

2. Factorise the expression  $ 7{p^2} + 21{q^2} $ .

Ans: To factorise the expression  $ 7{p^2} + 21{q^2} $ write  $ 7{p^2} $  and  $ 21{q^2} $  as a product.

 $ 7{p^2} $ can be written as  $ 7 \times p \times p $ .

 $ 21{q^2} $ can be written as  $ 3 \times 7 \times q \times q $ .

Substitute  $ 7 \times p \times p $  for  $ 7{p^2} $  and  $ 3 \times 7 \times q \times q $  for  $ 21{q^2} $  in expression  $ 7{p^2} + 21{q^2} $ .

 $ 7{p^2} + 21{q^2} = \left( {7 \times p \times p} \right) + \left( {3 \times 7 \times q \times q} \right) $ 

Take 7 as a common factor from right hand side of expression  $ 7{p^2} + 21{q^2} = \left( {7 \times p \times p} \right) + \left( {3 \times 7 \times q \times q} \right) $ .

 $ 7{p^2} + 21{q^2} = 7\left[ {\left( {p \times p} \right) + \left( {3 \times q \times q} \right)} \right] $ 

 $ 7{p^2} + 21{q^2} = 7\left( {{p^2} + 3{q^2}} \right) $ 

Therefore  $ 7{p^2} + 21{q^2} $  can be factorized as  $ 7\left( {{p^2} + 3{q^2}} \right) $ .

3.  Factorise the expression  $ 2{x^3} + 2x{y^2} + 2x{z^2} $ .

Ans: To factorise the expression  $ 2{x^3} + 2x{y^2} + 2x{z^2} $ write  $ 2{x^3} $ ,  $ 2x{y^2} $  and  $ 2x{z^2} $  as a product.

 $ 2{x^3} $ can be written as  $ 2 \times x \times x \times x $ .

 $ 2x{y^2} $ can be written as  $ 2 \times x \times y \times y $ .

 $ 2x{z^2} $ can be written as  $ 2 \times x \times z \times z $ .

Substitute  $ 2 \times x \times x \times x $ for  $ 2{x^3} $ ,  $ 2 \times x \times y \times y $  for  $ 2x{y^2} $  and  $ 2 \times x \times z \times z $  for  $ 2x{z^2} $  in expression  $ 2{x^3} + 2x{y^2} + 2x{z^2} $ .

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

Take  $ 2x $  as a common factor from right hand side of expression  $ 2{x^3} + 2x{y^2} + 2x{z^2} = \left( {2 \times x \times x \times x} \right) + \left( {2 \times x \times y \times y} \right) + \left( {2 \times x \times z \times z} \right) $ .

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

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

Therefore  $ 2{x^3} + 2x{y^2} + 2x{z^2} $  can be factorized as  $ 2x\left( {{x^2} + {y^2} + {z^2}} \right) $ .

4.  Factorise the expression  $ a{m^2} + b{m^2} + a{n^2} + b{n^2} $ .

Ans: To factorise the expression  $ a{m^2} + b{m^2} + a{n^2} + b{n^2} $ write  $ a{m^2} $ ,  $ b{m^2} $ ,  $ a{n^2} $  and  $ b{n^2} $  as a product.

 $ a{m^2} $ can be written as  $ a \times m \times m $ .

 $ b{m^2} $ can be written as  $ b \times m \times m $ .

 $ a{n^2} $ can be written as  $ a \times n \times n $ .

 $ b{n^2} $ can be written as  $ b \times n \times n $ .

Thus expression formed is  $ a{m^2} + b{m^2} + a{n^2} + b{n^2} = \left( {a \times m \times m} \right) + \left( {b \times m \times m} \right) + \left( {a \times n \times n} \right) + \left( {b \times n \times n} \right) $ 

Take  $ {m^2} $  as a common factor from  $ a \times m \times m $ ,  $ b \times m \times m $  and  $ {n^2} $  as a common factor from  $ a \times n \times n $ ,  $ b \times n \times n $  from right hand side of expression  $ a{m^2} + b{m^2} + a{n^2} + b{n^2} = \left( {a \times m \times m} \right) + \left( {b \times m \times m} \right) + \left( {a \times n \times n} \right) + \left( {b \times n \times n} \right) $ .

 $ a{m^2} + b{m^2} + a{n^2} + b{n^2} = {m^2}\left[ {\left( a \right) + \left( b \right)} \right] + {n^2}\left[ {\left( a \right) + \left( b \right)} \right] $ 

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

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

Therefore  $ a{m^2} + b{m^2} + a{n^2} + b{n^2} $  can be factorized as  $ \left( {a + b} \right)\left( {{m^2} + {n^2}} \right) $ .

5. Factorise the expression  $ \left( {lm + l} \right) + m + 1 $ .

Ans: To factorise the expression  $ \left( {lm + l} \right) + m + 1 $ take  $ l $  as a common factor from  $ \left( {lm + l} \right) $  and simplify.

 $ \left( {lm + l} \right) + m + 1 = l\left( {m + 1} \right) + \left( {m + 1} \right) $ 

 $ \left( {lm + l} \right) + m + 1 = \left( {l + 1} \right)\left( {m + 1} \right) $ 

Therefore  $ \left( {lm + l} \right) + m + 1 $  can be factorized as  $ \left( {l + 1} \right)\left( {m + 1} \right) $ .

6. Factorise the expression  $ y\left( {y + z} \right) + 9\left( {y + z} \right) $ .

Ans: To factorise the expression  $ y\left( {y + z} \right) + 9\left( {y + z} \right) $ take  $ \left( {y + z} \right) $  as a common factor from the expression and simplify.

 $ y\left( {y + z} \right) + 9\left( {y + z} \right) = \left( {y + z} \right)\left( {y + 9} \right) $ 

Therefore  $ y\left( {y + z} \right) + 9\left( {y + z} \right) $  can be factorized as  $ \left( {y + z} \right)\left( {y + 9} \right) $ .

7. Factorise the expression  $ 5{y^2} - 20y - 8z + 2yz $ .

Ans: To factorise the expression  $ 5{y^2} - 20y - 8z + 2yz $ take  $ 5y $  as a common factor from  $ 5{y^2} $ ,  $ 20y $  and  $ 2z $  as a common factor from  $  - 8z $ ,  $ 2yz $  from the expression  $ 5{y^2} - 20y - 8z + 2yz $ and simplify.

 $ 5{y^2} - 20y - 8z + 2yz = 5y\left( {y - 4} \right) + 2z\left( { - 4 + y} \right) $ 

 $ 5{y^2} - 20y - 8z + 2yz = 5y\left( {y - 4} \right) + 2z\left( {y - 4} \right) $ 

 $ 5{y^2} - 20y - 8z + 2yz = \left( {y - 4} \right)\left( {5y + 2z} \right) $ 

Therefore  $ 5{y^2} - 20y - 8z + 2yz $  can be factorized as  $ \left( {y - 4} \right)\left( {5y + 2z} \right) $ .

8. Factorise the expression  $ 10ab + 4a + 5b + 2 $ .

Ans: To factorise the expression  $ 10ab + 4a + 5b + 2 $ take  $ 2a $  as a common factor from  $ 10ab $ ,  $ 4a $  and 1 as a common factor from  $ 5b $ , 2 from the expression  $ 10ab + 4a + 5b + 2 $ and simplify.

 $ 10ab + 4a + 5b + 2 = 2a\left( {5b + 2} \right) + 1\left( {5b + 2} \right) $ 

 $ 10ab + 4a + 5b + 2 = \left( {5b + 2} \right)\left( {2a + 1} \right) $ 

Therefore  $ 10ab + 4a + 5b + 2 $  can be factorized as  $ \left( {5b + 2} \right)\left( {2a + 1} \right) $ .

9. Factorise the expression  $ 6xy - 4y + 6 - 9x $ .

Ans: To factorise the expression  $ 6xy - 4y + 6 - 9x $ take  $ 2y $  as a common factor from  $ 6xy $ ,  $ 4y $  and 3 as a common factor from  $ 9x $ , 6 from the expression  $ 6xy - 4y + 6 - 9x $ and simplify.

 $ 6xy - 4y + 6 - 9x = 2y\left( {3x - 2} \right) + 3\left( {2 - 3x} \right) $ 

 $ 6xy - 4y + 6 - 9x = 2y\left( {3x - 2} \right) - 3\left( {3x - 2} \right) $ 

 $ 6xy - 4y + 6 - 9x = \left( {3x - 2} \right)\left( {2y - 3} \right) $ 

Therefore  $ 6xy - 4y + 6 - 9x $  can be factorized as  $ \left( {3x - 2} \right)\left( {2y - 3} \right) $ .

4. Factorise the following.

1. Factorise the expression  $ {a^4} - {b^4} $ .

Ans: To factorise $ {a^4} - {b^4} $ rewrite the expression as the square terms.

 $ {a^4} $ can be written as  $ {\left( {{a^2}} \right)^2} $ .

 $ {b^4} $ can be written as  $ {\left( {{b^2}} \right)^2} $ .

Substitute  $ {\left( {{a^2}} \right)^2} $  for  $ {a^4} $  and  $ {\left( {{b^2}} \right)^2} $  for  $ {b^4} $  in expression  $ {a^4} - {b^4} $ .

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

Simplify  $ {a^4} - {b^4} = {\left( {{a^2}} \right)^2} - {\left( {{b^2}} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Again simplify  $ \left( {{a^2} - {b^2}} \right) $  using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Therefore  $ {a^4} - {b^4} $  can be factorized as  $ \left( {{a^2} + {b^2}} \right)\left( {a - b} \right)\left( {a + b} \right) $ .

2. Factorise the expression  $ {p^4} - 81 $ .

Ans: To factorise $ {p^4} - 81 $ rewrite the expression as the square terms.

 $ {p^4} $ can be written as  $ {\left( {{p^2}} \right)^2} $ .

81 can be written as  $ {\left( {{3^2}} \right)^2} $ .

Substitute  $ {\left( {{p^2}} \right)^2} $  for  $ {p^4} $  and  $ {\left( {{3^2}} \right)^2} $  for 81 in expression  $ {p^4} - 81 $ .

 $ {p^4} - 81 = {\left( {{p^2}} \right)^2} - {\left( {{3^2}} \right)^2} $ 

Simplify  $ {p^4} - 81 = {\left( {{p^2}} \right)^2} - {\left( {{3^2}} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Again simplify  $ \left( {{p^2} - {3^2}} \right) $  using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

 $ {p^4} - 81 = \left( {{p^2} + 9} \right)\left( {p - 3} \right)\left( {p + 3} \right) $ 

Therefore  $ {p^4} - 81 $  can be factorized as  $ \left( {{p^2} + 9} \right)\left( {p - 3} \right)\left( {p + 3} \right) $ .

3. Factorise the expression  $ {x^4} - {\left( {y + z} \right)^4} $ .

Ans: To factorise $ {x^4} - {\left( {y + z} \right)^4} $ rewrite the expression as the square terms.

 $ {x^4} $ can be written as  $ {\left( {{x^2}} \right)^2} $ .

 $ {\left( {y + z} \right)^4} $ can be written as  $ {\left( {{{\left( {y + z} \right)}^2}} \right)^2} $ .

Substitute  $ {\left( {{x^2}} \right)^2} $  for  $ {x^4} $  and  $ {\left( {{{\left( {y + z} \right)}^2}} \right)^2} $  for  $ {\left( {y + z} \right)^4} $  in expression  $ {x^4} - {\left( {y + z} \right)^4} $ .

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

Simplify  $ {x^4} - {\left( {y + z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {y + z} \right)}^2}} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Again simplify  $ \left( {{x^2} - {{\left( {y + z} \right)}^2}} \right) $  using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

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

Therefore  $ {x^4} - {\left( {y + z} \right)^4} $  can be factorized as  $ \left( {{x^2} + {{\left( {y + z} \right)}^2}} \right)\left( {x - y - z} \right)\left( {x + y + z} \right) $ .

4. Factorise the expression  $ {x^4} - {\left( {x - z} \right)^4} $ .

Ans: To factorise $ {x^4} - {\left( {x - z} \right)^4} $ rewrite the expression as the square terms.

 $ {x^4} $ can be written as  $ {\left( {{x^2}} \right)^2} $ .

 $ {\left( {x - z} \right)^4} $ can be written as  $ {\left( {{{\left( {x - z} \right)}^2}} \right)^2} $ .

Substitute  $ {\left( {{x^2}} \right)^2} $  for  $ {x^4} $  and  $ {\left( {{{\left( {x - z} \right)}^2}} \right)^2} $  for  $ {\left( {x - z} \right)^4} $  in expression  $ {x^4} - {\left( {x - z} \right)^4} $ .

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

Simplify  $ {x^4} - {\left( {x - z} \right)^4} = {\left( {{x^2}} \right)^2} - {\left( {{{\left( {x - z} \right)}^2}} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

Again simplify  $ \left( {{x^2} - {{\left( {x - z} \right)}^2}} \right) $  using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

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

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

Therefore  $ {x^4} - {\left( {x - z} \right)^4} $  can be factorized as  $ z\left( {2{x^2} - 2xz + {z^2}} \right)\left( {2x - z} \right) $ .

5. Factorise the expression  $ {a^4} - 2{a^2}{b^2} + {b^4} $ .

Ans: To factorise $ {a^4} - 2{a^2}{b^2} + {b^4} $ rewrite the expression as the square terms.

Rewrite  $ {a^4} - 2{a^2}{b^2} + {b^4} $ as  $ {\left( {{a^2}} \right)^2} - \left( {2{a^2}{b^2}} \right) + {\left( {{b^2}} \right)^2} $ .

$ {\left( {{a^2}} \right)^2} - \left( {2{a^2}{b^2}} \right) + {\left( {{b^2}} \right)^2} $ can be written as  $ {\left( {{a^2} - {b^2}} \right)^2} $  using algebraic identity  $ {\left( x \right)^2} - 2xy + {\left( y \right)^2} = {\left( {x - y} \right)^2} $ .

Simplify  $ {a^4} - 2{a^2}{b^2} + {b^4} = {\left( {{a^2} - {b^2}} \right)^2} $ using algebraic identity  $ {x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) $ .

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

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

Therefore  $ {a^4} - 2{a^2}{b^2} + {b^4} $  can be factorized as  $ {\left( {a - b} \right)^2}{\left( {a + b} \right)^2} $ .

5. Factorise the following.

1. Factorise the expression  $ {p^2} + 6p + 8 $ .

Ans: To factorise $ {p^2} + 6p + 8 $ , split the middle term such that the product of terms is equal to the product of the first term and the last term of expression and the sum is equal to the middle term of expression.

Write  $ 6p $  as  $ 4p + 2p $  because  $ 4p + 2p $  gives  $ 6p $  and  $ 4p \times 2p $ gives  $ 8{p^2} $ .

 $ {p^2} + 6p + 8 = {p^2} + 4p + 2p + 8 $ 

Take  $ p $  as a common factor from  $ {p^2} $ ,  $ 4p $  and 2 as a common factor from  $ 2p $ , 8 in expression  $ {p^2} + 6p + 8 = {p^2} + 4p + 2p + 8 $ and simplify.

 $ {p^2} + 4p + 2p + 8 = p\left( {p + 4} \right) + 2\left( {p + 4} \right) $ 

 $ {p^2} + 6p + 8 = \left( {p + 4} \right)\left( {p + 2} \right) $ 

Therefore  $ {p^2} + 6p + 8 $ can be factorized as  $ \left( {p + 4} \right)\left( {p + 2} \right) $ .

2. Factorise the expression  $ {q^2} - 10q + 21 $ .

Ans: To factorise $ {q^2} - 10q + 21 $ split the middle term such that the product of terms is equal to the product of the first term and the last term of expression and the sum is equal to the middle term of expression.

Write  $  - 10q $  as  $  - 7q - 3q $  because  $  - 7q - 3q $  gives  $  - 10q $  and  $ \left( { - 7q} \right) \times \left( { - 3q} \right) $ gives  $ 21{q^2} $ .

 $ {q^2} - 10q + 21 = {q^2} - 7q - 3q + 21 $ 

Take  $ q $  as a common factor from  $ {q^2} $ ,  $ 7q $  and  $  - 3 $  as a common factor from  $  - 3q $ , 21 in expression  $ {q^2} - 10q + 21 = {q^2} - 7q - 3q + 21 $ and simplify.

 $ {q^2} - 7q - 3q + 21 = q\left( {q - 7} \right) - 3\left( {q - 7} \right) $ 

 $ {q^2} - 10q + 21 = \left( {q - 7} \right)\left( {q - 3} \right) $ 

Therefore  $ {q^2} - 10q + 21 $ can be factorized as  $ \left( {q - 7} \right)\left( {q - 3} \right) $ .

3. Factorise the expression  $ {p^2} + 6p - 16 $ .

Ans: To factorise $ {p^2} + 6p - 16 $ split the middle term such that the product of terms is equal to the product of the first term and the last term of expression and the sum is equal to the middle term of expression.

Write  $ 6p $  as  $ 8p - 2p $  because  $ 8p - 2p $  gives  $ 6p $  and  $ 8p \times \left( { - 2p} \right) $ gives  $  - 16{p^2} $ .

 $ {p^2} + 6p - 16 = {p^2} + 8p - 2p - 16 $ 

Take  $ p $  as a common factor from  $ {p^2} $ ,  $ 8p $  and  $  - 2 $  as a common factor from  $  - 2p $ ,  $  - 16 $  in expression  $ {p^2} + 6p - 16 = {p^2} + 8p - 2p - 16 $ and simplify.

 $ {p^2} + 8p - 2p - 16 = p\left( {p + 8} \right) - 2\left( {p + 8} \right) $ 

 $ {p^2} + 6p - 16 = \left( {p + 8} \right)\left( {p - 2} \right) $ 

Therefore  $ {p^2} + 6p - 16 $ can be factorized as  $ \left( {p + 8} \right)\left( {p - 2} \right) $ .

NCERT Solutions for Class 8 Maths Chapter 12 Factorisation Exercise 12.2

Opting for the NCERT solutions for Ex 12.2 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.2 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Factorisation textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 12 Exercise 12.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 12 Exercise 12.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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