# RELATIONSHIP BETWEEN THE ZEROS (ROOTS) AND FACTORS OF POLYNOMIALS

• PRACTICE (online exercises and printable worksheets)

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 $\,x-2\,$ is a factor of a polynomial $\,P(x)\,$.
Then, $\,P(x) = (x-2)(\text{stuff})\,$.
Thus, $\,P(2) \ =\ (2-2)(\text{stuff}) \ =\ 0\cdot\text{stuff} \ =\ 0\,$.
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$x-c$ alternate namefor $\,x-c$ note $2$ $x-2$ $\displaystyle\frac 12$ $\displaystyle x-\frac 12$ $\displaystyle\frac 12(2x-1)$ 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 $\,P(x) = 0$
• $x-c\,$ is a factor of $\,P(x)\,$
• $x-c\,$ goes into $\,P(x)\,$ evenly (remainder is zero)
• $P(x) = (x-c)(\text{stuff})$

## 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 $\,x-4\,$ is a factor at this point, we have $\,P(x) = (x+3)(x-2)(x-4)\,$, 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: $$P(x) = (x+3)(x-2)^2(x-4)^2$$ or $$P(x) = 5(x+3)(x-2)(x-4)(x+1)(x-7)$$ There are many other correct answers. both $\,2\,$ and $\,3\,$are roots of $\,P\,$ $x = 5\,$ is a solutionof 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 $\,P(x) = (x-2)(x-3)(x-5)\,$, 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 $\,P(x) = (x-1)(x)(x+3)(x-7)\,$, which has degree $\,4$ the only possible remaining factor is a constant factor $\,K$ $\,P(x) = K(x-1)(x)(x+3)(x-7)\,$ $\,P(2) = K(2-1)(2)(2+3)(2-7) = -50K\,$ Since $\,P(2) = 3\,$,     $\,-50K = 3\,$,     $\displaystyle\,K = -\frac{3}{50}$ EXACTLY ONE polynomial satisfies the stated conditions $\displaystyle P(x) = -\frac{3}{50}x(x-1)(x+3)(x-7)$
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
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
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