Question
Factor polynomial $$ 324y^{3}-196y $$
Answer
$$ 324y^{3}-196y = 4y( 9y-7 )( 9y+7 ) $$
Explanation
Step 1 :
After factoring out $ 4y $ we have:
$$ 324y^{3}-196y = 4y ( 81y^{2}-49 ) $$Step 2 :
Rewrite $ 81y^{2}-49 $ as:
$$ 81y^{2}-49 = (9y)^2 - (7)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9y $ and $ II = 7 $ , we have:
$$ 81y^{2}-49 = (9y)^2 - (7)^2 = ( 9y-7 ) ( 9y+7 ) $$This page was created using
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