Factor polynomial $$ 16x^{4}-81 $$
Answer
$$ 16x^{4}-81 = ( 2x-3 )( 2x+3 )( 4x^{2}+9 ) $$
Explanation
Step 1 :
Rewrite $ 16x^{4}-81 $ as:
$$ 16x^{4}-81 = (4x^{2})^2 - (9)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 4x^{2} $ and $ II = 9 $ , we have:
$$ 16x^{4}-81 = (4x^{2})^2 - (9)^2 = ( 4x^{2}-9 ) ( 4x^{2}+9 ) $$Step 2 :
Rewrite $ 4x^{2}-9 $ as:
$$ 4x^{2}-9 = (2x)^2 - (3)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2x $ and $ II = 3 $ , we have:
$$ 4x^{2}-9 = (2x)^2 - (3)^2 = ( 2x-3 ) ( 2x+3 ) $$This page was created using
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