Question
Factor polynomial $$ p^{8}-81 $$
Answer
$$ p^{8}-81 = ( p^{2}-3 )( p^{2}+3 )( p^{4}+9 ) $$
Explanation
Step 1 :
Rewrite $ p^{8}-81 $ as:
$$ p^{8}-81 = (p^{4})^2 - (9)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = p^{4} $ and $ II = 9 $ , we have:
$$ p^{8}-81 = (p^{4})^2 - (9)^2 = ( p^{4}-9 ) ( p^{4}+9 ) $$Step 2 :
Rewrite $ p^{4}-9 $ as:
$$ p^{4}-9 = (p^{2})^2 - (3)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = p^{2} $ and $ II = 3 $ , we have:
$$ p^{4}-9 = (p^{2})^2 - (3)^2 = ( p^{2}-3 ) ( p^{2}+3 ) $$This page was created using
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