Factorization using Identities







Factorization using Identities :There are some identities and using that the factorization is much easier.

A number of expressions to be factorized are of the form or can be put into the form : a2 + 2ab + b2, a2 – 2ab + b2, a2 – b2 and x2 + (a + b) + ab. These expressions can be easily factorized using Identities I, II, III and IV

In this section we will learn Factorization using Identities one by one.

1) a2 + 2ab + b2 = (a + b)2

2) a2 – 2ab + b2 = (a – b)2

3) a2 – b2 = (a + b) (a – b)

4) x2 + (a + b) x + ab = (x + a) (x + b)

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Factorization using Identities :
In the 1st identity, a2 + 2ab + b2 = (a + b)2,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is positive.

Examples on 1st Identity of Factorization :

1) 9a2 + 12ab + 4b2

Solution :
9a2 + 12ab + 4b2

= (3a)2 + 2 . (3a) . (2b) + (2b)2

= (3a + 2b)2 [ since a = 3a and b = 2b; a2 + 2ab + b2 = (a + b)2]

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2) x4 + 2 + 1/x4

Solution :
x4 + 2 + 1/x4

= (x2)2 + 2.(x2).1/x2 + (1/x2)2

= (x2 + 1/x2)2 [ since a = x2 and b = 1/x2 ]

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In the 2nd identity, a2 - 2ab + b2 = (a - b)2,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is negative.

Examples on 2nd Identity of Factorization :

1) 4p2 - 20pq + 25q2

Solution :
4p2 - 20pq + 25q2

= (2p)2 - 2 . (2p) . (5q) + (5q)2

= (2p - 5q)2 [ since a = 2p and b = 5q; a2 + 2ab - b2 = (a - b)2]

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2)1 - 16x2 + 64x4

Solution :
= (1)2 - 2 . (2) . (8x2) + (8x2)2

= (1 - 8x2)2 [ since a = 1 and b = 8x2; a2 - 2ab + b2 = (a - b)2]

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Some quadratic polynomials will be missing the middle term. Often these polynomials are the difference of two squares.
These polynomials come from multiplying the sum and difference of binomials, such as (a+b)(a-b)= a2-b2 when simplified.


Examples on 3rd Identity of Factorization :

1) 16x2 - 9y2

Solution :
16x2 - 9y2

= (4x)x2 - (3y)x2

= (4x + 3y)(4x - 3y)[ since a = 4x and b = 3y; a2 - b2 = (a + b)(a - b)]

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2) x4 - x4y4

Solution :
x4 - x4y4

= (x2)x2 - [(xy)2]x2

= (x2 + x2y2)(x2 - x2y2)[ since a = x2 and b = (xy)2; a2 - b2 = (a + b)(a - b)]

In the 2nd parenthesis(bracket), we can apply the 3rd identity of Factorization again

= (x2 + x2y2)(x + xy)(x - xy)[ since a = x and b = xy;
a2 - b2 = (a + b)(a - b)]

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Consider x2 + 5x + 6, Observe that this expressions are not of the type
(a + b)2 or (a – b)2, i.e., they are not perfect squares.
For example, in x22 + 5x + 6, the term 6 is not a perfect square. This expressions obviously also do not fit the type (a2 – b2) either. They, however, seem to be of the type
x2 + (a + b) x + a b. We may therefore, try to use Identity 4.

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Factoring

Factorization by common factor
Factorization by Grouping
Factorization using Identities
Factorization of Cubic Polynomial
Solved Examples on Factorization

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