Factor polynomial $$ 81x^{4}-256 $$
Answer
$$ 81x^{4}-256 = ( 3x-4 )( 3x+4 )( 9x^{2}+16 ) $$
Explanation
Step 1 :
Rewrite $ 81x^{4}-256 $ as:
$$ 81x^{4}-256 = (9x^{2})^2 - (16)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9x^{2} $ and $ II = 16 $ , we have:
$$ 81x^{4}-256 = (9x^{2})^2 - (16)^2 = ( 9x^{2}-16 ) ( 9x^{2}+16 ) $$Step 2 :
Rewrite $ 9x^{2}-16 $ as:
$$ 9x^{2}-16 = (3x)^2 - (4)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 3x $ and $ II = 4 $ , we have:
$$ 9x^{2}-16 = (3x)^2 - (4)^2 = ( 3x-4 ) ( 3x+4 ) $$This page was created using
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