Question
Factor polynomial $$ 729w^{3}-9w $$
Answer
$$ 729w^{3}-9w = 9w( 9w-1 )( 9w+1 ) $$
Explanation
Step 1 :
After factoring out $ 9w $ we have:
$$ 729w^{3}-9w = 9w ( 81w^{2}-1 ) $$Step 2 :
Rewrite $ 81w^{2}-1 $ as:
$$ 81w^{2}-1 = (9w)^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9w $ and $ II = 1 $ , we have:
$$ 81w^{2}-1 = (9w)^2 - (1)^2 = ( 9w-1 ) ( 9w+1 ) $$This page was created using
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