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# Factorisation using identities

In the same way, algebraic expressions also have factors that divide the expression exactly. Given an algebraic expression, the factors of that algebraic expression are two or more expressions whose product is the given expression.

Factorisation using identities

A factor of a number is the number that exactly divides a given number without leaving any remainder. For example, the factors of 4 are 1, 2 and 4; the factors of 12 are 1, 2, 3, 4, 6 and 12. And, the number can be expressed as product of its factors such as 12 = 1 × 12 or 2 × 6 or 3 × 4.

In the same way, algebraic expressions also have factors that divide the expression exactly. Given an algebraic expression, the factors of that algebraic expression are two or more expressions whose product is the given expression. For example, x2 y = x × x ×y , hence the factors of the algebraic expression x2 y are x, x and y.

Consider the algebraic expression x2 4 .

Now, x 2 4 = x2 22 (x + 2)(x − 2) [because a 2 b 2 =(a + b)(a b) ]

Therefore, (x +2) and (x − 2) are the factors of x2 4 .

Obviously, 1 is also a factor of x2 4.

We know that, x2 = x × x. Similarly, (x + 5)2 =(x + 5)×(x +5) .

Thus, the factors of (x + 5)2 are (x + 5) and (x + 5).

The process of writing an algebraic expression as the product of its factors is called factorisation.

We shall learn more to factorise algebraic expressions, using the basic identities.

Example 3.5

Express the following algebraic expressions as the product of its factors:

(i) ab2   �(ii) 3pq3   (iii) 12m2n2 p

Solution:

(i) ab2 = a ×b×b .

(ii) − 3 pq3 =− 3 ×p× q × q ×q .

(iii) 12 m2n2p = 4 × 3 × m × m × n × n × p = 2 × 2× 2 × 3 × m × m × n × n × p.

Example 3.6

Factorise by using the identity: a2 b2= (a +b)(ab)

(i) a2 − 1   (ii) 9k2 25 (iii) 64x2 − 81y2

Solution:

(i) a2 1 =a2 12 = (a + 1)(a − 1) [since 12=1×1=1]

Therefore, the factors of a2 − 1 are (a + 1) and (a − 1).

(ii) 9k2 − 25 = (32 × k2 ) − 52 = (3k)2 − 52  [since am × b m =(a×b)m ]

= (3k+ 5)(3k − 5)  [ since, a2 b2 = (a + b)(a b)]

Therefore, the factors 9k2 − 25 are (3k+5)and(3k-5).

(iii) 64x2 81y2 =(8 2 ×x 2 )− (9 2 × y2 )= (8x)2 (9 y)2

= (8x +9 y)(8x 9 y)

Therefore, the factors of 64x 2 − 81y2 are (8x + 9y) and (8x – 9y).

Example 3.7

Factorise 9x2 + 30xy +25y2

Solution:

9x2+30xy+25y2=(32×x2)+(2×3×5) ×xy+(52×y2)

=(3x)2+2×(3x) ×(5y) ×(5y)2

This expression is in the form of identity a2 + 2ab + b2 = (a+b)2.

Hence, (3x)2 +2× (3x)×(5y) +(5y)2 = (3x +5y)2

=(3x+5y)(3x+5y).

Therefore, the factors of 9x2 + 30xy +25y2 are (3x+5y) and (3x+5y).

Example 3.8

Factorise 4x2 4 xy + y2

Solution:

4x2 4 xy + y2 = 2 2 ×x 2 (2×2)× xy + y2

= (2x )2 2× (2x)×y + y2

This expression is in the form of identity a2 – 2ab + b2 = (a–b)2.

Say 2x= a and y =b, thus we have,

(2x )2 2× (2x)×y + y2 = (2x y)2

= (2xy)(2xy)

Therefore,are the factors of 4x2 4xy + y2 are (2xy) and (2xy).

Think

Can we factorise the following expressions using any basic identities? Justify your answer.

(i) x2 + 5 x + 4

Solution:

(i) x2 + 5x + 4 = x2 + (1 + 4)x + (1 × 4)

Which is of the form x2 + (a + b)x + ab

= (x + a) (x + b)

x2 + (1+4)x +(1 × 4) = (x + 1) (x + 4)

x2 + 5x + 4 = (x + 1) (x + 4)

(ii) x2 5 x + 4

x2  – 5x + 4 = x2 + ((–1) + (–4))x + (–1)(–4)

Which is of the form x2 + (a + b) x + ab

= (x + a) (x + b)

x2 + ((–1) + (–4))x + ((–1)(–4)) = (x + (–1)) (x + (–4)) = (x –1) (x – 4)

x2 – 5x + 4 = (x – 1) (x – 4))

Example 3.9

Factorise: x 2 2 xy + y2 z2

Solution:

x 2 2 xy + y2 z2 = ( x 22xy + y2 ) z2 =( x y )2 z2 [by identity-3]

Let, (xy) a and z = b.

Therefore, ( x y )2 z 2 = a 2 b2

We know that, a2 b2 = ( a + b)(ab) [Identity-4]

Hence, x2 − 2 xy + y2 −z2 = (x − y + z)(x − y − z).

Therefore, the factors of x2 – 2xy +y2 − z2  are (xy + z) and (xyz).

Activity

Here is an interesting number game to thrill your friends. Follow the steps.

1. Ask your friend to think a number in mind. Let the number be in between 1 to 10. (for example, let the number be 5).

2. Double the number and add to its square. (So, 10 + 25 = 35).

3. Add one to the result. (So, 35 + 1 = 36).

4. Ask the final resultant number.

5. The number will be a perfect square. Consider its base in the exponent form and reduce 1 from it. The result is the number in your friend’s mind. ( 36 = 62 6 is the base; so, 6 1 =5 ).

Remember the perfect square numbers and their corresponding base numbers 11, 4 22, 9 32, 16 42, 25 52, 36 62, 49 72, 64 82, 8192, 100 102

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