# How to solve this rational expression?

• Jul 30th 2012, 11:14 AM
Ashir
How 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 AM
Plato
Re: How to solve this rational expression?
Quote:

Originally Posted by Ashir
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.

$\displaystyle 2x^2-7x-15=(2x+3)(x-5)$
• Jul 30th 2012, 12:26 PM
Ashir
Re: How to solve this rational expression?
Thank you. How did you get that?
• Jul 30th 2012, 12:31 PM
Plato
Re: How to solve this rational expression?
Quote:

Originally Posted by Ashir
How did you get that?

Practice, practice, practice.
Multiply the answer out to see how it works.
• Jul 30th 2012, 12:41 PM
Ashir
Re: 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 PM
Plato
Re: How to solve this rational expression?
Quote:

Originally Posted by Ashir
I don't understand; what method did you use to arrive at that answer?

There is no method. I just looked at and knew the answer. I have done enough of these to have that skill.
• Jul 30th 2012, 12:58 PM
Ashir
Re: How to solve this rational expression?
(2x+2)(x-5) =
2x^2-10x+2x-10
2x^2-8x-10
• Jul 30th 2012, 01:05 PM
Plato
Re: How to solve this rational expression?
Quote:

Originally Posted by Ashir
(2x+2)(x-5) =
2x^2-10x+2x-10
2x^2-8x-10

But it is $\displaystyle (2x+3)(x-5)$
• Jul 30th 2012, 01:39 PM
Ashir
Re: How to solve this rational expression?
You posted (2x+2) though?
• Jul 30th 2012, 08:13 PM
Prove It
Re: How to solve this rational expression?
Quote:

Originally Posted by Ashir
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.

\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 AM
emakarov
Re: 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 AM
Ashir
Re: How to solve this rational expression?
Quote:

Originally Posted by Prove It
\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*}

Got it, thanks.