Factor polynomial $$ 121x^{4}-9y^{4} $$
Answer
$$ 121x^{4}-9y^{4} = (11x^{2}+3y^{2})(11x^{2}-3y^{2}) $$
Explanation
Rewrite $ 121x^4-9y^4 $ as:
$$ \color{blue}{ 121x^4-9y^4 = (11x^2)^2 - (3y^2)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 11x^2 $ and $ II = 3y^2 $ , we have:
$$ 121x^4-9y^4 = (11x^2)^2 - (3y^2)^2 = ( 11x^2-3y^2 ) ( 11x^2+3y^2 ) $$This page was created using
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