Factor polynomial $$ 125m^{4}-20n^{4} $$
Answer
$$ 125m^{4}-20n^{4} = 5(5m^{2}+2n^{2})(5m^{2}-2n^{2}) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 5 } $:
$$ 125m^4-20n^4 = 5 ( 25m^4-4n^4 ) $$Step 2 :
Rewrite $ 25m^4-4n^4 $ as:
$$ \color{blue}{ 25m^4-4n^4 = (5m^2)^2 - (2n^2)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 5m^2 $ and $ II = 2n^2 $ , we have:
$$ 25m^4-4n^4 = (5m^2)^2 - (2n^2)^2 = ( 5m^2-2n^2 ) ( 5m^2+2n^2 ) $$This page was created using
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