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