Factor polynomial $$ 4v^{4}-64v^{2} $$
Answer
$$ 4v^{4}-64v^{2} = 4v^{2}( v-4 )( v+4 ) $$
Explanation
Step 1 :
After factoring out $ 4v^{2} $ we have:
$$ 4v^{4}-64v^{2} = 4v^{2} ( v^{2}-16 ) $$Step 2 :
Rewrite $ v^{2}-16 $ as:
$$ v^{2}-16 = (v)^2 - (4)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = v $ and $ II = 4 $ , we have:
$$ v^{2}-16 = (v)^2 - (4)^2 = ( v-4 ) ( v+4 ) $$This page was created using
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