Question
Factor polynomial $$ w^{6}-64w^{2} $$
Answer
$$ w^{6}-64w^{2} = w^{2}( w^{2}-8 )( w^{2}+8 ) $$
Explanation
Step 1 :
After factoring out $ w^{2} $ we have:
$$ w^{6}-64w^{2} = w^{2} ( w^{4}-64 ) $$Step 2 :
Rewrite $ w^{4}-64 $ as:
$$ w^{4}-64 = (w^{2})^2 - (8)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = w^{2} $ and $ II = 8 $ , we have:
$$ w^{4}-64 = (w^{2})^2 - (8)^2 = ( w^{2}-8 ) ( w^{2}+8 ) $$This page was created using
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