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) a
^{2} + 2ab + b
^{2} = (a + b)
^{2} = (a + b) (a + b)
(ii) a
^{2}  2ab + b
^{2} = (a  b)
^{2} = (a  b) (a  b)
Solved
examples on factorization of perfect square:
1.
Factorize the perfect
square completely:
(i) 4x
^{2} + 9y
^{2} + 12xy
Solution:
First we arrange the given expression 4x
^{2} + 9y
^{2} + 12xy in the form of a
^{2} + 2ab + b
^{2}.
4x
^{2} + 12xy + 9y
^{2}
= (2x)
^{2} + 2 (2x) (3y) + (3y)
^{2}
Now applying the formula of a
^{2} + 2ab + b
^{2} = (a + b)
^{2} then we get,
= (2x + 3y)
^{2}
= (2x + 3y) (2x + 3y)
(ii) 25x
^{2} – 10xz + z
^{2}
Solution:
We can express the given expression 25x
^{2} – 10xz + z
^{2} as a
^{2}  2ab + b
^{2}
= (5x)
^{2} – 2 (5x) (z) + (z)
^{2}
Now we will apply the formula of a
^{2} 2ab + b
^{2} = (a  b)
^{2} then we get,
= (5x – z)
^{2}
= (5x – z)(5x – z)
(iii) x
^{2} + 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.
= x
^{2} + 6x + 8 + 1  1
= x
^{2} + 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:
(i) 4m
^{4} + 1
Solution:
4m
^{4} + 1
To get the above expression in the form of a
^{2} + 2ab + b
^{2} we need to add 4m
^{2} and to keep the expression same we also need to subtract 4m
^{2} at the same time so that the expression remain same.
= 4m
^{4} + 1 + 4m
^{2}  4m
^{2}
= 4m
^{4} + 4m
^{2} + 1 – 4m
^{2}, rearranged the terms
= (2m
^{2})
^{2} + 2 (2m
^{2}) (1) + (1)
^{2} – 4m
^{2}
Now we apply the formula of a
^{2} + 2ab + b
^{2} = (a + b)
^{2}
= (2m
^{2} + 1)
^{2}  4m
^{2}
= (2m
^{2} + 1)
^{2}  (2m)
^{2}
= (2m
^{2} + 1 + 2m) (2m
^{2} + 1 – 2m)
= (2m
^{2} + 2m + 1) (2m
^{2} – 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 a
^{2} + 2ab + b
^{2}.
Here, a = x + 2y and b = 3y – x
Now we will apply the formula of a
^{2} + 2ab + b
^{2} = (a + b)
^{2} then we get,
[(x + 2y) + (3y – x)]
^{2}
= [x + 2y + 3y – x]
^{2}
= [5y]
^{2}
= 25y
^{2}
8th Grade Math Practice
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