RELATIONSHIP BETWEEN THE ZEROS (ROOTS)
AND FACTORS OF POLYNOMIALS

LESSON READ-THROUGH
by Dr. Carol JVF Burns (website creator)
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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, $\,\cssId{s6}{P(2)} \cssId{s7}{\ =\ (2-2)(\text{stuff})} \cssId{s8}{\ =\ 0\cdot\text{stuff}} \cssId{s9}{\ =\ 0}\,.$
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:

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 name
for $\,x-c$
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.

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: $$\cssId{s101}{P(x) = (x+3)(x-2)^2(x-4)^2}$$ or $$\cssId{s103}{P(x) = 5(x+3)(x-2)(x-4)(x+1)(x-7)}$$ There are many other correct answers.
  • both $\,2\,$ and $\,3\,$
    are roots of $\,P\,$
  • $x = 5\,$ is a 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 $\,P(x) = (x-2)(x-3)(x-5)\,,$
    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:
1 2 3 4 5 6 7 8 9 10 11
AVAILABLE MASTERED IN PROGRESS

(MAX is 11; there are 11 different problem types.)