1. ## equation

i've got to factorise $\displaystyle 6x^2 + 5x - 6$ and i've used the steps i've been shown so far to get to this: multiply 6 and -6 to get -36. find two numbers that add together to make 5 and multiply to make -36. i came to -4 and 9. i put them into the middle of the equation while removing the 5x as instructed so i came to $\displaystyle 6x^2 -4x + 9x - 6$ grouped them and then factored them $\displaystyle x(6x - 4) 3(3x-2)$. just wondering where i go from here. any detailed steps would be appreciated

2. you're almost there
$\displaystyle x(6x - 4)+3(3x-2) = 2x(3x-2)+3(3x-2)= (2x+3)(3x-2)$

3. i've got to factorise
$\displaystyle 6x^{2}+5x-6=(2x+3)(3x-2)$

just play around with the integers that make up 6 namely 1,2,3 and 6. We see that the final term has a minus sign attached to it so we know we need to incorporate that

$\displaystyle 3\times 3x-2x\times 2=9x-4x=5x$

4. If you have x(6x - 4) 3(3x-2) --> you forgot the plus sign here,
the next step would be 2x(3x-2)+3(3x-2). Both factors need to be the same.
For (6x-4) you can write 2(3x-2)

Now, if you have 2x(3x-2)+3(3x-2), you can factor to (2x+3)(3x-2), which is 6x^2+5x-6

5. anyone have any info on what to do when there is no constant and only 2 numbers? ie how would i go about $\displaystyle 5x-2x^2$? thanks

6. Then $\displaystyle x(5-2x)$ is the factorization, it's that simple.