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Factorization of Perfect Square
In factorization of perfect square we will learn how to factor different types of algebraic expressions using the following identities.
(i) a2 + 2ab + b2 = (a + b)2 = (a + b) (a + b)(ii) a2 - 2ab + b2 = (a - b)2 = (a - b) (a - b)
Solved examples on factorization of perfect square:
1.
Factorize the perfect
square completely:
Solution:
First we arrange the given expression 4x2 + 9y2 + 12xy in the form of a2 + 2ab + b2.
4x2 + 12xy + 9y2
= (2x)2 + 2 (2x) (3y) + (3y)2
Now applying the formula of a2 + 2ab + b2 = (a + b)2 then we get,
= (2x + 3y)2
= (2x + 3y) (2x + 3y)
(ii) 25x2 – 10xz + z2
Solution:
We can express the given expression 25x2 – 10xz + z2 as a2 - 2ab + b2
= (5x)2 – 2 (5x) (z) + (z)2
Now we will apply the formula of a2- 2ab + b2 = (a - b)2 then we get,
= (5x – z)2
= (5x – z)(5x – z)
(iii) x2 + 6x + 8
Solution:
We can that the given expression is not a perfect square. To get the expression as a perfect square we need to add 1 at the same time subtract 1 to keep the expression unchanged.
= x2 + 6x + 8 + 1 - 1= x2 + 6x + 9 – 1
= [(x)2 + 2 (x) (3) + (3)2] – (1)2
= (x + 3)2 - (1)2
= (x + 3 + 1)(x + 3 - 1)
= (x + 4)(x + 2)
2. Factor using the identity:
Solution:
4m4 + 1
To get the above expression in the form of a2 + 2ab + b2 we need to add 4m2 and to keep the expression same we also need to subtract 4m2 at the same time so that the expression remain same.
= 4m4 + 1 + 4m2 - 4m2
= 4m4 + 4m2 + 1 – 4m2, rearranged the terms
= (2m2)2 + 2 (2m2) (1) + (1)2 – 4m2
Now we apply the formula of a2 + 2ab + b2 = (a + b)2
= (2m2 + 1)2 - 4m2
= (2m2 + 1)2 - (2m)2
= (2m2 + 1 + 2m) (2m2 + 1 – 2m)
= (2m2 + 2m + 1) (2m2 – 2m + 1)
(ii) (x + 2y)2 + 2(x + 2y) (3y – x) + (3y - x)2
Solution:
We see that the given expression (x + 2y)2 + 2(x + 2y) (3y – x) + (3y - x)2 is in the form of a2 + 2ab + b2.
Here, a = x + 2y and b = 3y – x
Now we will apply the formula of a2 + 2ab + b2 = (a + b)2 then we get,
[(x + 2y) + (3y – x)]2
= [x + 2y + 3y – x]2
= [5y]2
= 25y2
8th Grade Math Practice
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