**Level – 01**

1 ) Factorise : x^{2} + 7x + 12

Solution : x^{2} + 7x + 12

= x^{2} + (4 + 3 ) x + 12

= x^{2} + 4x + 3x + 12

= x ( x + 4 ) + 3 ( x + 4 )

= ( x + 4 ) ( x + 3 )

2 ) Factorise : x^{2} – x – 12

Solution : x^{2} – x – 12

= x^{2} – ( 4 – 3 )x – 12

= x^{2} – 4x + 3x- 12

= x ( x – 4 ) + 3 ( x – 4 )

= ( x + 4 ) ( x + 3 )

3 ) Factorise : x^{2} + 7x + 6

Solution : x^{2} + 7x + 6

= x^{2} + ( 6 + 1 )x + 6

= x^{2} + 6x + x + 6

= x ( x + 6 ) + 1 ( x + 6 )

= ( x + 6 ) ( x + 1 )

4 ) Factorise : x^{2} + 9x + 8

Solution : x^{2} + 9x + 8

= x^{2} + ( 8 + 1 )x + 8

= x^{2} + 8x + x + 8

= x ( x + 8 ) + 1 ( x + 8 )

= ( x + 8 ) ( x + 1 )

5 ) Factorise : x^{2} – 2x – 8

Solution : x^{2} – 2x – 8

= x^{2} – ( 4 – 2 ) x – 8

= x^{2} – 4x + 2x – 8

= x ( x – 4 ) + 2 ( x – 4 )

= ( x – 4 ) ( x + 2 )

6 ) Factorise : x^{2} + 7x – 8

Solution : x^{2} + 7x – 8

= x^{2} + ( 8 – 1 )x – 8

= x^{2} + 8x – x – 8

= x ( x – 8 ) – 1 ( x – 8 )

= ( x – 8 ) ( x – 1 )

7 ) Factorise : x^{2} – 7x – 8

Solution : x^{2} – 7x – 8

= x^{2} – ( 8 – 1 ) x – 8

= x^{2} – 8x + x – 8

= x ( x – 8 ) + 1 ( x – 8 )

= ( x – 8 ) ( x + 1 )

8 ) Factorise : x^{2} + 40x – 129

Solution : x^{2} + 40x – 129

= x^{2} + ( 43 – 3 )x – 129

= x^{2} + 43x -3x – 129

= x ( x + 43 ) – 3 ( x + 43 )

= ( x + 43 ) ( x – 3 )

9 ) Factorise : x^{2} + 46x + 129

Solution : x^{2} + 46x + 129

= x^{2} + ( 43 + 3 ) x + 129

= x^{2} + 43x + 3x + 129

= x ( x + 43 ) + 3 ( x + 43 )

= ( x + 43 ) ( x + 3 )

10 ) Factorise : x^{2} + 7x + 12

Solution : x^{2} + 7x + 12

= x^{2} + ( 4 +3 ) x + 12

= x^{2} +4x + 3x + 12

= x ( x + 4 ) + 3 ( x + 4 )

= ( x + 4 ) ( x + 3 )

11 ) Factorise : x^{2} + 12x + 35

Solution : x^{2} + 12x + 35

= x^{2} + ( 7 + 5 )x + 35

= x^{2} + 7x + 5x + 35

= x ( x + 7 ) + 5 ( x + 7 )

= ( x + 7 ) ( x + 5 )

12 ) Factorise : x^{2} + 2x – 35

Solution : x^{2} + 2x – 35

= x^{2} + ( 7 – 5 )x – 35

= x^{2} + 7x – 5x – 35

= x ( x + 7 ) – 5 ( x + 7 )

= ( x + 7 ) ( x – 5 )

13 ) Factorise : x^{2} – 2x – 35

Solution : x^{2} – 2x – 35

= x^{2} – ( 7 – 5 )x – 35

= x^{2} – 7x + 5x – 35

= x ( x – 7 ) + 5 ( x – 7 )

= ( x- 7 ) ( x + 5 )

14 ) Factorise : x^{2} + 36x + 35

Solution : x^{2} + 36x + 35

= x^{2} + ( 35 + 1 )x + 35

= x^{2} + 35x + x + 35

= x ( x + 35 ) + 1 ( x + 35 )

= ( x + 35 ) ( x + 1 )

15 ) Factorise : x^{2} – 36x + 35

Solution : x^{2} – 36x + 35

= x^{2} – ( 35 + 1 )x + 35

= x^{2} – 35x – x + 35

= x ( x – 35 ) – 1 ( x – 35 )

= ( x – 35 ) ( x – 1 )

**Level – 02**

**Level – 02**

**1 ) ( a + b ) ^{2} – 5a – 5b + 6**

= ( a + b )^{2} – 5(a + b) + 6

Let ( a + b ) = m

Then, the given algebraic expression becomes,

m^{2} -5m + 6

= m^{2} – ( 3 + 2 )m + 6

= m^{2} – 3m – 2m + 6

= m ( m – 3 ) – 2( m – 3 )

= ( m – 3 ) ( m – 2 )

Now, putting m = ( a + b ), we get

= ( a + b -3 ) ( a + b – 2 )

**2 ) ( x ^{2 }– 2x )^{2} + 5 ( x^{2} – 2x )^{2} – 36**

Let ( x^{2 }– 2x )^{2} = p, then the given algebraic expression becomes,

p^{2} + 5p – 36

= p^{2} + ( 9 – 4 )p – 36

=p^{2} + 9p – 4p – 36

= p( p + 9 ) – 4( p + 9 )

= ( p + 9 ) ( p – 4 )

Now, putting p = ( x^{2 }– 2x )^{2} , we get

= ( x^{2 }– 2x + 9 ) ( x^{2 }– 2x – 4 )

**3 ) ( p ^{2} – 3q^{2} ) ^{2} – 16 ( p^{2} – 3q^{2} ) + 63**

Let p^{2} – 3q^{2} = a, then the given algebraic expression becomes

a^{2} – 16 a + 63

= a^{2} – ( 7+ 9 ) a + 63

= a^{2} – 7a – 9a + 63

= a ( a – 7 ) – 9 ( a – 7 )

= ( a – 7 ) ( a – 9 )

Putting a = p^{2} – 3q^{2} , we get

= ( p^{2} – 3q^{2} – 7 ) ( p^{2} – 3q^{2} – 9 )

**Level – 03**

**4 ) a ^{4} + 4a^{2} – 5**

= a^{4} + ( 5 – 1 )a^{2} – 5

= a^{4} + 5a^{2} – a^{2 }– 5

= a^{2} ( a^{2 }+ 5 ) – 1( a^{2} + 5 )

= ( a^{2 }+ 5 )( a^{2} – 1 )

= ( a^{2 }+ 5 ){(a)^{2} – ( 1 )^{2}}

= ( a^{2} + 5 ) ( a + 1 ) ( a – 1 )

**5 ) a ^{6} + 3a^{3}b^{3} – 40b^{6}**

= a^{6} + ( 8 – 5 ) a^{3}b^{3} – 40b^{6}

= a^{6} + 8a^{3}b^{3} – 5a^{3}b^{3} – 40b^{6}^{}

= a^{3} ( a^{3} + 8b^{3} ) – 5b^{3} ( a^{3} – 8b^{3} )

= ( a^{3} + 8 ) ( a^{3} – 5b^{3} )

= {(a)^{3} + (2 )^{3} } ( a^{3} -5b^{3} )

= ( a + 2 ){ (a)^{2} -(a) (2) + (2)^{2}} ( a^{3} – 5b^{3} )

= ( a + 2 )( a^{2 }– 2a + 4 )( a^{3} – 5b^{3} )

**Level – 04**

**6 ) ( x + 1 )( x + 3 )( x – 4 )( x – 6 ) + 24**

= { ( x + 1 )( x – 4 )}{( x + 3 )(x – 6 )} + 24

= ( x^{2} – 4x + x – 4 )( x^{2} + 3x – 6x – 18 ) + 24

= ( x^{2} – 3x – 4 ) ( x^{2} – 3x – 18 ) + 24

Let x^{2} – 3x = a, then

( a – 4 ) ( a – 18 ) + 24

= a^{2} – 4a – 18a + 72 + 24

= a^{2} – 22 a + 96

= a^{2} – ( 16 + 6 ) a + 96

= a ^{2} – 16a – 6a + 96

= a( a – 16 ) – 6 ( a – 16)

= ( a – 16 ) ( a – 6 )

Putting a = x^{2} – 3x, we get

= ( x^{2} – 3x – 16 ) ( x^{2} – 3x – 6 )

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