How do you factor completely $9 x y^{2} - 24 x y + 15 x$ $9xy^2 - 24xy + 15x$ ?

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How do you factor completely $9 x y^{2} - 24 x y + 15 x$ $9xy^2 - 24xy + 15x$ ?

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Answer 1 · 2015-11-26T15:43:25

$3 x (3 y - 5) (y - 1)$ $3x(3y-5)(y-1)$

Explanation:

$x$ $x$ is a factor of everything so we can take that out. Also 3 is likewise a factor of everything so we can take that out as well!

$3 x (3 y^{2} - 8 y + 5)$ $3x(3y^2-8y+5)$

Both 3 and 5 are prime number so we can only use 1 and themselves as whole number factors. Factors of $5 \to {1, 5}$ $5->{1,5}$
Factors of $3 \to {1, 3}$ $3->{1,3}$ . So it a matter of experimenting with the order to see if we can reproduce $3 y^{2} - 8 y + 5$ $3y^2-8y+5$

Strait off there is no choice other than $(3 y + ?) (y + ?)$ $(3y+?)(y+?)$ to give $3 y^{2}$ $3y^2$

$3 x (3 y - 5) (y - 1)$ $3x(3y-5)(y-1)$

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How do you factor completely $9 x y^{2} - 24 x y + 15 x$ $9xy^2 - 24xy + 15x$ ?

1 Answer

Explanation:

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How do you factor completely $9 x y^{2} - 24 x y + 15 x$ $9xy^2 - 24xy + 15x$ ?