Factor polynomial $$ 32v^{2}-8 $$
Answer
$$ 32v^{2}-8 = 8( 2v-1 )( 2v+1 ) $$
Explanation
Step 1 :
After factoring out $ 8 $ we have:
$$ 32v^{2}-8 = 8 ( 4v^{2}-1 ) $$Step 2 :
Rewrite $ 4v^{2}-1 $ as:
$$ 4v^{2}-1 = (2v)^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2v $ and $ II = 1 $ , we have:
$$ 4v^{2}-1 = (2v)^2 - (1)^2 = ( 2v-1 ) ( 2v+1 ) $$This page was created using
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