Question

How to do factorization with algebraic identities ? Give each examples from every identities

Answer 1

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)

_____________________________________________________________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.