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Factorisation

Factorisation of Algebraic Expressions by the Method of Common Factors

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 numbers that fulfil these two conditions are 3 and 5.

Hence, we can write x2 + 8x + 15 as

x2 + (5 + 3) x + 5 × 3

= (x2 + 5x) + (3 × x + 5 × 3)

= x(x + 5) + 3(x + 5)

= (x + 5) (x + 3)         {Taking a common factor from each group}

Let us practise some more questions based on this concept.

Example 1:

Factorise the expression x2x − 42.

Solution:

On comparing x2x − 42 with x2 + (a + b) x + ab, we obtain a + b = −1 and ab = −42.

Here, we have to find two numbers, a and b, such that their sum is −1 and their product is −42.

Since the product (−1) of the numbers a and b is negative and their sum (−42) is also negative, we have to choose two numbers such that the bigger number is negative and the smaller number is positive.

The numbers that fulfil these conditions are −7 and 6.

Thus, obtain a + b = −1 and ab = − 42.
Hence,
x2x − 42 = x2 − 7x + 6x − 42
                   = x (x − 7) + 6 (x − 7)
                   = (x − 7) (x + 6)

Example 2:

Factorise the expression y213y + 36.

Solution:

On comparing the expression y2 − 13y + 36 with y2 + (a + b) y + ab, we obtain

a + b = −13 and ab = 36.

Since the product of the numbers a and b is positive and their sum is negative, we have to choose two negative numbers.

The numbers that fulfil these conditions are −9 and −4.

y2 − 13y + 36 = y2 − 4y − 9y + 36
                           = y (y − 4) − 9 (y − 4) = (y − 4) (y − 9)

 

Example 3:

Factorise the expression 7(x2 x)(2x2 2x 1) 42.

 

Solution:

We have

7(x2 x)(2x2 2x 1) 42

= 7(x2 x){2(x2 x) 1)} 42

Let x2 x = m

Thus, 7(x2 x){2(x2 x) 1)} 42 = 7m(2m 1) 42

                                                        = 14m2 7m 42

                                                        = 7(2m2 m 6)

                                                        = 7(2m2 4m + 3m 6)

                                                        = 7{2m (m 2) + 3(m 2)}

                                                        = 7(2m + 3)(m 2)

On re-substituting the value of m, we obtain

7(x2 x){2(x2 x) 1)} 42 = 7{2(x2 x) + 3)}( x2 x 2)

                                               = 7(2x2 2x + 3)( x2 x 2)


We know how to divide a given polynomial by a monomial. But how do we go about dividing a polynomial by another polynomial. The given video will help you understand this type of division.

Let us discuss some more examples, in which we will not only divide binomials, but also divide other types of polynomials.

Example 1:

Factorise the following expressions and divide as directed.

(i) 6ab (9a2 16b2) ÷ 2ab (3a + 4b)

(ii) (x214x − 32) ÷ (x + 2)

(iii) 36abc (5a − 25) (2b − 14) ÷ 24(a − 5) (b − 7)

Solution:

(i) We can factorise the given expressions as

6ab (9a2 − 16b2) = 2 × 3 × a × b [(3a)2 − (4b)2]

= 2 × 3 × a × b [(3a + 4b) (3a − 4b)] [a2b2 = (a + b)(ab)]

And, 2ab (3a + 4b) = 2 × a × b × (3a + 4b)

= 3 × (3a − 4b)

= 9a − 12b

∴ 6ab (9a2 − 16b2) ÷ 2ab (3a + 4b) = 9a − 12b

(ii) We can factorise the given expression as

x2 − 14x − 32 = x2 − (16 − 2) x − 32

= x2 − 16x + 2x − 32

= x (x − 16) + 2 (x − 16)

= (x − 16) (x + 2)

= (x + 2) = (x + 2)

= x − 16

∴ (x2 − 14x − 32) ÷ (x + 2) = (x − 16)

(iii) We can factorise the given expression as

Hence, 36abc (5a − 25) (2b − 14) ÷ 24(a − 5) (b − 7) = 15abc

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 identit…

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