Question
Factor polynomial $$ 72y^{4}-18y^{2} $$
Answer
$$ 72y^{4}-18y^{2} = 18y^{2}( 2y-1 )( 2y+1 ) $$
Explanation
Step 1 :
After factoring out $ 18y^{2} $ we have:
$$ 72y^{4}-18y^{2} = 18y^{2} ( 4y^{2}-1 ) $$Step 2 :
Rewrite $ 4y^{2}-1 $ as:
$$ 4y^{2}-1 = (2y)^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2y $ and $ II = 1 $ , we have:
$$ 4y^{2}-1 = (2y)^2 - (1)^2 = ( 2y-1 ) ( 2y+1 ) $$This page was created using
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