Factor polynomial $$ 9a^{2}b^{4}-4c^{2} $$
Answer
$$ 9a^{2}b^{4}-4c^{2} = (3ab^{2}+2c)(3ab^{2}-2c) $$
Explanation
Rewrite $ 9a^2b^4-4c^2 $ as:
$$ \color{blue}{ 9a^2b^4-4c^2 = (3ab^2)^2 - (2c)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 3ab^2 $ and $ II = 2c $ , we have:
$$ 9a^2b^4-4c^2 = (3ab^2)^2 - (2c)^2 = ( 3ab^2-2c ) ( 3ab^2+2c ) $$This page was created using
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