• Jun 1st 2009, 05:23 AM
Aaron01424
Hello, trying to factorise:

x^2-5x-6

So far i've got:

AC = -6, so x^2+x-6x-6 = x(x+1)-6(x+1)

Now getting to that point all makes sense to me, however in my textbook it says 'x+1 is a factor of both terms, so take that outside the bracket', leaving the answer as (x+1)(x-6). I don't understand how they got from x(x+1)-6(x+1) to (x+1)(x-6). Any help would be appreciated, thanks.
• Jun 1st 2009, 05:39 AM
mr fantastic
Quote:

Originally Posted by Aaron01424
Hello, trying to factorise:

x^2-5x-6

So far i've got:

AC = -6, so x^2+x-6x-6 = x(x+1)-6(x+1)

Now getting to that point all makes sense to me, however in my textbook it says 'x+1 is a factor of both terms, so take that outside the bracket', leaving the answer as (x+1)(x-6). I don't understand how they got from x(x+1)-6(x+1) to (x+1)(x-6). Any help would be appreciated, thanks.

Can you factorise xA - 6A ?
• Jun 1st 2009, 06:14 AM
Showcase_22
Since both terms are multiplied by (x+1), you can just take x+1 outside the brackets to give $x^2-5x-6=(x-6)(x+1)$.
• Jun 1st 2009, 06:39 AM
Aaron01424
Quote:

Originally Posted by mr fantastic
Can you factorise xA - 6A ?

Yeah, A(x-6).

Quote:

Originally Posted by Showcase_22
Since both terms are multiplied by (x+1), you can just take x+1 outside the brackets to give $x^2-5x-6=(x-6)(x+1)$.

Yeah, that's what it says in the book. Maybe i'm overcomplicating it, but I don't understand what you do when you take it out of the brackets.
• Jun 1st 2009, 06:52 AM
mr fantastic
Quote:

Originally Posted by Aaron01424
Yeah, A(x-6).

Yeah, that's what it says in the book. Maybe i'm overcomplicating it, but I don't understand what you do when you take it out of the brackets.

OK. Now replace A with (x + 1).
• Jun 1st 2009, 09:13 AM
Aaron01424
Aha. I see, great. Cheers.