1. ## Division of Polynomials?

Hi there.
I have got used to the standard division of polynomials ie divide x^3 + 2x^2 + 2x + 4 = 0 by (x-2).
This seems to flow well and the answer is easy to find. BUT, what happens if there is no coefficient for x^2.

My question is:
show that (x-2) is a factor of x^3 - 7x + 6 (which I have done).

It then goes on to say solve the equation f(x) = 0 ??????

Normally I would divide f(x) by long division and then factorise the answer to get something like (x-2)(x-?)(x+?)=0

Am I missing something very simple here?

Any help would be appreciated. Rob

2. Hi rsmith!

If there is no $x^2$, you can just insert a $0x^2$ into the equation for yourself.

So it becomes:
$(x^3+0x^2-7x+6)$ / $(x-2)$

Can you go on from here? Or do you want me to go further with it?

3. Thanks for that. I'll have a go and you can check it if that's ok with you?

PS: where do you type all of the maths symbols, it's a pain if they're not used.

4. x^2 + 2x - 3 I think that's right. Thanks.

5. No problem!

And the maths stuff comes from what's called Latex, I'm onlt getting used to it myself to be honest, but this is what I use when I'm typing out stuff:
http://www.mathhelpforum.com/math-he...-tutorial.html

Just click on the link in MathMan's first post, and it'll bring up all the Latex rules in Adobe for you

6. That's what I got anyway!!