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

Nov 28, 2016

$2 {x}^{3} {y}^{5} \left(6 {x}^{4} {y}^{4} + 3 x {y}^{2} - 5\right)$

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

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

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

$2 {x}^{3} {y}^{5} \left(6 {x}^{7 - 3} {y}^{9 - 5} + 3 {x}^{4 - 3} {y}^{7 - 5} - 5 {x}^{3 - 3} {y}^{5 - 5}\right) \implies$

$2 {x}^{3} {y}^{5} \left(6 {x}^{4} {y}^{4} + 3 {x}^{1} {y}^{2} - 5 {x}^{0} {y}^{0}\right) \implies$

$2 {x}^{3} {y}^{5} \left(6 {x}^{4} {y}^{4} + 3 x {y}^{2} - 5 \cdot 1 \cdot 1\right) \implies$

$2 {x}^{3} {y}^{5} \left(6 {x}^{4} {y}^{4} + 3 x {y}^{2} - 5\right)$