Question
Factor polynomial $$ 81b^{2}-64c^{2} $$
Answer
$$ 81b^{2}-64c^{2} = (9b+8c)(9b-8c) $$
Explanation
Rewrite $ 81b^2-64c^2 $ as:
$$ \color{blue}{ 81b^2-64c^2 = (9b)^2 - (8c)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9b $ and $ II = 8c $ , we have:
$$ 81b^2-64c^2 = (9b)^2 - (8c)^2 = ( 9b-8c ) ( 9b+8c ) $$This page was created using
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