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Factorisation

Factorisation of Algebraic Expressions by the Method of Common Factors

We know the identities

(i). a2 + 2ab + b2 = (a + b)2

(ii). a2 − 2ab + b2 = (ab)2

(iii). a2b2 = (a + b) (ab)

We can use these identities to factorise algebraic expressions as well. Let us discuss each identity one by one.

Application of the identity a2 + 2ab + b2 = (a + b)2 to factorise an algebraic expression

Let us factorise the expression x2 + 6x + 9.

In this expression, the first term is the square of x, the last term is the square of 3, and the middle term is positive and is twice the product of x and 3.

Thus, x2 + 6x + 9 can be written as

x2 + 6x + 9 = (x)2 + 2 × x × 3 + (3)2

The right hand side of this expression is in the form of a2 + 2ab + b2, where a = x and b = 3.

We know the identity a2 + 2ab + b2 = (a + b)2.

(x)2 + 2 × x × 3 + (3)2 = (x + 3)2

Thus, x2 + 6x + 9 = (x + 3)2.

Application of the identity a2 − 2ab + b2 = (ab)2 to factorise an algebraic expression

Let us factorise the expression 9y2 − 12y + 4.

In this expression, the first term is the square of 3y, the last term is the square of 2, and the middle term is negative and is twice of the product of 3y and 2.

Thus, 9y2 − 12y + 4 can be written as

9y2 − 12y + 4 = (3y)2 − 2 × 3y × 2 + (2)2

The right hand side of this expression is in the form of a2 − 2ab + b2, where a = 3y and

b = 2.

We know the identity a2 − 2ab + b2 = (ab)2.

(3y)2 − 2 × 3y × 2 + (2)2 = (3y − 2)2

Thus, 9y2 − 12y + 4 = (3y − 2)2.

Application of the identity a2b2 = (a + b) (ab) to factorise an algebraic expression

We use this identity when an expression is given as the difference of two squares.

Let us factorise the expression x2 − 25.

We can write it as

x2 − 25 = (x)2 − (5)2

The right hand side of this expression is in the form of a2b2, where a = x and b = 5

On using the identity a2b2 = (a + b) (a b), we obtain

x2 − 25 = (x + 5) (x − 5)

Thus, (x + 5) and (x − 5) are the factors of x2 − 25.

To factorise an algebraic expression, we have to observe the given expression. If it has a form that fits the left hand side of one of the identities mentioned in the beginning, then the expression corresponding to the right hand side of the identity gives the desired factorisation.

Let us discuss some more examples based on what we have discussed so far.

Example 1:

Factorise the given expressions.

  1. 25x2 + 40xy + 16y2

  1. 81x3 + x − 18x2

  1. (p + 1)2 − (p − 1)2

  1. 16a2 − 25b2 + 60bc − 36c2

  1. al2bm2am2 + bl2

  1. 81x4 − 256 y4

  1. 16x4 − (3a + 5c)4

Solution:

(1)The given expression is 25x2 + 40xy + 16y2.

25x2 + 40xy + 16y2 = (5x)2 + 2 × 5x × 4y + (4y)2

= (5x + 4y)2 [a2 + 2ab + b2 = (a + b)2]

25x2 + 40xy + 16y2 = (5x + 4y)2

(2) The given expression is 81x3 + x − 18x2.

Here, x is a factor common to all terms in the expression.

81x3 + x − 18x2 = x (81x2 + 1 − 18x)

= x [(9x)2 + (1)2 − 2 × 9x ×1]

= x [9x − 1]2 [a2 − 2ab + b2 = (ab)2)

81x3 + x − 18x2 = x (9x − 1)2

(3) The given expression is (p + 1)2 − (p − 1)2.

On using the identity a2b2 = (a + b) (ab), we obtain

(p + 1)2 − (p − 1)2 = {(p + 1) + (p − 1)} {(p + 1) − (p − 1)}

= (p + 1 + p − 1) (p + 1 − p + 1)

= (2p) (2) = 4p

(p + 1)2 − (p − 1)2 = 4p

(4) The given expression is 16a2 − 25b2 + 60bc − 36c2.

16a2 − 25b2 + 60bc − 36c2

= 16a2 − (25b2 − 60bc + 36c2)

= (4a)2 − {(5b)2 − 2 (5b) (6c) + (6c)2}

= (4a)2 − (5b − 6c)2 [Using the identity a2 − 2ab + b2 = (ab)2]

= {(4a) + (5b − 6c)}{(4a) − (5b − 6c)}[Using the identity a2b2 = (a + b) (ab)]

= (4a + 5b − 6c) (4a − 5b + 6c)

(5) The given expression is al2bm2am2 + bl2.

al2bm2am2 + bl2

= al2am2 + bl2bm2 {Regrouping the terms}

= a (l2m2) + b (l2m2)

= (l2m2) (a + b)

= (l + m) (lm) (a + b) [Using the identity a2b2 = (a + b) (ab)]

(6) The given expression is 81x4 − 256y4.

81x4 − 256y4

(7) The given expression is 16x4 − (3a + 5c)4.

16x4 − (3a + 5c)4

How can we factorise the algebraic expression x2 + 8x + 15?

Note that we cannot express this expression as (a + b)2, since 15 is not the square of any natural number. What do we do in such a case?

In this case, we can use the identity x2 + (a + b) x + ab = (x + a) (x + b).

If we compare x2 + 8x + 15 with x2 + (a + b) x + ab, then we obtain a + b = 8 and ab = 15.

Hence, we need to find two numbers, a and b, such that their sum is 8 and their product is 15.

The only num…

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