Factor polynomial $$ 80x^{4}y^{2}-5y^{2} $$
Answer
$$ 80x^{4}y^{2}-5y^{2} = 5y^{2}(4x^{2}+1)(2x+1)(2x-1) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 5y^2 } $:
$$ 80x^4y^2-5y^2 = 5y^2 ( 16x^4-1 ) $$Step 2 :
Rewrite $ 16x^4-1 $ as:
$$ \color{blue}{ 16x^4-1 = (4x^2)^2 - (1)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x^2 $ and $ II = 1 $ , we have:
$$ 16x^4-1 = (4x^2)^2 - (1)^2 = ( 4x^2-1 ) ( 4x^2+1 ) $$Step 3 :
Rewrite $ 4x^2-1 $ as:
$$ \color{blue}{ 4x^2-1 = (2x)^2 - (1)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2x $ and $ II = 1 $ , we have:
$$ 4x^2-1 = (2x)^2 - (1)^2 = ( 2x-1 ) ( 2x+1 ) $$This page was created using
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