Factor polynomial $$ 4y^{2}-121 $$
Answer
$$ 4y^{2}-121 = ( 2y-11 )( 2y+11 ) $$
Explanation
Rewrite $ 4y^{2}-121 $ as:
$$ 4y^{2}-121 = (2y)^2 - (11)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 2y $ and $ II = 11 $ , we have:
$$ 4y^{2}-121 = (2y)^2 - (11)^2 = ( 2y-11 ) ( 2y+11 ) $$This page was created using
Polynomial Factoring Calculator