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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)

Division is exactly the opposite of multiplication. For example, if 4 × 5 = 20, then it is also correct to say that 20 ÷ 5 = 4 and 20 ÷ 4 = 5.

We can use the same concept to divide algebraic expressions.

Let us try to divide the expression 3x by 3 and x.

We can factorize 3x as 3 × x.

This means that 3x is a product of 3 and x.

∴ 3x ÷ 3 = x and 3x ÷ x = 3

Each of the expressions i.e., 3, x, and 3x is a monomial. Hence, these were examples of division of monomials by monomials.

When we divide a monomial by another monomial, we first need to factorise each monomial. Next, we divide the monomial by cancelling the common factors.

To understand the concept better, look at the following video.

How do we divide a polynomial by a monomial? For this, we have two methods.

To understand both the methods, look at the following video.

Let us discuss some more examples based on the two methods we just discussed.

Example 1:

Divide the following expressions:

(i). 27x2y2z ÷ 27xyz

(ii). 144pq2r ÷ (− 48qr)

Solution:

(i). 27x2y2z ÷ 27xyz

Dividend = 27x2y2z

= 3 × 3 × 3 × x × x × y × y × z

Divisor = 27xyz

= 3 × 3 × 3 × x × y × z = xy

∴ 27x2y2z ÷ 27xyz = xy

(ii). 144pq2r ÷ (− 48qr)

Dividend = 144 pq2r

= 2 × 2 × 2 × 2 × 3 × 3 × p × q × q × r

Divisor = − 48 qr

= − 2 × 2 × 2 × 2 × 3 × q × r ∴ 144pq2r ÷ (− 48qr) = − 3pq

Example 2:

Carry out the following divisions:

(i). (x3y6x6y3) ÷ x3y3

(ii). 26xy (x + 5) ÷ 13xy

(iii). 27 (− a2bc + ab2cabc2) ÷ (− 3abc)

Solution:

(i). (x3y6x6y3) ÷ x3y3

Dividend = x3y6x6y3

= x3y3 (y3x3)

Divisor = x3y3 = y3x3

Another method of simplifying this expression is ∴ (x3y6x6y3) ÷ x3y3 = y3x3

(ii). Dividend = 26xy (x + 5)

= 2 × 13 × x × y × (x + 5)

Divisor = 13xy

= 13 × x × y

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