This section covers:

**Review of Polynomials****Polynomial Graphs****Polynomial Characteristics and Sketching Graphs****End Behavior of Polynomials****Zeros (Roots) and Multiplicity****Writing Equations for Polynomials****Finding Roots of Polynomials****Synthetic Division****Rational Root Test****Putting it All Together: Finding all Factors and Roots of a Polynomial Function****Factor and Remainder Theorems****DesCartes’ Rule of Signs****Conjugate Zeros Theorem Problem****Solving Polynomial Inequalities****Polynomial Applications****More Practice**

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**).

(Note that we will also review Finding Roots of Polynomials in the **Solving by 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).

# 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

**), we call that point an**

*y*= 0**,**

*x*-intercept**root**,

**zero**,

**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, the degree is 4, and the leading coefficient is 5; for, 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 2

*x*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 , but . 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 intercepts**,

**end 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

**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:**

*x*## 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**),

**(root**

*x*– 4**4 with multiplicity**

**2**), and

**(root –8 with multiplicity**

*x*+ 8**3**). We can ignore the leading coefficient 2, since it doesn’t have an

**in it. Also, for just plain**

*x***, it’s just like the factor**

*x***. So the total of all the**

*x*– 0**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**. We are only talking about

*x*axis**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

**= the factors of the highest degree coefficient (in our case,**

*q**a*), then the

**possible rational zeros**or

**roots**are where

**are all the**

*p***factors**of

*above, and*

**d****are all the factors of**

*q**above.*

**a**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**:

- Take out any Greatest Common Factors (
**GCF**s) of the polynomial, and you’ll have to set those to 0 too, to get any extra roots. For example, if you take anout, you’ll add a root of “0”.*x* - 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. - 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(or whatever variable you are using) to see if you end up with a*x*or of*y***0**; if you do, it’s a 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. - 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 +**is a root, then*i***3 –**is also a root (*i***Conjugates Zeros Theorem**). You will see this from the Quadratic Formula. - (Optional) Use the
**DesCartes’ Rule of Signs**(see below) to determine the number of positive and negative real roots. - 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

**) and also be able to quickly tell if a given number is a root.**

*y*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 (

**–**

*x***) is a factor.**

*a*The **Remainder Theorem** is a little less obvious and pretty cool! It says that if you evaluate a polynomial with ** a**, the answer (

*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*

**y****, the value for or**

*x**will be the same as the*

**y****remainder**of dividing by (

*x*– 3).

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

Here are some questions that you might see over the **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 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.

###### (a) Write a polynomial 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 has the greatest volume. Round to 2 decimal places.

*x*

###### (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 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

**is the number (in thousands) of kits produced. It costs the makeup company $15 to make each kit.**

*x*(a) Write a function of the company’s profit ** P** by subtracting the total cost to make

*kits from the total revenue (in terms of*

**x***).*

**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

**thousand**

*x***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

*thousand kits is 15*

**x***. So the total profit of is \(P\left( x \right)=\left( 40-4{{x}^{2}} \right)\left( x \right)-15x=40x-4{{x}^{3}}-15x=-4{{x}^{3}}+25x\).*

**x**(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:

**Learn these rules, and practice, practice, practice!**

Hit Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the **Mathway** site, where you can register for the **full version** (steps included) of the software. You can even get math worksheets.

You can also go to the **Mathway** site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to **Solving by Factoring** – you are ready!

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,

Lisa

thank you very much.

i appreciate your efforts in this site , and for all of the usefull informations and excellent explanation.

thanks a lot

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 🙂

Lisa

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!

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

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

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.

Probably the simplest function would be y = x^2. Lisa

I love your website! I teach at an all girls school and I can’t wait to share your site with them!

Just found a typo, and wanted to share though :(….in the second example in the “Zeros (Roots) and Multiplicity” section in the graph you note the y-intercept is (-16,0), when in fact it’s (0,-16). Thanks so much again!

Thanks so much for writing and finding the typo! Please look for more 😉 And I’m glad you like She Loves Math; let me know if there’s anything that’s missing, or how I can make it better. Thanks again! Lisa

This website is a really good review for my Algebra 2 class, and it is a lifesaver for homework. There is one thing that we went over in class that I didn’t fully get, and I didn’t see it on here. I’d like to know how write a polynomial from its roots, and more than that, what to do when imaginary numbers are involved, such as 1+i? Thanks so much for the great website, Sean.

Thanks for writing and thanks so much for pointing this out! I do have a section that addresses this some

here, but not with imaginary numbers, except for what I havehere>I definitely need to add more problems about this, though! Let me know what else is missing 😉

Oh and to answer your question, if 1+i is a root, because of the conjugates zero theorem, so is 1-i, so two of the factors are (x – (i+1))(x – (1-i)), so multiplying these together we get x^2-2x+2, so that’s one of the factors. Does that help? Lisa

wonderful work really she loves math I agree.

Awesome job! I use your site this month to help some of my Algebra 2 students and they felt more confident with the roots of polynomial functions.

Thanks so much for using SheLovesMath.com! I really appreciate your comment! Happy New Year! Lisa