1. ## Factorise and solve...

Factorise and solve:

2x^2 - 7x - 4

Also, how do you factorise using cross multiplication? Somebody told me to do that but I didn't know what that meant. Anyway...if anyone could show me that would be great also!

2. Sorry...I meant:

2x^2 - 7x - 4 = 0

...I forgot how to edit a post.

3. Originally Posted by Mr Rayon
Factorise and solve:

2x^2 - 7x - 4

Also, how do you factorise using cross multiplication? Somebody told me to do that but I didn't know what that meant. Anyway...if anyone could show me that would be great also!
You don't need to use the cross method. I hate it and the way I do it is more systematic.

Multiply your a and c values. In this case it would be $\displaystyle 2\times -4 = -8$

Now find 2 numbers that multiply to become the number you've just found (ac) and add to become b.

So you need two numbers that multiply to become -8 and add to become -7. They are -8 and +1.

So break up $\displaystyle -7x$ into $\displaystyle -8x + x$

Now you have

$\displaystyle 2x^2 - 7x - 4$

$\displaystyle = 2x^2 - 8x + x - 4$

$\displaystyle = 2x(x - 4) + 1(x - 4)$

[tex] = (2x + 1)(x - 4).

I assume if you're trying to solve the original quadratic that it was set equal to 0...

$\displaystyle 2x^2 - 7x - 4 = 0$

$\displaystyle (2x + 1)(x - 4) = 0$

By the null factor law, $\displaystyle 2x + 1 = 0$ or $\displaystyle x - 4 = 0$.

So $\displaystyle x = -\frac{1}{2}$ or $\displaystyle x = 4$.

4. Thanks...that method seems easier than cross multiplication!