Relationship Between the Zeros (Roots) and Factors of Polynomials
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 $\,x2\,$ is a factor of a polynomial $\,P(x)\,.$
Then: $$P(x) = (x2)(\text{stuff})$$
Thus:
$$\begin{align} \cssId{s6}{P(2)} &\cssId{s7}{\ =\ (22)(\text{stuff})}\cr &\cssId{s8}{\ =\ 0\cdot\text{stuff}}\cr &\cssId{s9}{\ =\ 0} \end{align} $$
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 alternative names for linear factors of the form $\,x  c\,$:
$c$  Linear Factor $xc$ 
Alternative 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:
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.
 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: $$\cssId{s101}{P(x) = (x+3)(x2)^2(x4)^2}$$ or $$\cssId{s103}{P(x) = 5(x+3)(x2)(x4)(x+1)(x7)}$$ 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
$\,K\,$:
$$P(x) = K(x1)(x)(x+3)(x7)$$
$$\begin{align} P(2) &= K(21)(2)(2+3)(27)\cr &= 50K \end{align} $$

Since $\,P(2) = 3\,$: $$ \begin{gather} \cssId{s129}{50K = 3\quad }\cr\cr \cssId{s130}{K = \frac{3}{50}} \end{gather} $$
Exactly one polynomial satisfies the stated conditions.
$$\cssId{s132}{P(x) = \frac{3}{50}x(x1)(x+3)(x7)}$$