Type 1: Factorisation by taking out the common factor from each term.
Type 2 : Factorisation by taking out the common binomial factor from each term
Type 3 : Factorisation by grouping
Type 4 : Factorisation using identities
Type 5 : Factorisation of the expression (ax2 + bx + c)

**Factorisation**

Expressing
any number as the product of two or more numbers is called as **factorisation.** The number 12 can be expressed
as the product of prime factors like 12 = 2 × 2 × 3. This is called prime factorisation.
How will you factorise an algebraic expression? Yes. Expressing an algebraic expression
as the product of two or more expressions is called the Factorisation.

**Note**

A number which is divisible by 1 and itself (or) A number which has
only 2 factors are called prime numbers. Example: 2, 3, 5, 7, 11, ...

A number which has more than 2 factors are called composite numbers.Example:
4, 6, 8, 9, 10, 12, ...

For example,
(i) *a* ^{2} −
*b*^{2} =
(*a* + *b*)(*a* − *b*)
Here, (*a* +
*b*) and (*a* − *b*) are the two factors
of *a* ^{2} −
*b*^{2}

(ii) 5*y* + 30 = 5( *y* + 6) , Here 5 and ( *y*
+
6) are the factors of 5*y* + 30

Any expression
can be Factorised as (1) × (expression)

For example,
*a*^{2} −
*b*^{2} can also be factorised as

(1)
×
(
*a* ^{2} −
*b*^{2} ) or
(
−1)
×
(*b*^{2} − *a*^{2}
)

because ‘1’
is a factor for all numbers and expressions

So, when
we factorise the expressions, follow the suitable type of factorisation given below
to get two or more factors other than 1. Stop doing the factorisation process once
you have taken out all the common factors from the expression and then list out
the factors.

**Type 1: Factorisation by taking out the common factor from
each term.**

**Type 2 : Factorisation by taking out the common binomial
factor from each term**

**Type 3 : Factorisation by grouping**

**Type 4 : Factorisation using identities **

**Type 5 : Factorisation of the expression (ax**2** + bx + c)**

** **

__Type 1:
Factorisation by taking out the common factor from each term.__

**Example 3.18**

Factorise:
4*x* ^{2} *y* + 8*xy*

*Solution:*

We have,
4*x* ^{2} *y* + 8*xy* This can be written
as,

= (2 × 2 × *x*
×
*x* × *y*
) +
(2 ×
2 ×
2 ×
*x* × *y*)

Taking out
the common factor 2, 2, *x* , *y* , we get

= 2 × 2 × *x*
×
*y*(*x*
+
2)

= 4*xy*(*x* + 2)

** **

__Type 2
: Factorisation by taking out the common binomial factor from each term__

**Example 3.19**

**(i) Factorise:
(2 x + 5)(x − y ) +
(4 y)(x −**

*Solution:*

We have (2*x* + 5)(*x* − *y* ) +
(4 *y*)(*x* − *y*)

Taking out
the common binomial factor (*x* −
*y*)

We get, (*x* − *y*)(
2*x* + 5 + 4 *y*)

**(ii) Factorise
3 n( p − 2) + 4(2 − p**

*Solution:*

*We have 3n( p *− *2) + 4(2 *− *p) (taking – as *common)

3*n*( *p*
−
2) −
4( *p* − 2)

Taking out
the common binomial factor ( *p* −
2)

We get, (
*p* − 2)(3*n* − 4)

** **

__Type 3
: Factorisation by grouping__

Sometimes,
the terms of a given expression are grouped suitably in such a way that they have
a common factor so that the factorisation is easy to take out common factor from
those terms.

**Example 3.20**

Factorise
: *x*^{2} +
*yz* + *xy* + *xz*

*Solution:*

We have,
*x*^{2} +
*yz* + *xy* + *xz*

Group the
terms suitably as,

= ( *x* ^{2} +
*xy* ) + ( *yz + xz*)

*= x *(*x
*+* y *)*
*+* z*(* y + x*)

*= x *(*x
*+* y *)*
*+* z*(*x + y*)* *(addition is
commutative)* *

= (*x* + *y*)[
*x* + *z*] [taking out the common factor (*x* + *y*) ]

** **

__Type 4
: Factorisation using identities__

(i) (*a* + *b*)^{2}
=
*a* ^{2} +
2*ab* + *b*^{2}

(ii) (*a* − *b*)^{2}
=
*a* ^{2} −
2*ab* + *b*^{2}

(iii) *a* ^{2} −
*b*^{2} =
(*a* + *b*)(*a* − *b*)

**Example 3.21**

Factorise
: *x* ^{2} +
8*x* +16

*Solution:*

Now, *x* ^{2} +
8*x* +16 can be written as
*x* ^{2} +
8*x* + 4^{2}

Comparing
this with *a* ^{2} +
2*ab* + *b*^{2}
=
(*a* + *b*)^{2}
we get *a *=* x *;* b *=* *4

(*x*^{2}) +
2(*x*)(4) +
(4)^{2} = ( *x* + 4)^{2}

*x *^{2}* *+* *8*x *+* *16* *=* *(* x *+* *4)^{2}

(*x+*4), (*x+*4) are the two factors.

**Example 3.22**

Factorise
49*x* ^{2} −
84*xy* + 36 *y*^{2}

*Solution:*

Now, 49*x* ^{2} −
84*xy* + 36 *y*^{2}

7^{2}
*x*^{2} −
84*xy* + 6^{2} *y*^{2} *=* (7*x*)^{2} −
84*xy* + (6 *y*)^{2}

Comparing
this with *a*^{2} −
2*ab* + *b*^{2
}= (*a* − *b*)^{2} we get *a* = 7*x* *,* *b* = 6 *y*^{}

(7*x*)^{2} −
2(7*x*)(6 *y*) + (6 *y*)^{2 }= (7*x* − 6 *y*)^{2}

∴
49*x*^{2} −
84*xy* + 36 *y*^{2}= (7*x* − 6 *y*)^{2}

(7*x*–6*y*),
(7*x*–6*y*) are the two factors.

**Try these**

Find the factors

**Example 3.23**

Factorise
: 49*x*^{2} −
64 *y*^{2}

*Solution:*

Now, 49*x*^{2} −
64 *y*^{2}

7^{2}
*x*^{2} −
8^{2} *y*^{2} = (7*x*)^{2} −
(8*y*)^{2}

Comparing
this with *a*^{2} −
*b*^{2} =
(*a* + *b*)(*a* − *b*)
we get *a* = 7*x* ,
*b* = 8*y*

(7*x*)^{2} −
(8 *y*)^{2} =
(7*x* + 8 *y*)(7*x* −
8 *y*)

(7*x*+8*y*),
(7*x*–8*y*) are the two factors.

** **

__Type 5
: Factorisation of the expression ( ax^{2}
+ bx + c)__

**Example 3.24**

Factorise
*x*^{2}+ 8*x*+15

*Solution:*

Given *x*^{2}* *+* *8*x *+15

This is in
the form of *ax*^{2} +
*bx* + *c*

We get *a* = 1,*b* = 8, *c* =
15

Now, the
product =
*a* × *c*
and sum =
*b b *=* *8

=1×15

= 15

= *x*^{2} +
8*x* +15

= *x*^{2}* *+* *3*x *+* *5*x *+* *15* *(the middle term 8*x
*can be written as 3*x*+5*x*)

= (*x*^{2} +
3*x*) + (5*x* +15)

= *x*(*x
*+* *3)* *+* *5(*x *+* *3)* *taking out the common
factor* x*+3 )

* x*^{2}* + *8*x
+ *15 = (* x + *3)(*x + *5)

Therefore,
(*x*+*3*), (*x*+5) are the two factors.

**Think**

* x*^{2}–4(*x*–2) = (*x*^{2}–4)(*x*–2)

Is this correct? If not correct it.

**Solution:**

Not correct

(3*a*)^{2} = 3^{2}*a*^{2} − 9*a*^{2}

*x*^{2} − 4(*x* − 2) = *x*^{2} – 4*x* + 8

**Example 3.25**

Factorise
7*c*^{2} +
2*c* – 5

*Solution:*

Given

This is in
the form of *ax*^{2}* *+* bx
*+* c*

We get *a* = 7, *b* = 2, *c* = −5

Now, the
product =
*a* × *c*
=
7 ×
(−5)
= –35 and sum *b* = 2

= 7*c* ^{2} +
2*c* – 5

= 7*c* ^{2} – 5*c* + 7*c*
– 5
(the middle term 2*c* can be written as –5*c*+7*c*)

= (7*c* ^{2} − 5*c*) + (7*c* − 5)

= *c* (7*c* − 5) + 1(7*c* − 5) (taking out the common
factor *7c*–5 )

= (7*c* − 5)(*c* + 1)

Therefore,
(*7c–*5), (*c*+1) are the two factors.

**Try these**

Factorise the following :

1) 3*y* + 6

2) 10*x*
^{2} +15*y*^{2 }

3) 7*m*(*m* − 5) + 1(5 − *m*)

4)
64 − *x*^{2}

5) *x*^{2}–3*x*+2

6) *y*^{2}–4*y*–32

7) *p*^{2}+2*p*–15

8) *m*^{2}+14*m*+48

9) *x*^{2}–*x*–90

10) 9*x*^{2}–6*x*–8

**Solution:**

**1.**** 3 y + 6**

3*y* + 6 = 3* × y* + 2 × 3

Taking out the common factor 3 from each term we get 3 (*y*
+ 2)

∴ 3*y* + 6 = 3(*y* + 2)

**2. ****10 x^{2} + 15y^{2}**

10*x*^{2} + 15*y*^{2 }= (2 × 5 × *x*
× *x*) + (3 × 5 × *y* × *y*)

Taking out the common factor 5 we have

10*x*^{2} + 15*y*^{2 }= 5(2*x*^{2} + 3*y*^{2})

**3. ****7 m (m − 5) + (15 – m)**

7*m* (*m* − 5) + (15 – *m*) = 7 *m* (*m*
− 5) + (−1 )(−5 + *m*)

= 7*m* (*m* − 5) − 1(*m* − 5)

Taking out the common binomial factor (*m* − 5) = (*m*
− 5) (7 *m* − 1)

**4.**** 64 − x^{2}**

64 − *x*^{2 }= 8^{2} − *x*^{2}

This is of the form *a*^{2} − *b*^{2}

Comparing with *a*^{2} − *b*^{2 }we
have *a* = 8, *b* = *x*

*a*^{2}− *b*^{2} = (*a + b*) (*a − b*)

64 − *x*^{2} = (8 + *x*) (8 − *x*)

**5. ***x*^{2}** – 3 x + 2**

*x*^{2} – 3*x* + 2** = ***x*^{2} − 2*x* − *x*
+ 2

= *x* (*x* − 2) − (*x* − 2)

= (*x* − 2) (*x* − l)

**6. ***y*^{2}** – 4 y – 32**

** ***y*^{2} – 4*y* – 32** **= *y*^{2} – 8*y* + 4*y*
− 32

= *y*(*y* − 8) + 4(*y* − 8)

= (*y* − 8) (*y* + 4)

**7. ***p*^{2}** + 2 p – 15**

*p*^{2} + 2*p* – 15 = *p*^{2}
+ 5*p* −3*p* − 15

= *p*(*p* + 5) −3 (*p* + 5)

= (*p* + 5) (*p* − 3)

**8. ***m*^{2}** + 14 m + 48**

*m*^{2} + 14 *m* + 48 = *m*^{2} + 8 *m* + 6 *m*
+ 48

= *m* (*m* + 8) + 6 (*m* + 8)

= (*m* + 6) (*m* + 8)

**9. ***x*^{2}** − x – 90**

*x*^{2} − *x* – 90** **= *x*^{2} − 10*x* + 9*x*
− 90

= *x* (*x* − 10) + 9 (*x − *10)

= (*x* + 9) (*x* − 10)

**10. ****9 x^{2} − 6x – 8**

** **9*x*^{2} − 6*x* – 8** = **9*x*^{2}
− 12*x* + 6*x* − 8

= 3*x* (3*x* − 4) + 2(3*x* − 4)

= (3*x* + 2) (3*x − *4)

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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