Question
Factor polynomial $$ 16a^{3}bc-49ab^{3}c $$
Answer
$$ 16a^{3}bc-49ab^{3}c = abc(4a+7b)(4a-7b) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ abc } $:
$$ 16a^3bc-49ab^3c = abc ( 16a^2-49b^2 ) $$Step 2 :
Rewrite $ 16a^2-49b^2 $ as:
$$ \color{blue}{ 16a^2-49b^2 = (4a)^2 - (7b)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4a $ and $ II = 7b $ , we have:
$$ 16a^2-49b^2 = (4a)^2 - (7b)^2 = ( 4a-7b ) ( 4a+7b ) $$This page was created using
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