# Thread: Factorisation

1. ## Factorisation

Hi,

I've a question on my assignment paper I just want to check to make sure I'm doing it right.

Q: Factorise the numerator and denominator of each of the following and then cancel out any common factors:

x(squared) + 2x - 3
-------------------
x(squared) + 5x + 6

So what I did was get the factors of both equations -

x(squared) + 2x - 3
factors are (x-1)(x+3)

x(squared) + 5x +6
factors are (x+2)(x+3)

so:
(x-1)(x+3)
-----------
(x+2)(x+3)

x(-1+3)
---------
x(2+3)

cancel the x's.

-1+3
------
2+3

= 2
---
5

Is that correct?

2. No, it is not. You can not bring out an x like you done in the end.

$\displaystyle \frac{(x-1)(x+3)}{(x+2)(x+3)}$

The x+3 cancels.

3. Okay, thank you.

4. I'm having a problem with another question.

3x(squared) +6x +3
-------------------
9x(squared) - 27x +18

I'm stuck with factorising the top line. I'm using the -b formula to factorise and get

-6 +/- 0
----------
6

Is that correct? I also tried factorising the other way and got (x+3)(x+3)?

5. You're going about this wrong. You should not get integers as a solution if you are merely factoring and not solving for x.

$\displaystyle 3x^{2}+6x+3=3(x+1)^{2}$

$\displaystyle 9x^{2}-27x+18=9(x-2)(x-1)$

$\displaystyle \frac{3(x+1)^{2}}{9(x-2)(x-1)}$

How are you getting this 6 thing?. Are you using the quadratic formula?. If so, don't. You are factoring, not solving. I least, I think.

The solutions are 2, 1, -1 if you need those. There is no 6 involved.

6. Originally Posted by galactus

$\displaystyle \frac{3(x+1)^{2}}{9(x-2)(x-1)}$
Is that the same as (3x+3)(x+1)
------------
(9x-9)(x-2)

That's what I got when I factorised again, without using the -b formula.

7. Yes, to see expand it back out.

Except, factor a 9 out of the bottom and a 3 out of the top

8. Last question!!!

x(squared) + x - 1
------------------
2x - 2

I'm unsure how to factorise the top equation. The bottom equation is 2(x-), right?

9. The top one does not factor nice and neat.

It's roots are $\displaystyle \frac{-\sqrt{5}-1}{2}, \;\ \frac{\sqrt{5}-1}{2}$

This is the related to the Golden Ratio and this is the closed form quadratic that represents it. You didn't need to know that, but.............

The bottom is simply 2(x-1)

Are you sure you did not mean $\displaystyle x^{2}+2x+1$ or

$\displaystyle x^{2}-2x+1$

10. No, it's x(squared) + x - 1. So I just have to leave it so.

Thanks very much for your help.

11. Are there factors for the top line? As the question asks to factorise and then cancel??

12. You can always solve x^2+x-1 by the quadratic formula.
roots are -2, 3 by it.