Factor polynomial $$ 32x^{2}-18y^{2} $$
Answer
$$ 32x^{2}-18y^{2} = 2(4x+3y)(4x-3y) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 2 } $:
$$ 32x^2-18y^2 = 2 ( 16x^2-9y^2 ) $$Step 2 :
Rewrite $ 16x^2-9y^2 $ as:
$$ \color{blue}{ 16x^2-9y^2 = (4x)^2 - (3y)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x $ and $ II = 3y $ , we have:
$$ 16x^2-9y^2 = (4x)^2 - (3y)^2 = ( 4x-3y ) ( 4x+3y ) $$This page was created using
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