# Graphing and Finding Roots of Polynomial Functions

Note that the Intermediate Value Theorem (IVT) is addressed here in the Limits and Continuity section.

We learned what a Polynomial is here in the Introduction to Multiplying Polynomials section. Remember that polynomial is just a collection of terms with coefficients and/or variables, and none have variables in the denominator (if they do, they are Rational Expressions). Many times, we’re given a polynomial in Standard Form, and we need to find the zeros or roots. We typically do this by factoring, like we did with Quadratics in the Solving Quadratics by Factoring and Completing the Square section. We also did more factoring in the Advanced Factoring section.

## Review of Polynomial Functions

As a review, here are some polynomials, their names, and their degrees. Remember that the degree of the polynomial is the highest exponent of one of the terms (add exponents if there are more than one variable in that term). Note again that a function such as $\displaystyle f\left( x \right)=\frac{5}{{{{x}^{2}}}}$ is not a polynomial, because of the ${{x}^{2}}$ in the denominator.

## Polynomial Graphs and Roots

We learned that a Quadratic Function is a special type of polynomial with degree 2; these have either a cup-up or cup-down shape, depending on whether the leading term (one with the biggest exponent) is positive or negative, respectively. Think of a polynomial graph of higher degrees (degree at least 3) as quadratic graphs, but with more twists and turns.

Remember that when a quadratic crosses the $x$-axis (when $y=0$), we call that point an $x$-intercept, rootzero, solution, value, or just “solving the quadratic”. Also remember that not all of “solutions” are real; this is when the quadratic graph never touches the $x$-axis. We learned about those Imaginary (Non-Real) and Complex Numbers here. Non-real solutions are still called roots or zeros, but not $x$-intercepts.

The same is true with higher order polynomials. If we can factor polynomials, we want to set each factor with a variable in it to 0, and solve for the variable to get the roots. This is because any factor that becomes 0 makes the whole expression 0. (This is the zero product property: if $ab=0$, then $a=0$ and/or $b=0$).

So, to get the roots (zeros) of a polynomial, we factor it and set the factors to 0. If $x-c$ is a factor, then $c$ is a root (more generally, if $ax-b$ is a factor, then $\displaystyle \frac{b}{a}$ is a root.)

Note these things about polynomials:

• Every polynomial can be factored completely if complex (non-real roots) are allowed; the total number of real/non-real roots is its degree. (Fundamental Theorem of Algebra)
• The maximum number of real roots is its degree.
• The maximum number of turning points of a polynomial is one less than its degree.

For example, a polynomial of degree 3, like $y=x\left( {x-1} \right)\left( {x+2} \right)$, has at most 3 real roots and at most 2 turning points, as you can see:

Notice for local maximums, the graph “concaves down” and for local minimums, the graph “concaves up”.  (We’ll talk about this in Calculus and Curve Sketching). Pretty cool!

The polynomial $y=x\left( {x-1} \right)\left( {x+2} \right)$ is in Factored Form, since we can “see all the factors”: $x,\,\left( {x-1} \right)$, and $\left( {x+2} \right)$. If we were to multiply it out, it would become $y=x\left( {{{x}^{2}}+x-2} \right)={{x}^{3}}+{{x}^{2}}-2x$; this is called Standard Form since it’s in the form $f\left( x \right)=a{{x}^{n}}+b{{x}^{{n-1}}}+c{{x}^{{n-2}}}+….\,d$.

Here is an example of a polynomial graph that is degree 4 and has 3 “turns”. Notice that we have 3 real solutions, two of which pass through the $x$-axis, and one “touches” it or “bounces” off of it (because of the ${{\left( {x+1} \right)}^{2}}$, actually):

## End Behavior and Leading Coefficient Test

There are certain rules for sketching polynomial functions, like we had for graphing rational functions. Let’s first talk about the characteristics we see in polynomials, and then we’ll learn how to graph them.

Again, the degree of a polynomial is the highest exponent if you look at all the terms (you may have to add exponents, if you have a factored form). The leading coefficient of the polynomial is the number before the variable that has the highest exponent (thus, the degree).

As examples, for $y=-x-2x+5{{x}^{4}}+2x-8$, the degree is 4, and the leading coefficient is 5; for $y=-5x{{\left( {x+2} \right)}^{2}}\left( {x-8} \right){{\left( {2x+3} \right)}^{3}}$, the degree is 7 (add exponents since the polynomial isn’t multiplied out and don’t forget the $x$ to the first power), and the leading coefficient is $-5{{\left( 2 \right)}^{3}}=-40$.

Note: In factored form, sometimes you have to factor out a negative sign. Note though, as an example, that \begin{align}{{\left( {3-x} \right)}^{{\text{odd power}}}}&={{\left( {-\left( {x-3} \right)} \right)}^{{\text{odd power}}}}\\&=-{{\left( {x-3} \right)}^{{\text{odd power}}}}\end{align}, but \begin{align}{{\left( {3-x} \right)}^{{\text{even power}}}}&={{\left( {-\left( {x-3} \right)} \right)}^{{\text{even power}}}}\\&={{\left( {x-3} \right)}^{{\text{even power}}}}\end{align}. So be careful if the factored form contains a negative $x$.

**Notes on End Behavior: To get the end behavior of a function, we just look at the smallest and largest values of $x$, and see which way the $y$ is going. Not all functions have end behavior defined; for example, “periodic functions” (those that go up and down) don’t have end behaviors.

Most of the time, our end behavior looks something like this:$\displaystyle \begin{array}{l}x\to -\infty \text{, }\,y\to \,\,?\\x\to \infty \text{, }\,\,\,y\to \,\,?\end{array}$ and we have to fill in the $y$ part. For example, the end behavior for a line with a positive slope is: $\begin{array}{l}x\to -\infty \text{, }\,y\to -\infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$, and the end behavior for a line with a negative slope is: $\begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to -\infty \end{array}$. One way to think of end behavior is that for $\displaystyle x\to -\infty$, we look at what’s going on with the $y$ on the left-hand side of the graph, and for $\displaystyle x\to \infty$, we look at what’s happening with $y$ on the right-hand side of the graph.

The end behavior of the polynomial can be determined by looking at the degree and leading coefficient. The shape of the graphs can be determined by the $\boldsymbol{x}$ and $\boldsymbol{y}$interceptsend behavior, and multiplicities of each factor. We’ll talk about end behavior and multiplicity of factors next.

The table below shows how to find the end behavior of a polynomial (which way the $y$ is “heading” as $x$ gets very small and $x$ gets very large). Sorry; this is something you’ll have to memorize, but you always can figure it out by thinking about the parent functions given in the examples:

## Zeros (Roots) and Multiplicity

Each factor in a polynomial has what we call a multiplicity, which just means how many times it’s multiplied by itself in the polynomial (its exponent). If there is no exponent for that factor, the multiplicity is 1 (which is actually its exponent!) And remember that if you sum up all the multiplicities of the polynomial, you will get the degree!

For example, for the factored polynomial $y=2x{{\left( {x-4} \right)}^{2}}{{\left( {x+8} \right)}^{3}}$, the factors are $x$ (root 0 with multiplicity 1), $x-4$ (root 4 with multiplicity 2), and $x+8$ (root –8 with multiplicity 3). We can ignore the leading coefficient 2, since it doesn’t have an $x$ in it. Also, for just plain $x$, it’s just like the factor $x-0$. The total of all the multiplicities of the factors is 6, which is the degree. Remember that $x-4$ is a factor, while 4 is a root (zero, solution, $x$-intercept, or value).

Now we can use the multiplicity of each factor to know what happens to the graph for that root – it tells us the shape of the graph at that root. Factors with odd multiplicity go through the $x$-axis, and factors with even multiplicity bounces or touches the $x$-axis. We are only talking about real roots; imaginary roots have similar curve behavior, but don’t touch the $x$-axis.

Also note that sometimes we have to factor the polynomial to get the roots and their multiplicity. We may even have to factor out any common factors and then do some “unfoiling” or other type of factoring; for example, this polynomial has a difference of squares: $y=-{{x}^{4}}+{{x}^{2}}=-{{x}^{2}}\left( {{{x}^{2}}-1} \right)\,\,=-{{x}^{2}}\left( {x-1} \right)\left( {x+1} \right)$.

Here are the multiplicity behavior rules and examples:

Now, let’s put it all together to sketch graphs; let’s find the attributes and graph the following polynomials. Notice that when you graph the polynomials, they are sort of “self-correcting”; if you’ve done it correctly, the end behavior and bounces will “match up”. Also note that you won’t be able to determine how low and high the curves are when you sketch the graph; you’ll just want to get the basic shape. Now you can sketch any polynomial function in factored form!

## Writing Equations for Polynomials

You might have to go backwards and write an equation of a polynomial, given certain information about it:

## Conjugate Zeros Theorem (Conjugate Root Theorem)

We don’t always have real roots, or when we have real roots, they may be irrational (numbers that can’t be expressed as the ratio of two integers; see types of numbers here).

The Conjugate Zeros Theorem and (also called Conjugate Root Theorem, Conjugate Pair Theorem, or Irrational Conjugate Theorem), states that if $a+b\sqrt{c}$ is a root, then so is $a-b\sqrt{c}$. (You will see this from the Quadratic Formula.)

The complex form of this theorem, the Complex Conjugate Zeros Theorem, states that if $a+bi$ is a root, then so is $a-bi$; these are imaginary numbers. Here are the types of problems you may see:

## Synthetic Division

When we find the roots of Polynomial Functions, we need to learn how to do synthetic division. We learned Polynomial Long Division here in the Graphing Rational Functions section, and synthetic division does the same thing, but is much easier!

Remember again that if we divide a polynomial by “$x-c$” and get a remainder of 0, then “$x-c$” is a factor of the polynomial and “$c$” is a root, or zero.

Here is an example of Polynomial Long Division, where you can see how similar it is to “regular math” division:

Now let’s do the division on the right above using Synthetic Division:

It does get a little more complicated when performing synthetic division with a coefficient other than 1 in the linear factor. Remember that, generally, if $ax-b$ is a factor, then $\displaystyle \frac{b}{a}$ is a root. Here’s what we have to do:

Now we can use synthetic division to help find our roots! There are a couple more tests and theorems we need to discuss before we can start finding our polynomial roots!

## Rational Root Test

When we want to factor and get the roots of a higher degree polynomial using synthetic division, it can be difficult to know where to start! In the examples so far, we’ve had a root to start, and then gone from there. There’s this funny little rule that someone came up with to help guess the real rational roots (either an integer or fraction of integers) of a polynomial, and it’s called the rational root test (or rational zeros theorem):

For a polynomial function $f\left( x \right)=a{{x}^{n}}+b{{x}^{{n-1}}}+c{{x}^{{n-2}}}+….\,d$ with integers as coefficients (no fractions or decimals), if $p=$ the factors of the constant (in our case, $d$), and $q=$ the factors of the highest degree coefficient (in our case, $a$), then the possible rational zeros or roots are $\displaystyle \pm \frac{p}{q}$, where $p$ are all the factors of $d$ above, and $q$ are all the factors of $a$ above. Remember that factors are numbers that divide perfectly into the larger number; for example, the factors of 12 are 1, 2, 3, 4, 6, and 12.

Now this looks really confusing, but it’s not too bad; let’s do some examples. Notice how I like to organize the numbers on top and bottom to get the possible factors, and also notice how you don’t have repeat any of the quotients that you get:

The rational root test help us find initial roots to test with synthetic division, or even by evaluating the polynomial to see if we get 0. However, it doesn’t make a lot of sense to use this test unless there are just a few to try, like in the first case above.

## Factor and Remainder Theorems

There are a couple of theorems that you’ll learn about that will help you evaluate polynomials, and also be able to quickly tell if a given number is a root.

The Factor Theorem basically repeats something that we already know from above: $x-a$ is a factor of a polynomial if and only if $f(a)=0$ ($a$ is a root). For example, 3 is a root of $f\left( x \right)={{x}^{2}}-9$, $x-3$ is a factor, and $f\left( 3 \right)=0$. Also, as we see in the next theorem, a root of a polynomial has a remainder 0 with synthetic division.

The Remainder Theorem is a little less obvious and pretty cool! It says that if you evaluate a polynomial with $a$, the answer ($y$-value) will be the remainder if you were to divide the polynomial by $(x-a)$. For example, if you have the polynomial $f\left( x \right)=-{{x}^{4}}+5{{x}^{3}}+2{{x}^{2}}-8$, and if you put a number like 3 in for $x$, the value for $f(x)$ or $y$ will be the same as the remainder of dividing $-{{x}^{4}}+5{{x}^{3}}+2{{x}^{2}}-8$ by $(x-3)$. (And, of course, from the factor theorem, roots will produce remainders of 0). Let’s do the math for the non-root example; pretty cool, isn’t it?

$\displaystyle \begin{array}{l}f\left( 3 \right)=-{{\left( 3 \right)}^{4}}+5{{\left( 3 \right)}^{3}}+2{{\left( 3 \right)}^{2}}-8\\\,\,\,\,\,\,\,\,\,\,\,\,\,=-81\,\,+\,\,135\,\,+\,\,18\,-8\\\,\,\,\,\,\,\,\,\,\,\,\,\,=64\\\\\left. {\underline {\, {\,\,3\,\,} \,}}\! \right| \,\,-1\,\,\,\,\,\,\,5\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,0\,\,\,\,-8\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-3\,\,\,\,\,\,6\,\,\,\,\,\,\,24\,\,\,\,\,\,72\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1\,\,\,\,\,\,\,2\,\,\,\,\,\,8\,\,\,\,\,\,\,24\,\,\,\,\left| \!{\overline {\, {\,64\,\,} \,}} \right. \,\end{array}$

Here are some questions that you might see on Factor or Remainder Theorems:

## DesCartes’ Rule of Signs

There’s another really neat trick out there that you may not talk about in High School, but it’s good to talk about and pretty easy to understand. Yes, and it was named after a French guy!

The DesCartes’ Rule of Signs will tell you the number of positive and negative real roots of a polynomial $P\left( x \right)$ by looking at the sign changes of the terms of that polynomial. DesCartes’ Rule of Signs is most helpful if you’ve used the $\displaystyle \pm \frac{p}{q}$ method and you want to know which roots (positive or negative) to try first. It says:

• The number of positive real zeros equals the number of sign changes in $P\left( x \right)$ (or less by 2 down to 0 or 1 roots).
• The number of negative real zeros equals the number of sign changes in $P\left( {-x} \right)$ (or less by 2 down to 0 or 1 roots).

The best way is to show examples:

## Putting it All Together: Finding all Factors and Roots of a Polynomial Function

Here are some broad guidelines to find the roots of a polynomial function:

1. Take out any Greatest Common Factors (GCFs) of the polynomial, and you’ll have to set those to 0 too, to get any extra roots. For example, if you take an $x$ out, you’ll add a root of “0”. Always try to factor first if you have a polynomial of four terms or less.
2. If you have access to a graphing calculator, graph the function and determine if there are any rational zeros with which you can use synthetic division. If you don’t have a calculator, guess a possible rational zero using the $\displaystyle \pm \frac{p}{q}$ method above.
3. Perform synthetic division (or long division, if synthetic isn’t possible) to determine if that root yields a remainder of zero. You can also just evaluate the possible root: plug it everywhere there is an $x$ (or whatever variable you are using) to see if you end up with a $y$ or $f(x)$ of 0; if you do, it’s a root. To see if “1 is a factor, you can just add up all the coefficients and see if you get 0 (see how that works?)!
4. If the remainder is 0 (root), use synthetic division again, if necessary, with the bottom numbers (not the remainder), trying another possible root. Do this until you get down to the quadratic level. At that point, try to factor or use Quadratic Formula with what you have left. You may end up with imaginary roots.
5. Remember that if you end up with an irrational root or non-real root, the conjugate is also a root. For example, if $3+\sqrt{{17}}$ is a root, then $3-\sqrt{{17}}$ is also a root, or if $3+i$ is a root, then $3-i$ is also a root (Conjugates Zeros Theorem). You might want to find the factor represented for the two irrational and/or complex roots and then perform long division to get remaining roots.
6. (Optional) Use the DesCartes’ Rule of Signs to determine the number of positive and negative real roots.
7. Remember again that a polynomial with degree $n$ will have a total of $n$ roots.

Let’s first try some problems where we are given one root, as a start; this is a little easier: use synthetic division to find all the factors and real (not imaginary) roots of the following polynomials. In these examples, one of the factors or roots is given, so the remainder in synthetic division should be 0. Remember to take out a Greatest Common Factor (GFC) first, like in the second example. Notice that we can use synthetic division again by guessing another factor, as we do in the last problem:

Here are a few more with irrational and complex roots (using the Conjugate Zeros Theorem):

Now let’s try to find roots of polynomial functions without having a first root to try. Here are examples (assuming we can’t use a graphing calculator to check for roots). When you do these, make sure you have your eraser handy!

Now let’s see some examples where we end up with irrational and complex roots; remember to use the Quadratic Formula, if needed. Note that in the second example, we say that ${{x}^{2}}+4$ is an irreducible quadratic factor, since it can’t be factored any further (therefore has imaginary roots).

## Finding Polynomial Characteristics Using a Graphing Calculator

You might also be asked to find characteristics of polynomials, including roots, local and absolute minimums and maximums (extrema), and increasing and decreasing intervals; we can do this with a graphing calculator. We looked at Extrema and Increasing and Decreasing Functions here in the Advanced Functions: Compositions, Even and Odd, and Extrema section, and we also looked how to find the minimums or maximums (the vertex) in the Introduction to Quadratics section. Here’s the type of problem you might see:

## Solving Polynomial Inequalities

We worked with Linear Inequalities, Quadratic Inequalities, and Rational Inequalities earlier.

Now that we know how to solve polynomial equations (by setting everything to 0 and factoring, then setting factors to 0), we can work with polynomial inequalities. The reason we might need these inequalities is, for example, if we were taking the volume of something with $x$’s in each dimension, and we wanted the volume to be less than or greater than a certain amount. We can solve these inequalities either graphically or algebraically.

When solving algebraically, we set the polynomial to 0 as an equation (where the leading coefficient is positive; we may have to change the inequality sign), factor it, and solve for the critical values, which are the roots. We can then use a sign chart or sign pattern, which is a number line that is separated into intervals with “critical values” that you get by setting the quadratic to 0 and solving for $x$ (the roots). With sign charts, we pick the intervals by looking at the inequality and put pluses and minuses in the intervals, depending on what a sample value in that interval gives us. Then, we pick the intervals with plus signs for greater than, and intervals with minus signs for less than. Sign charts will alternate positive to negative and negative to positive unless we have factors with even multiplicities (“bounces”).

When we solve graphically, we actually don’t have to set the polynomial to 0, but it’s better to do this, so we can solve the polynomial and get the exact values for the critical values. (We could use a graphing calculator, for example, and the Intersect feature to get these roots). Then, we pick the intervals above the $x$-axis for greater than, and intervals below the $x$-axis for less than. We could also try test points between each critical value to see if the original inequality works or doesn’t to get our answer intervals.

We have to be careful to either include or not include the points on the $x$-axis, depending on whether or not we have inclusive ($\le$ or $\ge$) or non-inclusive ($<$ and  $>$) inequalities. When you graph the functions or work them algebraically, I’d suggest putting closed circles on the critical values for inclusive inequalities, and open circles for non-inclusive inequalities.

Let’s try some problems, and solve both graphically and algebraically:

Here’s one more example:

Here are a couple more that are tricky: one with a factor that can never be 0, and one where we have to use the Quadratic Formula.

## Polynomial Applications

Earlier we worked with Quadratic Applications, but now we can branch out and look at applications with higher level polynomials. Here are some problems:

Problem:

Shannon, a cabinetmaker, started out with a block of wood, and then she hollowed out the center of the block. The dimensions of the block and the cutout is shown below.

(a)  Write (as polynomials in standard form) the volume of the original block, and the volume of the hole. (Ignore units for this problem.)    (b)  Write the polynomial for the volume of the wood remaining.

Solution:

Problem:

A cosmetics company needs a storage box that has twice the volume of its largest box. Its largest box measures 5 inches by 4 inches by 3 inches. The larger box needs to be made larger by adding the same amount (an integer) to each to each dimension. Find the increase to each dimension.

Solution:

Maximum Volume Problem:

A piece of cardboard 30 inches by 15 inches is made into an open donut box by cutting out squares of side $x$ from each corner.

(a) Write a polynomial $V\left( x \right)$ that represents the volume of this open box in factored form, and then in standard form.  (b) What would be a reasonable domain for the polynomial? (Hint: Each side of the three-dimensional box has to have a length of greater than 0 inches).  (c) Find the value of $x$ for which $V\left( x \right)$ has the greatest volume. Round to 2 decimal places.  (d) What is that maximum volume? Round to 2 decimal places.  (e) What are the dimensions of the three-dimensional open donut box with that maximum volume? Round to 2 decimal places.

Solution:
If we were to fold up the sides, the new length of the box is $\left( {30-2x} \right)$, the new width of the box is $\left( {15-2x} \right)$, and the height up of the box is “$x$” (since the outside pieces are folded up). The volume is length times width times height, so the volume of the box is the polynomial $V\left( x \right)=\left( {30-2x} \right)\left( {15-2x} \right)\left( x \right)$. Let’s answer the questions with a little help from a graphing calculator:

Cost Revenue Profit Problem

The price $p$ that a makeup company can charge for a certain kit is $p=40-4{{x}^{2}}$, where $x$ is the number (in thousands) of kits produced. It costs the makeup company $15 to make each kit. (a) Write a function of the company’s profit$ P$by subtracting the total cost to make$ x$kits from the total revenue (in terms of$ x$). (b) Currently, the company makes 1.5 thousand (1500) kits and makes a profit of$24,000. Write an equation and solve to find a lesser number of kits to make and still make the same profit.

Solution:

(a) Profit is total revenue to make all $x$-thousand kits minus the cost to make all $x$-thousand kits. The total revenue is price per kit times the number of kits (in thousands), or $\left( 40-4{{x}^{2}} \right)\left( x \right)$. The cost to make $x$-thousand kits is $15x$. Thus, the total profit of is $P\left( x \right)=\left( 40-4{{x}^{2}} \right)\left( x \right)-15x=40x-4{{x}^{3}}-15x=-4{{x}^{3}}+25x$.

(b) Since the company makes 1.5 thousand kits and makes a profit of 24 thousand dollars, we know that $P\left( {1.5} \right)=24$, or $24=-4{{\left( 1.5 \right)}^{3}}+25\left( 1.5 \right)$. (From this, we know that 1.5 is a root or solution to the equation). Now we need to find a different root for the equation $P\left( x \right)=-4{{x}^{3}}+25x-24$. We could find the other roots by using a graphing calculator, but let’s do it without:

Learn these rules, and practice, practice, practice!

For Practice: Use the Mathway widget below to try a Graphing Polynomial problem. Click on Submit (the blue arrow to the right of the problem) and click on Graph to see the answer.

You can also type in your own problem, or click on the three dots in the upper right-hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Exponential Functions – you are ready!

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