Factoring Differences of Squares
How
to solve factoring differences of squares?
To factorize an algebraic expression expressible as the difference of two squares, we use the following identity a
^{2} - b
^{2} = (a + b) (a – b).
Solved examples on factoring differences of
squares:
1. Factorize the
following algebraic expressions:
(i) 64 - x
^{2}
Solution:
64 - x
^{2}
= (8)
^{2} - x
^{2}, since we know 64 = 8 times 8 which is 8
^{2}
Now by using the formula of a
^{2} - b
^{2} = (a + b)(a – b) to complete the factor fully.
= (8 + x)(8 - x).
(ii) 3a
^{2} - 27b
^{2}
Solution:
3a
^{2} - 27b
^{2}
= 3(a
^{2} – 9b
^{2}), here we took 3 as common.
=3[(a)
^{2} – (3b)
^{2}], since we know 9b
^{2} = 3b times 3b which is (3b)
^{2}
So, now we need to apply the formula of a
^{2} - b
^{2} = (a + b)(a – b) to complete the factor fully.
= 3(a + 3b)(a – 3b)
(iii) x
^{3} - 25x
Solution:
x
^{3} - 25x
= x(x
^{2} - 25), here we took x as common.
= x(x
^{2} - 5
^{2}), since we know, 25 = 5
^{2}
Now we can write x
^{2} – 5
^{2} as using the formula of a
^{2} - b
^{2} = (a + b)(a – b).
= x(x + 5)(x - 5).
2. Factor the expressions:
(i) 81a
^{2} - (b - c)
^{2}
Solution:
We can write 81a
^{2} - (b - c)
^{2} as a
^{2} - b
^{2}
= (9a)
^{2} - (b - c)
^{2}, since we know, 81a
^{2} = (9a)
^{2}
Now using the formula of a
^{2} – b
^{2} = (a + b) (a – b) we get,
= [9a + (b – c)] [9a - (b – c)]
= [9a + b – c] [9a - b + c ]
(ii) 25(x + y)
^{2} - 36(x - 2y)
^{2}.
Solution:
We can write 25(x + y)
^{2} - 36(x - 2y)
^{2} as a
^{2} - b
^{2}.
= {5(x + y)}
^{2} - {6(x - 2y)}
^{2}
Now using the formula of a
^{2} – b
^{2} = (a + b) (a – b) we get,
= [5(x + y) + 6(x - 2y)] [5(x + y) - 6(x - 2y)]
= [5x + 5y + 6x – 12y] [5x + 5y – 6x + 12y], (applying
distributive property)
Now we will arrange and then simplify it.
= (11x - 7y) (17y - x).
(iii) (x – 2)
^{2} – (x – 3)
^{2}
Solution:
We can express (x – 2)
^{2} – (x – 3)
^{2} using the formula of a
^{2} - b
^{2} = (a + b) (a – b)
= [(x - 2) + (x - 3)][(x - 2) - (x - 3)]
= [x – 2 + x - 3][x - 2 – x + 3]
Now we will arrange and then simplify it.
= [2x – 5][1]
= [2x – 5]
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8th Grade Math Practice
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