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