Question
Factor polynomial $$ 10x^{8}y^{2}+80x^{2}y^{5} $$
Answer
$$ 10x^{8}y^{2}+80x^{2}y^{5} = 10x^{2}y^{2}(x^{2}+2y)(x^{4}-2x^{2}y+4y^{2}) $$
Explanation
Step 1 :
Factor out common factor $ \color{blue}{ 10x^2y^2 } $:
$$ 10x^8y^2+80x^2y^5 = 10x^2y^2 ( x^6+8y^3 ) $$Step 2 :
To factor $ x^{6}+8y^{3} $ we can use sum of cubes formula:
$$ I^3 + II^3 = (I + II) (I^2 - I \cdot II + II^2)$$After putting $ I = x^2 $ and $ II = 2y $ , we have:
$$ x^{6}+8y^{3} = ( x^{2}+2y ) ( x^{4}-2x^{2}y+4y^{2} ) $$This page was created using
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