Factor polynomial $$ 121b^{2}-64c^{2} $$
Answer
$$ 121b^{2}-64c^{2} = (11b+8c)(11b-8c) $$
Explanation
Rewrite $ 121b^2-64c^2 $ as:
$$ \color{blue}{ 121b^2-64c^2 = (11b)^2 - (8c)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 11b $ and $ II = 8c $ , we have:
$$ 121b^2-64c^2 = (11b)^2 - (8c)^2 = ( 11b-8c ) ( 11b+8c ) $$This page was created using
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