How do you factor (3x^2+4x-15)/(2x^2+3x-9)?

Jun 16, 2017

$\frac{3 x - 5}{2 x - 3}$

Explanation:

We can break the top polynomial like this:

$3 {x}^{2} + 4 x - 15$

$3 {x}^{2} + 9 x - 5 x - 15$

And then we can simplify it like this:

$\left(3 {x}^{2} + 9 x\right) - \left(5 x + 15\right)$

$3 x \left(x + 3\right) - 5 \left(x + 3\right)$

$= \left(3 x - 5\right) \left(x + 3\right)$

We can also break the bottom polynomial similarly like this:

$2 {x}^{2} + 3 x - 9$

$2 {x}^{2} + 6 x - 3 x - 9$

$\left(2 {x}^{2} + 6 x\right) - \left(3 x + 9\right)$

$2 x \left(x + 3\right) - 3 \left(x + 3\right)$

$= \left(2 x - 3\right) \left(x + 3\right)$

Therefore, our factored expression becomes:

(3x^2 + 4x - 15)/(2x^2 + 3x - 9) = ((3x-5)(x+3))/((2x-3)(x+3)) = ((3x-5)cancel((x+3)))/((2x-3)cancel((x+3))

After cancelling out the top and bottom of the expression, we get the result:

$\frac{3 x - 5}{2 x - 3}$