Question
Factor polynomial $$ 50y^{50}-32z^{100} $$
Answer
$$ 50y^{50}-32z^{100} = 2(5y^{25}+4z^{50})(5y^{25}-4z^{50}) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 2 } $:
$$ 50y^50-32z^100 = 2 ( 25y^50-16z^100 ) $$Step 2 :
Rewrite $ 25y^50-16z^100 $ as:
$$ \color{blue}{ 25y^50-16z^100 = (5y^25)^2 - (4z^50)^2 } $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 5y^25 $ and $ II = 4z^50 $ , we have:
$$ 25y^50-16z^100 = (5y^25)^2 - (4z^50)^2 = ( 5y^25-4z^50 ) ( 5y^25+4z^50 ) $$This page was created using
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