Question
Factor polynomial $$ -36y^{6}+4y^{2} $$
Answer
$$ -36y^{6}+4y^{2} = 4y^{2}( -3y^{2}+1 )( 3y^{2}+1 ) $$
Explanation
Step 1 :
After factoring out $ 4y^{2} $ we have:
$$ -36y^{6}+4y^{2} = 4y^{2} ( -9y^{4}+1 ) $$Step 2 :
Rewrite $ -9y^{4}+1 $ as:
$$ -9y^{4}+1 = 1 -9x^4 = (1)^2 - (3y^{2})^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 1 $ and $ II = 3y^{2} $ , we have:
$$ -9y^{4}+1 = (1)^2 - (3y^{2})^2 = ( -3y^{2}+1 ) ( 3y^{2}+1 ) $$This page was created using
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