Factor polynomial $$ 121x^{2}-81 $$
Answer
$$ 121x^{2}-81 = ( 11x-9 )( 11x+9 ) $$
Explanation
Rewrite $ 121x^{2}-81 $ as:
$$ 121x^{2}-81 = (11x)^2 - (9)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 11x $ and $ II = 9 $ , we have:
$$ 121x^{2}-81 = (11x)^2 - (9)^2 = ( 11x-9 ) ( 11x+9 ) $$This page was created using
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