x^2-8x+15

___________

2x^2-7x-15

So far I've got for the numerator:

(x-3)(x-5)

What goes below, and how did you do it? Thank you.

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- Jul 30th 2012, 11:14 AMAshirHow to solve this rational expression?
x^2-8x+15

___________

2x^2-7x-15

So far I've got for the numerator:

(x-3)(x-5)

What goes below, and how did you do it? Thank you. - Jul 30th 2012, 11:41 AMPlatoRe: How to solve this rational expression?
- Jul 30th 2012, 12:26 PMAshirRe: How to solve this rational expression?
Thank you. How did you get that?

- Jul 30th 2012, 12:31 PMPlatoRe: How to solve this rational expression?
- Jul 30th 2012, 12:41 PMAshirRe: How to solve this rational expression?
I don't understand; what method did you use to arrive at that answer?

- Jul 30th 2012, 12:50 PMPlatoRe: How to solve this rational expression?
- Jul 30th 2012, 12:58 PMAshirRe: How to solve this rational expression?
(2x+2)(x-5) =

2x^2-10x+2x-10

2x^2-8x-10 - Jul 30th 2012, 01:05 PMPlatoRe: How to solve this rational expression?
- Jul 30th 2012, 01:39 PMAshirRe: How to solve this rational expression?
You posted (2x+2) though?

- Jul 30th 2012, 08:13 PMProve ItRe: How to solve this rational expression?
$\displaystyle \displaystyle \begin{align*} 2x^2 - 7x - 15 \end{align*}$

Multiply the a and c values to get -30. You need to look for two numbers that multiply to give -30 and add to give -7. They are -10 and 3. So break up the middle value and factorise by grouping.

$\displaystyle \displaystyle \begin{align*} 2x^2 - 7x - 15 &= 2x^2 - 10x + 3x - 15 \\ &= 2x(x - 5) + 3(x - 5) \\ &= (x - 5)(2x + 3) \end{align*}$ - Jul 31st 2012, 01:25 AMemakarovRe: How to solve this rational expression?
The most reliable way to factor a quadratic polynomial $\displaystyle ax^2+bx+c$ is by finding its roots using the quadratic formula. If the roots are $\displaystyle x_1$ and $\displaystyle x_2$, then $\displaystyle ax^2+bx+c=a(x-x_1)(x-x_2)$.

- Jul 31st 2012, 01:56 AMAshirRe: How to solve this rational expression?