Factor polynomial $$ 81y^{4}-625 $$
Answer
$$ 81y^{4}-625 = ( 3y-5 )( 3y+5 )( 9y^{2}+25 ) $$
Explanation
Step 1 :
Rewrite $ 81y^{4}-625 $ as:
$$ 81y^{4}-625 = (9y^{2})^2 - (25)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9y^{2} $ and $ II = 25 $ , we have:
$$ 81y^{4}-625 = (9y^{2})^2 - (25)^2 = ( 9y^{2}-25 ) ( 9y^{2}+25 ) $$Step 2 :
Rewrite $ 9y^{2}-25 $ as:
$$ 9y^{2}-25 = (3y)^2 - (5)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 3y $ and $ II = 5 $ , we have:
$$ 9y^{2}-25 = (3y)^2 - (5)^2 = ( 3y-5 ) ( 3y+5 ) $$This page was created using
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