There are different situations where you may need to solve a polynomial equation:
GENERAL PROCEDURE 
EXAMPLE: Solve the equation: $$\cssId{s67}{x^5  2 = 2x^3 + 2x^2 + 3x}$$ 

STEP 1: Write the equation in the form $P(x) = 0$. That is, get the number zero on the righthand side; the polynomial on the left is asigned the named $\,P(x)\,$. The zeros of $\,P\,$ are the solutions of the equation. 
Subtract $\,2x^3 + 2x^2 + 3x\,$ from both sides to get this equivalent equation: $$ \cssId{s74}{\underbrace{x^5  2x^3  2x^2  3x  2}_{P(x)} = 0} $$  
STEP 2: Use the Bounding the Real Roots of a Polynomial theorem to get a bound on the real zeros of $P$. Call this bound $\,K\,$. Use a graphing device (such as a graphing calculator or the WolframAlpha widget supplied below) to graph $\,P\,$ using the initial window $\,10\le y\le 10\,$ and $\,K\le x\le K\,$. You're looking for a ‘nice’ zero—preferably an integer. Use the behavior of the graph to guess the multiplicity of each zero. Find all ‘nice’ zeros, regraphing on tighter intervals as needed. 
$$
\begin{align}
\cssId{s84}{K}
&\cssId{s85}{= 1 + \frac{\text{max of absolute values of coefficients}}{\text{absolute value of leading coefficient}}}\cr
&\cssId{s86}{= 1 + \frac{3}{1} = 4}
\end{align}
$$
From the graph below (obtained using the WolframAlpha widget at left), it's clear that there are only two real zeros: $1$ and $2$ The number $\,2\,$ is clearly a simple zero. The number $\,1\,$ probably has multiplicity two. 

STEP 3: CHECK your (potential) zeros from Step 2. You don't want to think that (say) $\,2\,$ is a zero, when it's actually $\,1.9\,$! Use synthetic division and the remainder theorem to do the checks. 
Don't forget the $\,0\,$ from the missing $\,x^4\,$ term in $\,P(x)\,$!


STEP 4: Use the bottom row of the synthetic division to get the factor that remains after all factors corresponding to real zeros have been divided out. 
The $\ \ \color{green}{1}\ \ \ \color{green}{0}\ \ \ \color{green}{1}\ \ $ from the last row
of the synthetic division represents: $1x^2 + 0x + 1 = x^2 + 1$
Thus, we have: $$ \cssId{s101}{\frac{P(x)}{(x2)(x+1)(x+1)} = x^2 + 1} $$ or, equivalently, $$ \cssId{s103}{x^5  2x^3  2x^2  3x  2 = (x2)(x+1)^2(x^2 + 1)} $$ 

STEP 5: Set the last factor (if any) equal to zero, and solve if possible. 
$$ \begin{gather} \cssId{s106}{x^2 + 1 = 0}\cr \cssId{s107}{x^2 = 1}\cr \cssId{s108}{x = \pm\sqrt{1}}\cr \cssId{s109}{x = \pm i} \end{gather} $$  
STEP 6: Use the results from Step 5 to factor $\,P(x)\,$ as completely as possible. 
$$\cssId{s112}{P(x) = (x2)(x+1)^2(x  i)(x + i)}\,$$  
STEP 7: State the solutions to the original equation. 
The solutions of the original equation (where repetition indicates multiplicity) are: $1\,$, $1\,$, $2\,$, $i\,$, and $i$ 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
