Factor polynomial $$ 3m^{4}-48n^{2} $$
Answer
$$ 3m^{4}-48n^{2} = 3(m^{2}+4n)(m^{2}-4n) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 3 } $:
$$ 3m^4-48n^2 = 3 ( m^4-16n^2 ) $$Step 2 :
Rewrite $ m^4-16n^2 $ as:
$$ \color{blue}{ m^4-16n^2 = (m^2)^2 - (4n)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = m^2 $ and $ II = 4n $ , we have:
$$ m^4-16n^2 = (m^2)^2 - (4n)^2 = ( m^2-4n ) ( m^2+4n ) $$This page was created using
Polynomial Factoring Calculator