Factor 2x^3y^4-8x^2y^3+6xy^2

$2 x^{3} y^{4} - 8 x^{2} y^{3} + 6 x y^{2}$

Factor $2 x y^{2}$ out of $2 x^{3} y^{4} - 8 x^{2} y^{3} + 6 x y^{2}$ .

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Factor $2 x y^{2}$ out of $2 x^{3} y^{4}$ .

$2 x y^{2} (x^{2} y^{2}) - 8 x^{2} y^{3} + 6 x y^{2}$

Factor $2 x y^{2}$ out of $- 8 x^{2} y^{3}$ .

$2 x y^{2} (x^{2} y^{2}) + 2 x y^{2} (- 4 x y) + 6 x y^{2}$

Factor $2 x y^{2}$ out of $6 x y^{2}$ .

$2 x y^{2} (x^{2} y^{2}) + 2 x y^{2} (- 4 x y) + 2 x y^{2} (3)$

Factor $2 x y^{2}$ out of $2 x y^{2} (x^{2} y^{2}) + 2 x y^{2} (- 4 x y)$ .

$2 x y^{2} (x^{2} y^{2} - 4 x y) + 2 x y^{2} (3)$

Factor $2 x y^{2}$ out of $2 x y^{2} (x^{2} y^{2} - 4 x y) + 2 x y^{2} (3)$ .

$2 x y^{2} (x^{2} y^{2} - 4 x y + 3)$

Rewrite $x^{2} y^{2}$ as ${(x y)}^{2}$ .

$2 x y^{2} ({(x y)}^{2} - 4 (x y) + 3)$

Let $u = x y$ . Substitute $u$ for all occurrences of $x y$ .

$2 x y^{2} (u^{2} - 4 u + 3)$

Factor $u^{2} - 4 u + 3$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$2 x y^{2} ((u - 3) (u - 1))$

Factor.

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Replace all occurrences of $u$ with $x y$ .

$2 x y^{2} ((x y - 3) (x y - 1))$

Remove unnecessary parentheses.

$2 x y^{2} (x y - 3) (x y - 1)$

Factor 2x^3y^4-8x^2y^3+6xy^2

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