Factorization of Perfect Square Trinomials
In factorization of perfect square trinomials we will
learn how to solve the algebraic expressions using the formulas. To factorize an algebraic expression
expressible as a perfect square, we use 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)
Note: We will also learn to use two identities in the
same question, to factorize the expression.
Solved problems on factorization of perfect square trinomials:
1. Factorization when the given expression
is a perfect square:
(i) x
^{4}  10x
^{2}y
^{2} + 25y
^{4}
Solution:
We can express the given expression x4  10x
^{2}y
^{2} + 25y
^{4} as a
^{2}  2ab + b
^{2}
= (x
^{2})
^{2}  2 (x
^{2}) (5y
^{2}) + (5y
^{2})
^{2}
Now it’s in the form of the formula of a
^{2} + 2ab + b
^{2} = (a + b)
^{2} then we get,
= (x
^{2}  5y
^{2})
^{2}
= (x
^{2} – 5y
^{2}) (x
^{2} – 5y
^{2})
(ii) x
^{2}+ 6x + 9
Solution:
We can express the given expression x
^{2} + 6x + 9 as a
^{2} + 2ab + b
^{2}
= (x)
^{2} + 2 (x) (3) + (3)
^{2}
Now we will apply the formula of a
^{2} + 2ab + b
^{2} = (a + b)
^{2} then we get,
= (x + 3)
^{2}
= (x + 3) (x + 3)
(iii) x
^{4}  2x
^{2} y
^{2} + y
^{4}
Solution:
We can express the given expression x
^{4}  2x
^{2} y
^{2} + y
^{4} as a
^{2}  2ab + b
^{2}
= (x
^{2})
^{2}  2 (x
^{2}) (y
^{2}) + (y
^{2})
^{2}
Now we will apply the formula of a
^{2}  2ab + b
^{2} = (a  b)
^{2} then we get,
=(x
^{2} – y
^{2})
^{2}
=(x
^{2}  y
^{2}) (x
^{2} – y
^{2})
Now we will apply the formula of differences of two squares i.e. a
^{2}  b
^{2} = (a + b) (a – b) then we get,
= (x + y) (x y) (x + y) (x y)
2. Factorize using the identity:
(i) 25 – x
^{2}  2xy  y
^{2}
Solution:
25 – x
^{2}  2xy  y
^{2}
= 25  [x
^{2} + 2xy + y
^{2}], rearranged
Now we see that x
^{2} + 2xy + y
^{2} as in the form of a
^{2} + 2ab + b
^{2}.
= (5)
^{2} – (x + y)
^{2}
Now we will apply the formula of differences of two squares i.e. a
^{2}  b
^{2} = (a + b) (a – b) then we get,
= [5 + (x + y)] [5  (x + y)]
= (5 + x + y) (5 – x  y)
(ii) 1 2xy (x
^{2} + y
^{2})
Solution:
1 2xy (x
^{2} + y
^{2})
= 1  2xy  x
^{2}  y
^{2}
= 1  (x
^{2} + 2xy + y
^{2}), rearranged
= 1  (x + y )
^{2}
= (1)
^{2} – (x + y)
^{2}
= [1 + (x + y)] [1  (x + y)]
= [1 + x + y] [1  x  y]
Note:
We see that to solve the above problems
on factorization of perfect square trinomials we not only used perfect square
identities but we also used the difference of two squares identity in different
situations.
8th Grade Math Practice
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