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How would I express $f(x)=2x^3-5x^2+3$ in the form $f(x)=(x-c)q(x)+r$ for $c = -2$?

I have no idea where to even start. Could anyone help me out?

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Matt, it would be a good idea for you to learn to type your questions using at least some basic Latex commands. It helps people who are reading your questions a lot, and besides, why would other people have to be editing your questions all the time? It is really not hard to learn the basics and it will improve the readability of your questions a lot. I'm saying this because you have asked several questions by now and they had to be edited by other users. – Adrián Barquero May 1 '11 at 0:32
will do, sorry! I didn't know it was as easy as including a few $ haha – Matt May 1 '11 at 0:40
up vote 6 down vote accepted

Use polynomial long division to divide $2x^3 - 5x^2 + 3$ by $x - (-2) = x + 2$. That is, you want to set up $x + 2 \overline{)2x^3 - 5x^2 + 0x + 3}$.

If you're not sure how to do this, I would recommend watching the process on a YouTube video, such as the one here.

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Horner (i.e., synthetic division ) is also useful. – J. M. May 1 '11 at 1:47
@J. M.: Synthetic division can seem like magic, while explicit long division can at least be related to integer long division. For someone comfortable with the "how/why" of polynomial long division, I agree, synthetic division can be very useful as a shortcut. – Hans Parshall May 1 '11 at 1:50

You do it by polynomial long division, similar to arithmetic long division. There is a detailed example on Wikipedia

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Another way is this.

First note that $f$ is of degree 3, and hence $q$ must have degree 2. Thus we get the equation $$f(x)=(x-(-2))(Ax^2+Bx+C)+r$$ but $f(x)=2x^3-5x^2 +3$, hence multiplying out the right hand-side and identifying coefficients we get $$A=2,\quad B+2A = -5,\quad C+2B =0,\quad 2C+r=3 $$
which is easily solved.

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