Factoring.

Okay, so I phoned a friend and got that one sorted out. I have one more factorising question that I'm stuck on now. 2x^2+13x-7

Guest:

Okay, so I phoned a friend and got that one sorted out. I have one more factorising question that I'm stuck on now. 2x^2+13x-7

One of the answers is (2x - 1)(x + 7). How I got there, might be a bit tricky, I'll try to explain.

When you have an equation of the 2nd order you will always end up with (ax + c)(bx + d) from fx^2 + gx + h.

What you now instantly is that a*b=f, and that d*c = h. But these have multiple combinations, and what g will end up as is based on which of these combinations you use. ex. a*b=f, 4*1=4, 2*2=4, these two will give different values on g.

what you know about g is that g = (a*d + c*b)

Start from there and try to find a good strategy that work for yourself.

Well, that's all I can remember about factorisation right now. For me factorisation has always been a bit of a trial and error, but I have gotten highest grades in all of my math courses so far, so it should be sufficient.

x^2-y^2=(x-y)(x+y)
25p^2-100q^2=(5p-10q)(5p+10q)

I hope it is helpful

Guest:

x^2-y^2=(x-y)(x+y)
25p^2-100q^2=(5p-10q)(5p+10q)

I hope it is helpful

ax^2+bx+c

I assume that a=d*e and c=f*g.
if (d*g)+(e*f)=b, we can factorize the equation lie that (dx+f)(ex+g).

2x^2-13x-7
| |
V V
1 -7
2 1

2*(-7)+1*1=-13, so now we can write (2x-7)(x+1)