Thread: 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.

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.

Thank you. How did you get that?

Originally Posted by Ashir

How did you get that?

Practice, practice, practice.
Multiply the answer out to see how it works.

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

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.

(2x+2)(x-5) =
2x^2-10x+2x-10
2x^2-8x-10

Originally Posted by Ashir

(2x+2)(x-5) =
2x^2-10x+2x-10
2x^2-8x-10

But it is

You posted (2x+2) though?

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.

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.

The most reliable way to factor a quadratic polynomial is by finding its roots using the quadratic formula. If the roots are and , then .

Originally Posted by Prove It

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.

Got it, thanks.