# Graphing and Finding Roots of Polynomial Functions

This section covers:

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 (these are Rational Expressions).

# 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).

# Polynomial Graphs

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.  As a matter of fact, the maximum number of top and bottom (cup-up and cup-down) turning points (turns) of a polynomial just happens to be one more than its degree!  For example, a polynomial of degree 3, like , has 3 – 1 = 2 “turns”: . Notice that the negative part of the graph is more of a “cup down” and the positive is more of a “cup up”.  Pretty cool!

The polynomial is in Factored Form, since we can “see all the factors”:   If we were to multiply it out, it would become; this is called Standard Form since it’s in the form .

Now remember that we said 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”.  And also remember that not all of the “solutions” were real – when the quadratic graph never touched the x axis.  We learned about those Imaginary (Complex) Numbers here.

Here is an example of a polynomial graph that is degree 4 and has 3 “turns” (sort of 2 “cup up”s and 1 “cup down” in the middle).  Notice that we have 3 real solutions, two of which pass through the x axis, and one “touches” it or “bounces” off of it:

# Polynomial Characteristics and Sketching Graphs

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 (the highest degree).

So 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 $${{\left( {3-x} \right)}^{{\text{odd power}}}}={{\left( {-\left( {x-3} \right)} \right)}^{{\text{odd power}}}}=-{{\left( {x-3} \right)}^{{\text{odd power}}}}$$, but $${{\left( {3-x} \right)}^{{\text{even power}}}}={{\left( {-\left( {x-3} \right)} \right)}^{{\text{even power}}}}={{\left( {x-3} \right)}^{{\text{even power}}}}$$. So be careful if the factored form contains a negative x.

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 x and y interceptsend behavior, and multiplicities of each factor.  We’ll talk about end behavior and multiplicity of factors next.

## End Behavior of Polynomials

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!

So for example, for the factored polynomial , 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.  So 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 (this has a difference of squares):

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:

Here are a few problems where we use the Conjugate Zeroes Theorem and Complex Conjugate Zeroes Theorem (also called Conjugate Root Theorem or Conjugate Pair Theorem), which states that if  $$a+b\sqrt{c}$$  is a root, then so is  $$a-b\sqrt{c}$$.  The complex form of this theorem states that if  $$a+bi$$  is a root, then so is  $$a-bi$$.

# Finding Roots (Zeros) of Polynomials

Remember that when we factor, 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, than a = 0 and/or b = 0).

Also remember that when we factor to solve quadratics or any polynomials, we can never just divide by factors (with variables) on both sides to get rid of them.  If we do this, we may be missing solutions!

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 a little more factoring here, in the Rational Functions and Equations section.

For higher level polynomials, the factoring can be a bit trickier, but it can be sort of fun — like a puzzle!

## Synthetic Division

We learned Polynomial Long Division here in the Graphing Rational Functions section, but let’s revisit and then I’ll show you a nifty way to the same thing that’s much easier!

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:

Remember 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.

Let’s use synthetic division to find all the factors and real (not imaginary) roots of the following polynomials.  In these examples, one of the factors 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:

## 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 (either an integer or fraction of integers) roots of a polynomial, and it’s called the rational root test (or rational zeros theorem):

For a polynomial function  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  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.

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

Now let’s try to find roots of polynomial functions without having a first root to try.  Remember that if you get down to a quadratic that you can’t factor, you will have to use the Quadratic Formula to get the roots.

Also remember that you may end up with imaginary numbers as roots, like we did with quadratics.

Here are some 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”.
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  method above.
3. Perform synthetic division 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.
4. 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.
5. Remember that if you end up with an irrational root or non-real root, the conjugate is also a root.   For example, if   is a root, then  is also a root, or if  3 + i is a root, then 3 – i is also a root (Conjugates Zeros Theorem).  You will see this from the Quadratic Formula.
6. (Optional) Use the DesCartes’ Rule of Signs (see below) 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.

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.  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).

# Factor and Remainder Theorems

There are a couple of theorems that you’ll learn about that will help you evaluate polynomials (for a given x, find the y) 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: if a number is a root of a polynomial (like if 3 is a root of , which it is), then when you divide 3 into  (like with synthetic division), you get a remainder of 0.  Also, .    Also, if 3 if a root of , then (x – 3) is a factor.

So it’s really many ways to say the same thing: a root of a polynomial makes the remainder 0, and also produces 0 when you plug in that number into the polynomial.  And if a number a is a root of a polynomial, then (xa) is a factor.

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 .  For example, if you have the polynomial , 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   by (x – 3).

Let’s do the math; pretty cool, isn’t it?

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

 Problem Solution Find $$P\left( {-3} \right)$$ for$$P\left( x \right)=2{{x}^{4}}+6{{x}^{3}}+5{{x}^{2}}-45$$ What does the result tell us about the factor $$\left( {x+3} \right)$$? Let’s just evaluate the polynomial for $$x=-3$$: $$P\left( {-3} \right)=2{{\left( {-3} \right)}^{4}}+6{{\left( {-3} \right)}^{3}}+5{{\left( {-3} \right)}^{2}}-45=0$$Since $$P\left( {-3} \right)=0$$, we know by the Factor Theorem that  –3  is a root and $$\left( {x-\left( {-3} \right)} \right)$$ or $$\left( {x+3} \right)$$ is a factor. Find the value of $$k$$ for which $$\left( {x-3} \right)$$ is a factor of: $$P\left( x \right)={{x}^{5}}-15{{x}^{3}}-10{{x}^{2}}+kx+72$$    Note: Without the factor theorem, we could get the $$k$$ by setting the polynomial to 0 and solving for $$k$$ when $$x=3$$: \begin{align}{{x}^{5}}-15{{x}^{3}}-10{{x}^{2}}+kx+72&=0\\{{\left( 3 \right)}^{5}}-15{{\left( 3 \right)}^{3}}-10{{\left( 3 \right)}^{2}}+k\left( 3 \right)+72&=0\\243-405-90+3k+72&=0\\3k&=180\\k&=60\end{align} Let’s use the Factor Theorem and synthetic division and solve for $$k$$ to make the remainder 0. Remember to put a 0 in for the $${{x}^{4}}$$ position since it’s missing in the polynomial:\begin{array}{l}\left. {\underline {\, {\,\,3\,\,} \,}}\! \right| \,\,\,\,\,1\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,-15\,\,\,\,\,\,-10\,\,\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,72\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,-18\,\,\,\,\,\,\,\,-84\,\,\,\,\,\,\,\,\,\,3\left( {k-84} \right)\,\,\,\,\,\,\,\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,-6\,\,\,\,\,\,\,-28\,\,\,\,\,\,\,k-84\,\,\,\,\left| \!{\overline {\, {\,72\,+\,3\left( {k-84} \right)} \,}} \right. \end{array}Now let’s solve for $$k$$ to make the remainder 0:\displaystyle \begin{align}72+3\left( {k-84} \right)&=0\\72+3k-252&=0\\3k-180&=0\\k&=\,\,60\end{align}So the whole polynomial for which 3 is a factor is: $$P\left( x \right)={{x}^{5}}-15{{x}^{3}}-10{{x}^{2}}+60x+72$$ When $$P\left( x \right)$$ is divided by $$\left( {x+12} \right)$$, which is $$\left( {x-\left( {-12} \right)} \right)$$, the remainder is –100.  Find $$P\left( {-12} \right)$$ Remember that the  Remainder Theorem says that if you evaluate a polynomial with $$a$$, that number will be the remainder if you were to divide the polynomial by $$\left( {x-a} \right)$$. So, simply by this theorem, $$P\left( {-12} \right)=-100$$. If $$P\left( x \right)=2{{x}^{4}}+6{{x}^{3}}+k{{x}^{2}}-45$$,find $$k$$ for which $$P\left( {-3} \right)=9$$ Again, we can use the Remainder Theorem and synthetic division to solve for $$k$$ to make the remainder 9. Remember to put a 0 in for the $$x$$ position:\begin{array}{l}\left. {\underline {\, {\,\,-3\,\,} \,}}\! \right| \,\,\,\,\,2\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,-45\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-6\,\,\,\,\,\,\,\,0\,\,\,\,\,-3k\,\,\,\,\,\,\,\,\,\,\,9k\,\,\,\,\,\,\,\,\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,k\,\,\,\,\,-3k\,\,\,\,\,\left| \!{\overline {\, {\,-45+9k} \,}} \right. \end{array}Solve for $$k$$ to make the remainder 9:\begin{align}-45+9k&=9\\9k&=54\\k&=\,\,\,6\end{align}The whole polynomial for which $$P\left( {-3} \right)=9$$ is:.$$P\left( x \right)=2{{x}^{4}}+6{{x}^{3}}+6{{x}^{2}}-45$$

# 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  by looking at the sign changes of the terms of that polynomial.   DesCartes’ Rule of Signs is most helpful if you’ve used the  method and you want to know whether to hone in on the positive roots or negative roots to test roots.   It says:

• The number of positive real zeros equals the number of sign changes in  (or less by 2 down to 0  or 1 roots)
• The number of negative real zeros equals the number of sign changes in  (or less by 2 down to 0  or 1 roots)

The best way is to show examples:

# Conjugate Zeros Theorem Problem

We talked a little bit about the Complex Conjugate Zeros Theorem here when we talked about all the steps required to find all the factors and roots of a polynomial.  This is also called the Complex Conjugates Root Theorem.

Here’s a type of problem that you might see that requires using synthetic division using complex roots.  The problem is based on the Conjugate Zeros Theorem.

# Solving Polynomial Inequalities

We worked with Linear Inequalities and Quadratic Inequalities earlier.

Now that we know how to solve polynomial equations (by setting everything to 0 and factoring, and 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 number.

We can solve these inequalities either graphically or algebraically.    In both cases, we set the polynomial to 0 as an equation, factor it, and solve for the critical values, which are the roots.

When we solve inequalities, we want to get 0 on the right hand side, and get the leading coefficient (highest degree) of x positive on the left side; this way we can look at the inequality sign and decide if we want values below (if we have a less than sign) or above (if we have a greater than sign) the x-axis.    And when we’re solving to get 0 on the right hand side, don’t forget to change the sign if we multiply or divide by a negative number.   (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).

(Note that 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.  But if we used a graphing calculator, for example, we could just use the Intersect feature to get where the two sides of the polynomial intersect).

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.   So 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.

For graphing the polynomials, we can use what we know about end behavior.

For solving the polynomials algebraically, we can use sign charts.   Again, a sign chart or sign pattern is simply a number line that is separated into intervals with boundary points (called “critical values”) that you get by setting the quadratic to 0 (without the inequality) and solving for x (the roots).

With sign charts,  we pick that interval (or intervals) by looking at the inequality (where the leading coefficient is positive) and put pluses and minuses in the intervals, depending on what a sample value in that interval gives us.  Sign charts will alternate positive to negative and negative to positive unless we have factors with even multiplicities (“bounces”).

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

Here’s one more example:

Here’s one more where we can ignore a factor that can never be 0:

# Polynomial Applications

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

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.

###### (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 will be , the new width of the box will be , and the height up of the box will “x” (since the outside pieces are folded up).  The volume is length x width x height, so the volume of the box is the polynomial .

Now 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.  So 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  $$P\left( x \right)$$  when x is 1.5,  must be 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  $$P\left( x \right)=-4{{x}^{3}}+25x-24$$  (since  $$0=-4{{\left( 1.5 \right)}^{3}}+25\left( 1.5 \right)-24$$).   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:

# Revisiting Factoring to Solve Polynomial Equations

We first learned about factoring methods here in the Solving Quadratics by Factoring and Completing the Square section. Then we got into a little more complicated factoring here in the Rational Functions and Equations section. Now that we know how to solve by Finding Roots of Polynomial Functions, we can do fancier factoring, and thus find more roots. Remember that when we want to find solutions or roots, we set the equation to 0, factorset each factor to 0 and solve.

Here are some examples of factoring and solving polynomial equations; solve over the reals:

Now we can do more advanced solving, since we know how to factor higher degree polynomials.  Note that we use the techniques that we learned in the  Finding Roots of Polynomial Functions section.   Let’s solve over the real and complex numbers:

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!

## 2 thoughts on “Graphing and Finding Roots of Polynomial Functions”

1. 1st Problem: Anne is making a rectangular box. the width is 4 more than twice the height, and length is 10 more than the width. Find the volume of each section, if she will divide the box into 4x equal parts.

2nd Problem: the volume of rectangular solid is 12x(cube)+8x(squared)-3x-2. The length of the solid is 3x+2 and the width is 2x+1. Find the Height.

Note: cube and squared in the parenthesis represent exponents. 🙂