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).

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 Chart

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 Cubic Polynomial, has 3 – 1 = 2 “turns”: Cubic Graph. 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 polynomialCubic Polynomial is in Factored Form, since we can “see all the factors”: Factors of Polynomial  If we were to multiply it out, it would becomeMultiplied Polynomial; this is called Standard Form since it’s in the form Standard 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 Graph

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 forPolynomial 2, the degree is 4, and the leading coefficient is 5; forPolynomial, 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 –10 (you can tell by the –5 in front and the 2x in the factor with the highest exponent).

Note:  In factored form, sometimes you have to factor out a negative sign.  Note though, as an example,  that 3 - x to the odd even power, but 3 - x to the even odd power2.  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:

End Behavior of Polynomials

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 Example of Multiplicity, 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):  Factoring Polynomial

Here are the multiplicity behavior rules and examples:

Polynomial Multiplicity Chart

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.

Drawing Polynomial Graphs

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:

Writing Equations for Polynomials

Finding Roots (Zeros) of Polynomials

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:

Polynomial Long Division

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

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: 

Roots of Polynomials using Synthetic Division

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 Standard Form 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 p over 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:

Rational Root Test

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

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:

  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 P over q small 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 of 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 3 + square root of 17  is a root, then 3 - square root of 17 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!

Finding Roots of Polynomials

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).

Irrational and Complex Roots of Polynomials using Synthetic Division

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 x squared - 9, which it is), then when you divide 3 into x squared - 9 (like with synthetic division), you get a remainder of 0.  Also, Factor Theorem x squared - 9.    Also, if 3 if a root of x squared - 9, 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 x - a.  For example, if you have the polynomial Remainder Theorem Example, and if you put a number like 3 in for x, the value for f of x or y will be the same as the remainder of dividing Polynomial for Remainder Theorem  by (x – 3).

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

Remainder Theorem

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

Factor and 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 of x by looking at the sign changes of the terms of that polynomial.   DesCartes’ Rule of Signs is most helpful if you’ve used the P over q small 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 P of x (or less by 2 down to 0  or 1 roots)
  • The number of negative real zeros equals the number of sign changes in P of negative x (or less by 2 down to 0  or 1 roots)

The best way is to show examples:

Descartes Rule of Signs

Conjugate Zeros Theorem Problem

We talked a little bit about the Conjugate Zeros Theorem here when we talked about all the steps required to find all the factors and roots of a polynomial.

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.

Conjugate Zeros Theorem Problem

Polynomial Applications

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


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.



Polynomial Word 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.


Another Polynomial Word Problem

Maximum Volume Problem:

Donut Box

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 of x that represents the volume of this open box in factored form, and 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 at least 0 inches).
(c)  Find the value of x for which V of x 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.


If we were to fold up the sides, the new length of the box will be 30 - 2x, the new width of the box will be 15 - 2x, 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 Volume Polynomial.

Now let’s answer the questions with a little help from a graphing calculator:

Polynomial Application Problem

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.


(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:

Polynomial Profit Problem

Learn these rules, and practice, practice, practice!

On to Exponential Functions – you are ready!

13 thoughts on “Graphing and Finding Roots of Polynomial Functions

  1. Just stumbled upon your website and it’s great – I’m excited to pass it along to my Precalculus and Math 3 students as a resource. Thanks!

    • Erica,
      Thank you so much – comments like those make me want to keep writing and writing! Please let me know if there’s anything I can do to improve it :) Thanks again,

  2. thank you very much.
    i appreciate your efforts in this site , and for all of the usefull informations and excellent explanation.
    thanks a lot

  3. Hi, im a student in accelerated algebra 2, and i need some help. I know how to do most of the things listed but 1 thing my whole class and i are confused on is when none of the possible rational roots work in synthetic division and it wasn’t factorable from the start. A sample problem is x^2 + 6x + 11. We need to state the turns(1) sketch it, and state the x/y intercepts, which is what we can’t find. Please help i have a test on wednesday and thursday on this and have no clue what to do.

    • This is a great question! What I like to do to see if a quadratic can be factored is to look at the discriminant, b^2 – 4ac. If it’s a perfect square, it can be factored. If it’s negative (in this case), there are no real roots, and it never touches the x axis (so no x-intercepts). (If there the discriminant is +, there are 2 real roots (x-intercepts), if it’s 0, there is one 1 real root (x-intercept)). I don’t know if you’ve learned about complex roots yet, but you can use the Quadratic Formula to get the imaginary (complex) roots.

      So there will still be 1 turn (you take the degree of the polynomial, which is 2, and subtract 1. Now there will be a y-intercept – that’s when x = 0; this is (0, 11).

      So if you have a higher level polynomial, it’s hard to tell without graphing it on the calculator how many roots it is, but if it’s an even degree, it may never touch the x axis (have no real roots, or x-intercepts). But you can look for roots using the Rational Roots Theorem.

      This stuff is talked about in Graphing and Finding Roots of Polynomial Functions, but your question isn’t really covered. I think I’ll need to add it! Thanks!

      Let me know if you have more questions :)


      • Thank you soo much. This helped me out so much I finally feel more confident about the quiz. Just 1 more question. So you can tell if the graph is towards the 1st quadrant if it the discriminant is negative, and if its a positive non-perfect square it’s towards the 2nd quadrant? Since we can’t tell without the x-ints.

        • Great question again! No, you can’t tell which quadrant the graph is in by the discriminant; but remember that if the x^2 is positive, it goes UP, and if x^2 is negative, it goes down (and you can find end behavior similarly if it’s higher order) (see End Behavior of Polynomials).
          Also – you can always create a T-Chart and put in random points to see where it is.
          Good luck – you’ll do great!

  4. Sometimes when I pick a possible rational root it will equal zero but then for the next step I get an imaginary number. When I am later told the actual answer the one zero that I found is included but they started with a different number. I would just assume if it becomes imaginary that it is wrong but there are imaginary answers, so I don’t know how to tell if I have all of them or why it doesn’t work.

    • Kelly,
      Could you give me an example? If you get zero for a solution, and then continue with the synthetic division, you should get real roots before the imaginary ones (I think) – but I’d like to see an example to make sure. Thanks, Lisa

  5. Hello Lisa,
    Thanks for the wonderful website.

    I have some confusion with something in this page. It says :
    “p = the factors of the constant 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 a, and q are all the factors of d )).”

    You can please revise it it as I think the last sentence in opposite to first one and the real theorem.

    Thanks for your time

    • Thanks for writing! YES – you are absolutely correct; I had it wrong, and fixed it. Feel free to read the whole site and find more errors :)
      Thanks again, Lisa

  6. Give an example of a polynomial function (f) such that the function goes up on the left and up on the right and is symmetric about the y-axis.

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