Factor polynomial $$ t^{8}-16 $$
Answer
$$ t^{8}-16 = ( t^{2}-2 )( t^{2}+2 )( t^{4}+4 ) $$
Explanation
Step 1 :
Rewrite $ t^{8}-16 $ as:
$$ t^{8}-16 = (t^{4})^2 - (4)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = t^{4} $ and $ II = 4 $ , we have:
$$ t^{8}-16 = (t^{4})^2 - (4)^2 = ( t^{4}-4 ) ( t^{4}+4 ) $$Step 2 :
Rewrite $ t^{4}-4 $ as:
$$ t^{4}-4 = (t^{2})^2 - (2)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = t^{2} $ and $ II = 2 $ , we have:
$$ t^{4}-4 = (t^{2})^2 - (2)^2 = ( t^{2}-2 ) ( t^{2}+2 ) $$This page was created using
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