How do you factor $12 x^{7} y^{9} + 6 x^{4} y^{7} - 10 x^{3} y^{5}$ $12x^7y^9+6x^4y^7-10x^3y^5$ ?

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How do you factor $12 x^{7} y^{9} + 6 x^{4} y^{7} - 10 x^{3} y^{5}$ $12x^7y^9+6x^4y^7-10x^3y^5$ ?

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Answer 1 · 2016-11-28T22:04:14

$2 x^{3} y^{5} (6 x^{4} y^{4} + 3 x y^{2} - 5)$ $2x^3y^5(6x^4y^4 + 3xy^2 - 5)$

Explanation:

You need to pull the highest common value out of each variable and constant in the terms:

For the constants 12, 6 and 10 the highest common value is 2

For $x$ $x$ the highest common value is $x^{3}$ $x^3$

For $y$ $y$ the highest common value is $y^{5}$ $y^5$

So, we can rewrite this problem, using the rules for exponents as:

$2 x^{3} y^{5} (6 x^{7 - 3} y^{9 - 5} + 3 x^{4 - 3} y^{7 - 5} - 5 x^{3 - 3} y^{5 - 5}) \Rightarrow$ $2x^3y^5(6x^(7-3)y^(9-5)+ 3x^(4-3)y^(7-5) - 5x^(3-3)y^(5-5)) =>$

$2 x^{3} y^{5} (6 x^{4} y^{4} + 3 x^{1} y^{2} - 5 x^{0} y^{0}) \Rightarrow$ $2x^3y^5(6x^4y^4 + 3x^1y^2 - 5x^0y^0) =>$

$2 x^{3} y^{5} (6 x^{4} y^{4} + 3 x y^{2} - 5 \cdot 1 \cdot 1) \Rightarrow$ $2x^3y^5(6x^4y^4 + 3xy^2 - 5*1*1) =>$

$2 x^{3} y^{5} (6 x^{4} y^{4} + 3 x y^{2} - 5)$ $2x^3y^5(6x^4y^4 + 3xy^2 - 5)$

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How do you factor $12 x^{7} y^{9} + 6 x^{4} y^{7} - 10 x^{3} y^{5}$ $12x^7y^9+6x^4y^7-10x^3y^5$ ?

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Explanation:

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How do you factor 12x7y9+6x4y7−10x3y512x^7y^9+6x^4y^7-10x^3y^5?

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How do you factor $12 x^{7} y^{9} + 6 x^{4} y^{7} - 10 x^{3} y^{5}$ $12x^7y^9+6x^4y^7-10x^3y^5$ ?