Question

can u give some factorization using indentities problems

Answer 1

Examples on 1st Identity of Factorization :

1) 9a ² + 12ab + 4b ²

Solution :
9a ² + 12ab + 4b ²

= (3a) ² + 2 . (3a) . (2b) + (2b) ²

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

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2) x ⁴ + 2 + 1/x ⁴

Solution :
x ⁴ + 2 + 1/x ⁴

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

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

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In the 2nd identity, a ² - 2ab + b ² = (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) 4p ² - 20pq + 25q ²

Solution :
4p ² - 20pq + 25q ²

= (2p) ² - 2 . (2p) . (5q) + (5q) ²

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

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2)1 - 16x ² + 64x ⁴

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

= (1 - 8x ² ) ² [ since a = 1 and b = 8x ² ; a ² - 2ab + b ² = (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) 16x ² - 9y ²

Solution :
16x ² - 9y ²

= (4x)x ² - (3y)x ²

= (4x + 3y)(4x - 3y)[ since a = 4x and b = 3y; a ² - b ² = (a + b)(a - b)]

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2) x ⁴ - x ⁴ y ⁴

Solution :
x ⁴ - x ⁴ y ⁴

= (x ² )x ² - [(xy) ² ]x ²

= (x ² + x ² y ² )(x ² - x ² y ² )[ since a = x ² and b = (xy) ² ; a ² - b ² = (a + b)(a - b)]

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

= (x ² + x ² y ² )(x + xy)(x - xy)[ since a = x and b = xy;
a ² - b ² = (a + b)(a - b)]

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Consider x ² + 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 x ² 2 + 5x + 6, the term 6 is not a perfect square. This expressions obviously also do not fit the type (a ² – b ² ) either. They, however, seem to be of the type
x ² + (a + b) x + a b. We may therefore, try to use Identity 4.