In the prior section,
Introduction to Polynomials,
you began to see the beautiful relationship between
the zeros and factors of polynomials.
This section continues the discussion.
Suppose
[beautiful math coming... please be patient]$\,x2\,$ is a factor of a polynomial $\,P(x)\,$.
Then,
[beautiful math coming... please be patient]$\,P(x) = (x2)(\text{stuff})\,$.
Thus, $\,P(2) \ =\ (22)(\text{stuff}) \ =\ 0\cdot\text{stuff} \ =\ 0\,$.
Thus, $\,2\,$ is an input, whose corresponding output is zero.
Thus, $\,2\,$ is a zero of $\,P\,$.
So:
if $\,x\color{red}{2}\,$ is a factor, then $\color{red}{2}$ is a zero
The other direction is also true:
if $\color{red}{2}$ is a zero, then $\,x\color{red}{2}\,$ is a factor
Linear Factors
Factors of the form $\,ax + b\,$ are called linear factors.
Here, $\,a\,$ and $\,b\,$ are real numbers, with $\,a\ne 0\,$.
In a linear factor, you must have a variable (say, $\,x\,$), which can only be raised to the first power.
No $x^2$, no $x^3$, no $x$ in denominators, no $x$ under square roots, and so on.
The variable may be multiplied by a nonzero real number, and there may be a constant term.
Thus, all the following are examples of linear factors:
 $2x + 3$ ($a = 2$, $b = 3$)
 $2x  3$ ($a = 2$, $b = 3$)
 $\hphantom{2}x + 3$ ($a = 1$, $b = 3$)
 $\hphantom{2}x  3$ ($a = 1$, $b = 3$)
In particular, note that:
factors of the form $\,x  c\,$,
where $\,c\,$ is any real number,
are linear factors
Examples of Linear Factors of the form $\,x  c$
The table below also shows some alternate names for linear factors of the form $\,x  c\,$:
$c$ 
linear factor $xc$ 
alternate name for $\,xc$ 
note 
$2$ 
$x2$ 


$\displaystyle\frac 12$ 
$\displaystyle x\frac 12$ 
$\displaystyle\frac 12(2x1)$ 
factor out $\displaystyle\frac 12$ 
$2$ 
$x(2)$ 
$x + 2$ 
if $\,c\,$ is negative, then the factor takes the form ‘$x + (\text{positive #})$’ 
$\displaystyle \frac 53$ 
$\displaystyle x\frac 53$ 
$\displaystyle \frac 13(3x  5)$ 
factor out $\displaystyle\frac 13$ 
Equivalent Characterizations of a Zero of a Polynomial
The previous section, Introduction to Polynomials,
gave a list of
equivalent ways to characterize a zero of an arbitrary function.
For polynomials, we can extend the list:
ZEROS (ROOTS) OF POLYNOMIALS
A
zero ( or
root ) of a function is an input, whose
corresponding output is zero.
Let $\,P\,$ be a polynomial, and let $\,c\,$ be an input to $\,P\,$.
Then, the following are equivalent:
 $c\,$ is a zero of $\,P$
 $c\,$ is a root of $\,P$
 $c\,$ is an input, with corresponding output $\,0$
 $P(c) = 0$
 the point $\,(c,0)\,$ lies on the graph of $\,P\,$
 $P\,$ has an $\,x\,$intercept equal to $\,c$
 the graph of $\,P\,$ crosses the $x$axis at $\,c\,$
 $x = c$ is a solution of the equation $\,P(x) = 0$
 $xc\,$ is a factor of $\,P(x)\,$
 $xc\,$ goes into $\,P(x)\,$ evenly (remainder is zero)
 $P(x) = (xc)(\text{stuff})$
EXAMPLE
For each set of conditions given below, there is EXACTLY ONE, MORE THAN ONE, or NO polynomial $\,P\,$
that satisfies all the stated conditions.
 If no such polynomial exists, give a reason why.
 If exactly one polynomial exists, give a formula for the polynomial.
 If more than one polynomial is possible, give two different polynomials that satisfy
all the stated conditions.
SET OF CONDITIONS 
ANALYSIS 
ANSWER 
 the degree of $\,P\,$ is $\,5\,$
 $3\,$ is a zero of $\,P$
 $x  2\,$ is a factor of $\,P(x)$
 the graph of $\,P\,$ crosses the $x$axis at $\,x = 4$

The conditions are not necessarily used in the order given.
 since $3\,$ is a zero, $\,x + 3\,$ is a factor
 $\,x  2\,$ is a factor
 since the graph of $\,P\,$ crosses the $x$axis at $\,x = 4\,$,
the number $\,4\,$ is a zero,
so $\,x4\,$ is a factor
 at this point, we have $\,P(x) = (x+3)(x2)(x4)\,$,
which has degree $\,3\,$
 to get degree $\,5\,$, you can raise the existing factors to higher powers,
and/or put in additional factors
 you can also include a constant factor

MORE THAN ONE
polynomial is possible
For example:
$$P(x) = (x+3)(x2)^2(x4)^2$$
or
$$P(x) = 5(x+3)(x2)(x4)(x+1)(x7)$$
There are many other correct answers.

 both $\,2\,$ and $\,3\,$
are roots of $\,P\,$
 $x = 5\,$ is a solution
of the equation $\,P(x) = 0\,$
 the degree of $\,P\,$ is $\,2\,$

 since $2\,$ is a zero, $\,x  2\,$ is a factor
 since $3\,$ is a zero, $\,x  3\,$ is a factor
 since $\,P(5) = 0\,$, $\,x  5\,$ is a factor
 at this point, we have $\,P(x) = (x2)(x3)(x5)\,$,
which has degree $\,3\,$

NO polynomial
is possible
The stated
conditions force the polynomial to have at least degree $\,3\,$.

 $x  1\,$ goes into $\,P(x)\,$ evenly
 $P\,$ has degree $\,4$
 the points $\,(0,0)\,$ and $\,(3,0)\,$ are on the graph of $\,P\,$
 $P(2) = 3$
 $7\,$ is a zero of $\,P$

 $\,x  1\,$ is a factor
 since $\,0\,$ is a zero, $\,x\,$ is a factor
 since $\,3\,$ is a zero, $\,x + 3\,$ is a factor
 since $\,7\,$ is a zero, $\,x 7\,$ is a factor
 at this point, we have $\,P(x) = (x1)(x)(x+3)(x7)\,$, which has degree $\,4$
 the only possible remaining factor is a constant factor $\,K$
 $\,P(x) = K(x1)(x)(x+3)(x7)\,$
$\,P(2) = K(21)(2)(2+3)(27) = 50K\,$
Since $\,P(2) = 3\,$, $\,50K = 3\,$, $\displaystyle\,K = \frac{3}{50}$

EXACTLY ONE polynomial
satisfies the stated conditions
$\displaystyle P(x) = \frac{3}{50}x(x1)(x+3)(x7)$

Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
end behavior of polynomials