Factor polynomial $$ 18z^{2}-8y^{2} $$
Answer
$$ 18z^{2}-8y^{2} = 2(3z+2y)(3z-2y) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 2 } $:
$$ 18z^2-8y^2 = 2 ( 9z^2-4y^2 ) $$Step 2 :
Rewrite $ 9z^2-4y^2 $ as:
$$ \color{blue}{ 9z^2-4y^2 = (3z)^2 - (2y)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 3z $ and $ II = 2y $ , we have:
$$ 9z^2-4y^2 = (3z)^2 - (2y)^2 = ( 3z-2y ) ( 3z+2y ) $$This page was created using
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