Factor Theorem and Long Division For instance
Given $f(x) = 2x^3-7x^2+2x+3$, and $(x-3)$ is one of the factors, how do I obtain $2x^2-x-1$ as the quotient without using the long division. Is there any other method apart from long division?
Many thanks.
 A: Although there's already an answer linking to the method known as Ruffini's Rule, I detail its computation below, because it was very popular when I was in high school. It enables us to find easily the division of a polynomial by a linear term of the form $ x-x_0 $, even when the remainder $r(x)$ is not equal to $0 $. In your case $ x_0=3 $ and $r(x)=0$.
You want to find the unique second degree polynomial $ g (x) $ satisfying 
$$\frac{f(x)}{x-x_0}=\frac{2x^3-7x^2+2x+3}{x-{\color{blue}3}}=g (x).$$
Its coefficients can be found by Ruffini's Rule. Applied to this division it consists of the following algorithm:
\begin{array}{c|rrrl}
&  x^{3} & x^{2} & x^{1} & \phantom{-} x^{0}   \\ 
& 2 & -7 & 2 & \; \; \phantom{-}3 &\to \text{  coefficients of } f (x) \\
   x_0={ \color{blue}3} & \downarrow & 6 & -3 & \;\;-3 \\
              &    & {\color{gray}{=}}{\; \color{blue}3}{\color{gray}{\cdot 2}} &  {\color{gray}{=}}{\; \color{blue}3}{\color{gray}{ (-1)}} &  {\color{gray}{=}}{\; \color{blue}3}{\color{gray}{ (-1)}} \\
    \hline  & 2 & -1 & -1 & |\phantom {-}{\color{green}0}&\to \text{  coefficients of } g (x) \text{ and }{\color{green}{\text{remainder}}} \\
   & & { \color {gray} {=-7+6 } } &  { \color {gray}{=2-3} }  & | {   \color {gray}{=3-3 }} \\
 &  x^{2} & x^{1} & x^{0} & |\color{green}{\text{remainder}}
    \end{array}
which confirms that 
\begin{equation*}
g(x)=2x^{2}-x-1.
\end{equation*}
ADDED. In this answer I've posted another example of the application of Ruffini's rule.
A: You can simplify the process of long division by a linear factor using an algorithm known as synthetic division, also called Ruffini's Rule. You are working only with coefficients here plus you are dealing with addition rather than subtraction, which tends to be easier. 
You can read more here:
http://www.purplemath.com/modules/synthdiv.htm
https://en.m.wikipedia.org/wiki/Ruffini%27s_rule
A: You can set-up and solve simultaneous questions. 
You know that $2x^3-7x^2+2x+3$ will factorise as follows:
$$2x^3-7x^2+2x+3 \equiv (x-3)(ax^2+bx+c)$$
where $a$, $b$ and $c$ are constants you need to find. If you expand the right-hand side you get
$$2x^3-7x^2+2x+3 \equiv ax^3+(b-3a)x^2+(c-3b)x-3c$$
Comparing coefficients gives: $a=2$, $b-3a=-7$, $c-3b=2$ and $-3c=3$. You need to solve these simultaneously. It's clear that $a=2$ and $c=-1$ so we get $b=-1$ and hence
$$2x^3-7x^2+2x+3 \equiv (x-3)(2x^2-x-1)$$
I would recommend that you try to master long division. It is much easier when, for example, you are dividing a quartic by a quadratic and there is a remainder. For easy questions, the method above is fine, but for more complicated ones, division is the way to go.
A: To divide a polynomial $P(x)$ by another $D(x)$, you need to find a third polynomial $Q(x)$ such that 
$$P(x)=D(x)\times Q(x),$$ where the $\times$ denotes the product of polynomials. From this relation, we see that the degree of $P$ equals the sum of the degrees of $D$ and $Q$.
It turns out that in general this factorization is impossible, as is the case with integers. Anyway, the situation can be also rescued by considering a remainder polynomial $R(x)$ such that
$$P(x)=D(x)\times Q(x)+R(x),$$ and the degree of $R$ is smaller than that of $D$.
In the special case that $D(x)=x+a$, a binomial of the first degree, we have that the degree of $Q$ must be one less than that of $P$ and the degree of $R$ is $0$ ($R$ is a constant.)

The most efficient and easy way to obtain $Q$ is by long division.
A: Put
$$g(x) = \frac{2x^3 -7 x^2 + 2 x + 3 }{x-3}$$
To write $g(x)$ explicitly as a quadratic polynomial (where you remove the removable singularity at $x = 3$) can then be done in many different ways, long division, or equating it to an undetermined quadratic and solving for the coefficients as in Fly by Night's solution, or series expansion around x = infinity is equivalent to synthetic division mentioned by Deepak.
Another method is to use use Lagrange interpolation to find the quadratic polynomial. You then evaluate $y_i = g(x_i)$ at 3 different points $x_i$. You can then write:
$$g(x) = \sum_{i} y_{i}\prod_{j\neq i}\frac{x-x_j}{x_i-x_j}$$
