Question

4x^2 + 12x - 16/2x + 10/6x + 24/x^2 + 9x + 20. Basically, these are 2 fractions on top of each other. @phi

Answer 1

I know i need to start by separating, flipping the second fraction, and making the division multiplication. Then I need to factor.

Answer 2

I can factor a 4 out of 4x^2 + 12x - 16

Answer 3

4(x^2 + 3x - 4). This factors to...(4x - 1)(x + 4)? Or would it become (4x - 1)(x + 1), because we divide the 4 in the second parentheses by 4?

Answer 4

I think I was right with the (x + 4), b/c that would be congruent with the other factors I'm getting. I've factored the other 3 parts.

Answer 5

All of the factors on the right cancel; all I have left there is 6. On the left, I have (4x - 1)(x + 4)/2.

Answer 6

once you get 4(x^2+3x-4)
the 4 stays outside. you factor x^2+3x-4 into (x+4)(x-1)
and the final factoring is
4(x-1)(x+4)

Answer 7

Sweet, I did it right :)

Answer 8

Oh wait..not quite. I put the 4 inside. Sorry. I was thinking of the previous problem we did.

Answer 9

Okay, so 4/2 = 2/1, so we just get 2(x - 1)(x + 4) = 6.

Answer 10

2x^2 + 6x - 14 = 0

Answer 11

2(x^2 + 3x - 7) = 0

Answer 12

the original does not show an equation. Is there more to this?

Answer 13

Oh, you're right...still doing it like the other problem. Ugh! I hate these.

Answer 14

It needs to look like a fraction. Where did I start going wrong?

Answer 15

First, is this the problem?
\[ \frac{ \frac{4x^2+12x-16}{2x+10} } { \frac{6x+24} {x^2+9x+20} }\]

Answer 16

Yes.

Answer 17

so we re-write it as
\[ \frac{4x^2+12x-16}{2x+10} \cdot \frac{x^2+9x+20}{6x+24} \]

Answer 18

Yes, but I've done that up to the point where you last helped me - factoring the numerator on the left to 4(x - 1)(x + 4)

Answer 19

I think I may have done something wrong with cancelling, so let's go back to that. It was still a fraction then.

Answer 20

now just go slowly
\[ \frac{4(x-1)(x+4)}{2(x+5)} \cdot \frac{(x+4)(x+5)}{6(x+4)} \]