No, it is not. You can not bring out an x like you done in the end.
The x+3 cancels.
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?
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)?
You're going about this wrong. You should not get integers as a solution if you are merely factoring and not solving for x.
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.
The top one does not factor nice and neat.
It's roots are
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 or