by Dr. Carol JVF Burns (website creator)
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DEFINITION polynomial equations
A polynomial equation is an equation of the form $\,P(x) = Q(x)\,$, where $\,P\,$ and $\,Q\,$ are polynomials.

If $\,P\,$ has degree $\,m\,$ and $\,Q\,$ has degree $\,n\,$ for positive integers $\,m\,$ and $\,n\,$,
then the degree of the polynomial equation is $\,\max(m,n)\,$.

In other words, the degree of the polynomial equation is the highest power of $\,x\,$ that appears, when looking at both sides of the equation.

There are different situations where you may need to solve a polynomial equation:

  1. in a classroom, to demonstrate that you have learned appropriate solution techniques
  2. outside of a classroom (in a real-life application!) where you only need to get the solutions—you don't need to know how they were found
In the second situation, jump up to and type in something like this:

solve x^5 - 2 = 2x^3 + 2x^2 + 3x

Try it! You'll be presented with all the complex solutions (which of course include the real solutions), together with appropriate graphs.
Easy, quick, reliable, and free.

If you find yourself in the classroom situation, then this lesson is for you—and it will also be a great review, pulling together ideas from prior sections.

Solutions of Polynomial Equations: General Concepts

All polynomials in this lesson are assumed to have real number coefficients.

Therefore, instead of saying things like
‘Let $\,P(x) = ax + b\,$ where $\,a\,$ and $\,b\,$ are real numbers and $\,a\ne 0\,$’,
we can more simply say
‘Let $\,P(x) = ax + b\,$, where $\,a\ne 0\,$’.

An Approach for finding Exact Solutions of (some) Polynomial Equations of Degree $3^+$

Solve the equation:
$$\cssId{s67}{x^5 - 2 = 2x^3 + 2x^2 + 3x}$$

Write the equation in the form $P(x) = 0$.

That is, get the number zero on the right-hand 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: $$ \cssId{s74}{\underbrace{x^5 - 2x^3 - 2x^2 - 3x - 2}_{P(x)} = 0} $$

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, re-graphing 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.

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 $\,P(x)\,$!

$2$ $1$ $0$ $-2$ $-2$ $-3$ $-2$  
    $2$ $4$ $4$ $4$ $2$  
$-1$ $1$ $2$ $2$ $2$ $1$ $0$ $\leftarrow$ so $2$ is a zero
    $-1$ $-1$ $-1$ $-1$    
$-1$ $1$ $1$ $1$ $1$ $0$   $\leftarrow$ so $-1$ is a zero
    $-1$ $0$ $-1$      
  $\color{green}{1}$ $\color{green}{0}$ $\color{green}{1}$ $0$     $\leftarrow$ so $-1$ is a zero again

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: $$ \cssId{s101}{\frac{P(x)}{(x-2)(x+1)(x+1)} = x^2 + 1} $$ or, equivalently, $$ \cssId{s103}{x^5 - 2x^3 - 2x^2 - 3x - 2 = (x-2)(x+1)^2(x^2 + 1)} $$

Set the last factor (if any) equal to zero, and solve if possible.
$$ \begin{gather} \cssId{s106}{x^2 + 1 = 0}\cr \cssId{s107}{x^2 = -1}\cr \cssId{s108}{x = \pm\sqrt{-1}}\cr \cssId{s109}{x = \pm i} \end{gather} $$

Use the results from Step 5 to factor $\,P(x)\,$ as completely as possible.
$$\cssId{s112}{P(x) = (x-2)(x+1)^2(x - i)(x + i)}\,$$

State the solutions to the original equation.
The solutions of the original equation
(where repetition indicates multiplicity) are:

$-1\,$,   $-1\,$,   $2\,$,   $i\,$,   and   $-i$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
introduction to rational functions
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3 4 5 6 7 8 9

(MAX is 9; there are 9 different problem types.)