How would you factorise:
x Turn on thread page Beta
- Thread Starter
- 15-11-2008 20:16
- 15-11-2008 20:17
EDIT: Ignore.Last edited by LakHani_Hasnain; 15-11-2008 at 20:25.
- 15-11-2008 20:20
It's a quadratic so you're trying to write it in the form (ax+b)(cx+d). In particular you can see by multiplying out the factors that you want ac=6, bd=-2 (and ad+bc=1). Try some nice integer values...
- 15-11-2008 20:21
Take the co-efficient of x^2 and multiply it with the constant, so:
6 * -2 = -12
Now find two numbers that multiply to give that -12 and add to give the co-efficient of x (1). Let's say the first number is A and the second number is B. Then write is as
6x^2 + Ax + Bx - 2 and factorise the 6x^2 + Ax separately from Bx - 2. See how that works out.
- PS Helper
- 15-11-2008 20:25
6x^2 can be gotten by either using (6x)(x) or (3x)(2x)
To get -2, the only thing that can be done is (-2)(1) or (2)(-1)
this leaves us with
(3x-2)(2x+1)=6x^2-x-2 Incorrect,as the sign of x in the original equation is negative. Reversing the +ve and-ve signs solves this:
Hope that helpedLast edited by Ezraeil; 15-11-2008 at 20:37.
- 15-11-2008 20:27
Hmmm,you could do that sum and product thing........
Its a quadratic so it is in the form ax^2+bx+c
You times a by c so >6 x -2=-12
And then basically,you try to find two factors of -12 which will make 1(b).They can be 4 and -3.
You then split the equation up like this....
6x^2 +4x-3x-2 and then...you you factorise half of the equation at a time so.....
The stuff inside the bracket should be the same....so you take one of the brackets..
(3x+2) and then,the other bracket is just the stuff outside....so (2x-1)
you are left with(3x+2)(2x-1).
Hope this helps.
- 15-11-2008 20:31
- 15-11-2008 20:36
which two numbers multiply to make 6x^2
3x and 2x or 6x and x
Which two no.s add to make 1 and times to make -2
2 and -1 or -1 and 2
(3x + 2) (2x-1)