Factor polynomial $$ 16x^{2}-121 $$
Answer
$$ 16x^{2}-121 = ( 4x-11 )( 4x+11 ) $$
Explanation
Rewrite $ 16x^{2}-121 $ as:
$$ 16x^{2}-121 = (4x)^2 - (11)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x $ and $ II = 11 $ , we have:
$$ 16x^{2}-121 = (4x)^2 - (11)^2 = ( 4x-11 ) ( 4x+11 ) $$This page was created using
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