Factorize the following:

$2 x^{3} y^{2} - 4 x^{2} y^{3} + 8 x y^{4}$

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Solution

$2 x^{3} y^{2} - 4 x^{2} y^{3} + 8 x y^{4}$

HCF of 2, 4, 8 = 2

HCF of $x^{3}, x^{2}, x = 1$

and HCF of $y^{2}, y^{3}, y^{4} = y^{2}$

$∴ 2 x^{3} y^{2} - 4 x^{2} y^{3} + 8 x y^{4}$

= $2 x y^{2} (x^{2} - 2 x y + 4 y^{2})$

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