This section covers:
 Review of Polynomials
 Polynomial Graphs and Roots
 End Behavior of Polynomials and Leading Coefficient Test
 Zeros (Roots) and Multiplicity
 Writing Equations for Polynomials
 Conjugate Zeros Theorem
 Synthetic Division
 Rational Root Test
 Factor and Remainder Theorems
 DesCartes’ Rule of Signs
 Putting it All Together: Finding all Factors and Roots of a Polynomial Function
 Solving Polynomial Inequalities
 Polynomial Applications
 Revisiting Factoring to Solve Polynomial Equations
 More Practice
Note: 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.
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).
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 
Degree 
Number of Terms 
Name 
\(10{{x}^{3}}+4{{x}^{2}}+x4\) 
3 (from the \({{x}^{3}}\)) 
4  Cubic Polynomial 
\(t\left( {{{t}^{3}}+t} \right)={{t}^{4}}+{{t}^{2}}\) 
4 (from the \({{t}^{4}}\)) 
2  Quartic Binomial 
\(8\) 
0 (no variables) 
1  Constant Monomial 
\(\displaystyle \frac{{\left( {x+4} \right)}}{2}+\frac{{xy}}{{\sqrt{3}}}+3\) 
2 (from the \(xy\)) 
3  Quadratic Trinomial 
\(4{{x}^{3}}{{y}^{4}}+2{{x}^{2}}y+xy+3xy+x+y4\) 
7 (from the \({{x}^{3}}{{y}^{4}}\)) 
6 (since we can combine the \(xy\) and \(3xy\))  Polynomial of Degree 7 
\(x{{\left( {x+4} \right)}^{2}}{{\left( {x3} \right)}^{5}}\) 
8 (add up the exponents: \(1+2+5=8\)). 
(Difficult to say unless multiply out)  (Difficult to say unless multiply out)

Polynomial Graphs and Roots
We learned that a Quadratic Function is a special type of polynomial with degree 2; these have either a cupup or cupdown 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, root, zero, solution, value, or just “solving the quadratic”. Also remember that not all of the “solutions” were real – when the quadratic graph never touched the \(x\)axis. We learned about those Imaginary (NonReal) and Complex Numbers here. Nonreal solutions are still called roots or zeroes, 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 \(xc\) is a factor, then \(c\) is a root (more generally, if \(axb\) is a factor, then \(\displaystyle \frac{b}{a}\) is a root.)
Note these things about polynomials:
 The total number of real and nonreal roots is its degree
 The maximum number of real roots is its degree, and
 The maximum number of turning points is one less than its degree
For example, a polynomial of degree 3, like \(y=x\left( {x1} \right)\left( {x+2} \right)\), has at most 3 real roots and at most 2 turning points, as you can see:
Notice that when \(x<0\), the graph is more of a “cup down” and when \(x>0\), the graph is more of a “cup up”. (We’ll talk about this in Calculus and Curve Sketching). Pretty cool!
The polynomial \(y=x\left( {x1} \right)\left( {x+2} \right)\) is in Factored Form, since we can “see all the factors”: \(x,\,\left( {x1} \right)\), and \(\left( {x+2} \right)\). If we were to multiply it out, it would become\(y=x\left( {x1} \right)\left( {x+2} \right)=x\left( {{{x}^{2}}+x2} \right)={{x}^{3}}+{{x}^{2}}2x\); this is called Standard Form since it’s in the form \(f\left( x \right)=a{{x}^{n}}+b{{x}^{{n1}}}+c{{x}^{{n2}}}+….\,d\).
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:
Graph  Observations 
\(\boldsymbol{x}\)intercepts: Note that the \(x\)intercepts of the polynomial function are \((–4,0)\), \((–1, 0)\), and \((3, 0)\). Remember that the \(x\)intercepts are when \(y=0\), so these make sense when you look at the factored form of the polynomial: \(\displaystyle y=\left( {x+\text{4}} \right)\left( {x3} \right){{\left( {x+1} \right)}^{2}}\). These are also the roots.
Notice also that each factor has an odd exponent when the graph passes through the \(x\)axis and an even exponent when the function “bounces” off of the \(x\)axis. No coincidence here! (We’ll learn about this soon).
\(\boldsymbol{y}\)intercept: Note that the \(y\)intercept of the polynomial function (when \(x=0\)) is \((0,–12)\). Remember that there can only be one \(\boldsymbol{y}\)intercept; otherwise, it would not be a function (because of the vertical line test).
We see that the end behavior of the polynomial function is: \(\left\{ \begin{array}{l}x\to \infty ,\,\,y\to \infty \\x\to \infty ,\,\,\,\,\,y\to \infty \end{array} \right.\). Notice also that the degree of the polynomial is even, and the leading term is positive. No coincidence here either with its end behavior, as we’ll see.

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 (the highest degree).
For \(y=x2x+5{{x}^{4}}+2x8\), the degree is 4, and the leading coefficient is 5; for \(y=5x{{\left( {x+2} \right)}^{2}}\left( {x8} \right){{\left( {2x+3} \right)}^{3}}\), 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 \({{\left( {3x} \right)}^{{\text{odd power}}}}={{\left( {\left( {x3} \right)} \right)}^{{\text{odd power}}}}={{\left( {x3} \right)}^{{\text{odd power}}}}\), but \({{\left( {3x} \right)}^{{\text{even power}}}}={{\left( {\left( {x3} \right)} \right)}^{{\text{even power}}}}={{\left( {x3} \right)}^{{\text{even power}}}}\). 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 \(\boldsymbol{x}\)– and \(\boldsymbol{y}\)–intercepts, end 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!
So for example, for the factored polynomial \(y=2x{{\left( {x4} \right)}^{2}}{{\left( {x+8} \right)}^{3}}\), the factors are \(x\) (root 0 with multiplicity 1), \(x4\) (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 \(x0\). The total of all the multiplicities of the factors is 6, which is the degree.
Remember that \(x4\) 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): \(y={{x}^{4}}+{{x}^{2}};\,\,\,\,\,y={{x}^{2}}\left( {{{x}^{2}}1} \right);\,\,\,\,\,y={{x}^{2}}\left( {x1} \right)\left( {x+1} \right)\).
Here are the multiplicity behavior rules and examples:
Factor(s)  Root  Multiplicity  Behavior  Example Graph 
\(\left( {x+3} \right)\)  \(3\)  Odd: 1  Pass Through  
\({{\left( {x1} \right)}^{3}}\)  \(1\) 
Odd: 3, 5, … (the higher the odd degree, the flatter the “squiggle”) 
“Squiggle” Pass Through 

\({{x}^{2}}\)  \(0\)  Even: 2  Bounce or Touch  
\({{\left( {x+2} \right)}^{4}}\) 
\(2\) 
Even: 4, 6, … (the higher the even degree, the flatter the bounce) 
“Flatter” Bounce 
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 “selfcorrecting”; 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.
Graphing a Polynomial Example  Graphing a Polynomial Example  
\(y={{x}^{2}}\left( {x+2} \right)\left( {x1} \right)\)
\(y\)Intercept: Set \(x\) to 0 \(\begin{array}{c}y={{\left( 0 \right)}^{2}}\left( {0+2} \right)\left( {01} \right)=0\\\left( {0,0} \right)\end{array}\)
End Behavior: Leading Coefficient: Negative Degree: 4 (even) \(\begin{array}{l}x\to \infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\)

\(y=2\left( {x+2} \right){{\left( {x1} \right)}^{3}}\left( {x+4} \right)\)
\(y\)Intercept: Set \(x\) to 0 \(\begin{array}{c}y=2\left( {0+2} \right){{\left( {01} \right)}^{3}}\left( {0+4} \right)=2\left( 2 \right)\left( {1} \right)\left( 4 \right)=16\\(0,16)\end{array}\)
End Behavior: Leading Coefficient: Positive Degree: 5 (odd) \(\begin{array}{l}x\to \infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\)

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:
Problem  Solution 
Find a polynomial (Factored Form and Standard Form) with \(x\)intercepts of \((3,0)\), \((1,0)\) with multiplicity of 2; \(y\)intercept of \((0,2)\), and end behavior:
\(\begin{array}{c}x\to \infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\) 
To build the polynomial, start with the factors and their multiplicity. There will be a coefficient (positive or negative) at the beginning:
\(y=a\left( {x3} \right){{\left( {x+1} \right)}^{2}}\). The end behavior indicates that the polynomial has an odd degree with a positive coefficient; our polynomial above might work with \(a=1\).
But the \(y\)intercept is at \((0,2)\), so we have to solve for \(a\): \(\displaystyle 2=a\left( {03} \right){{\left( {0+1} \right)}^{2}};\,\,\,\,\,2=a\left( {3} \right)\left( 1 \right);\,\,\,\,\,\,\,a=\frac{{2}}{{3}}\,\,=\,\,\frac{2}{3}\)
The polynomial is \(\displaystyle y=\frac{2}{3}\left( {x3} \right){{\left( {x+1} \right)}^{2}}\). Multiply all the factors to get Standard Form: \(\displaystyle y=\frac{2}{3}{{x}^{3}}\frac{2}{3}{{x}^{2}}\frac{{10}}{3}x2\). (You can put all forms of the equations in a graphing calculator to make sure they are the same.) 
Find a polynomial equation in Factored Form for the graph’s function:

To build the polynomial, start with the factors and their multiplicity. It looks like \(x\)intercept \((3,0)\) has a multiplicity of 1; \((1,0)\) has a multiplicity of 2, and \((1,0)\) has a multiplicity of 3 (slight squiggle).
There will be a coefficient (positive or negative) at the beginning, so here’s what we have so far: \(y=a\left( {x+3} \right){{\left( {x+1} \right)}^{2}}{{\left( {x1} \right)}^{3}}\).
The end behavior indicates that the polynomial has an even degree and with a positive coefficient, so the degree is fine, and our polynomial will have a positive coefficient. Solving for \(a\) with our \(y\)intercept at \((0,6)\) should confirm that’s it’s positive: \(6=a\left( {0+3} \right){{\left( {0+1} \right)}^{2}}{{\left( {01} \right)}^{3}};\,\,\,\,6=a\left( 3 \right)\left( 1 \right)\left( {1} \right);\,\,\,\,\,\,a=2\)
The polynomial is \(y=2\left( {x+\,\,3} \right){{\left( {x+1} \right)}^{2}}{{\left( {x1} \right)}^{3}}\). 
Write a thirddegree polynomial \(P(x)\) in Standard Form (integer coefficients) with zeros at 0, \(\displaystyle \frac{{10}}{3}\), and \(\displaystyle \frac{{3}}{4}\).  In factored form, the polynomial would be \(\displaystyle P(x)=x\left( {x\frac{{10}}{3}} \right)\left( {x\frac{3}{4}} \right)\). Since we weren’t given a \(y\)intercept, we can take the liberty to write the polynomial with integer coefficients: \(P(x)=x\left( {3x+10} \right)\left( {4x3} \right)\). (See how we get the same zeros?) Multiplying out to get Standard Form, we get:
\(P(x)=12{{x}^{3}}+31{{x}^{2}}30x\). 
Conjugate Zeros Theorem
We don’t always have real roots, or when we have real roots, they may be rational (see types of numbers here).
The Conjugate Zeroes Theorem and (also called Conjugate Root Theorem or Conjugate Pair Theorem), states that if \(a+b\sqrt{c}\) is a root, then so is \(ab\sqrt{c}\). The complex form of this theorem, the Complex Conjugate Zeroes Theorem, states that if \(a+bi\) is a root, then so is \(abi\).
Here are the types of problems you may see:
Problem  Solution 
Find a polynomial (Factored Form and Standard Form) with \(x\)intercepts of \((4,0)\) and \(\left( {1\sqrt{3},\,\,0} \right)\), and \(y\)intercept of \(\left( {0,\,2} \right)\).  Since \(1\sqrt{3}\) is a root, by the conjugate pair theorem, so is \(1+\sqrt{3}\).
Let’s start building the polynomial: \(y=a\left( {x4} \right)\left( {x\left( {1\sqrt{3}} \right)} \right)\left( {x\left( {1+\sqrt{3}} \right)} \right)\). Note that this can be simplified to: \(\require{cancel} \begin{align}y&=a\left( {x4} \right)\left( {x1+\sqrt{3}} \right)\left( {x1\sqrt{3}} \right)\\&=a\left( {x4} \right)\left( {{{x}^{2}}x\cancel{{x\sqrt{3}}}x+1+\cancel{{\sqrt{3}}}+\cancel{{x\sqrt{3}}}\cancel{{\sqrt{3}}}3} \right)\end{align}\), which is \(y=a\left( {x4} \right)\left( {{{x}^{2}}2x2} \right)\)^{*} (distribute and multiply through the last two factors). Since the \(y\)intercept is at \((0,2)\), let’s solve for \(a\): \(\displaystyle 2=a\left( {04} \right)\left( {{{0}^{2}}2\left( 0 \right)2} \right);\,\,\,\,2=8a;\,\,\,\,a=\frac{1}{4}\).
The polynomial is \(\displaystyle y=\frac{1}{4}\left( {x4} \right)\left( {{{x}^{2}}2x2} \right)\). Multiply all the factors to get Standard Form: \(\displaystyle y=\frac{1}{4}{{x}^{3}}\frac{3}{2}{{x}^{2}}+\frac{3}{2}x+2\).
^{*}Note that there’s another (easier) way to find a factored form for a polynomial, given an irrational root (and thus its conjugate). Using the example above: \(1\sqrt{3}\) is a root, so let \(x=1\sqrt{3}\) or \(x=1+\sqrt{3}\) (both get same result). Then we have: \(\begin{array}{c}x=1\sqrt{3};\,\,\,\,x1=\left( {\sqrt{3}} \right);\,\,\,\,{{\left( {x1} \right)}^{2}}={{\left( {\sqrt{3}} \right)}^{2}}\\{{x}^{2}}2x+1=3;\,\,\,\,{{x}^{2}}2x2=0\end{array}\) The factor that represents these roots is \({{x}^{2}}2x2\). Pretty cool trick! 
Find a polynomial (Factored Form and Standard Form) with roots \(1,5,\) and \(\left( {2+3i} \right)\), that goes through the point \(\left( {1,\,160} \right)\) (\(f\left( 1 \right)=160\)) .  Since \(2+3i\) is a root, by the complex conjugate zeroes theorem, so is \(23i\).
Let’s start building the polynomial: \(\displaystyle y=a\left( {x+1} \right)\left( {x5} \right)\left( {x\left( {2+3i} \right)} \right)\left( {x\left( {23i} \right)} \right)\). Note that this can be simplified to: \(\displaystyle \begin{align}y&=a\left( {x+1} \right)\left( {x5} \right)\left( {x23i} \right)\left( {x2+3i} \right)\\&=a\left( {x+1} \right)\left( {x5} \right)\left( {{{x}^{2}}2x\cancel{{+3ix}}2x+4\cancel{{6i}}\cancel{{3ix}}\cancel{{+6i}}9{{i}^{2}}} \right)\end{align}\), which is \(\displaystyle y=a\left( {x+1} \right)\left( {x5} \right)\left( {{{x}^{2}}4x+13} \right)\). Since \(f\left( 1 \right)=160\), let’s find \(a\): \(\begin{array}{c}160=a\left( {1+1} \right)\left( {15} \right)\left( {{{1}^{2}}4\left( 1 \right)+13} \right)=a\left( 2 \right)\left( {4} \right)\left( {10} \right)\\160=80a;\,\,\,\,\,a=2\end{array}\)
The polynomial is \(\displaystyle y=2\left( {x+1} \right)\left( {x5} \right)\left( {{{x}^{2}}4x+13} \right)\). Multiply all the factors to get Standard Form: \(y=2{{x}^{4}}16{{x}^{3}}+46{{x}^{2}}64x130\).
^{*}Note that there’s another (easier) way to find a factored form for a polynomial, given a complex root (and thus its conjugate). Using the example above: \(2+3i\) is a root, so let \(x=2+3i\) or \(x=23i\) (both get same result). Then we have: \(\begin{array}{c}x=2+3i;\,\,\,\,x2=3i;\,\,\,\,{{\left( {x2} \right)}^{2}}={{\left( {3i} \right)}^{2}}\\{{x}^{2}}4x+4=9;\,\,\,\,{{x}^{2}}4x+13=0\end{array}\) The factor that represents these roots is \({{x}^{2}}4x+13\). 
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 “\(xc\)” and get a remainder of 0, then “\(xc\)” 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 \(axb\) is a factor, then \(\displaystyle \frac{b}{a}\) is a root.
Here’s what we have to do:
Synthetic Division when Coefficient of Divisor is Not 1  
Divide \(\displaystyle \frac{{12{{x}^{3}}5{{x}^{2}}5x+2}}{{3x2}}\)
Go down a level (subtract 1) with the exponents for the variables: \(4{{x}^{2}}+x1\). 
Since the coefficient of the divisor is not 1, we have to rewrite the fraction, and simplify to make this coefficient 1. To do this, I like to divide both the numerator (dividend) and denominator (divisor) by this coefficient; in our case, 3:
\(\displaystyle \begin{align}\frac{{12{{x}^{3}}5{{x}^{2}}5x+2}}{{3x2}}&=\frac{{\frac{{12{{x}^{3}}5{{x}^{2}}5x+2}}{3}}}{{\frac{{3x2}}{3}}}\\&=\frac{{4{{x}^{3}}\frac{5}{3}{{x}^{2}}\frac{5}{3}x+\frac{2}{3}}}{{x\frac{2}{3}}}\end{align}\)
Now, perform the synthetic division, using the fractional root (see left)!

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 (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 \(f\left( x \right)=a{{x}^{n}}+b{{x}^{{n1}}}+c{{x}^{{n2}}}+….\,d\) 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:
Polynomial  Possible Rational Roots 
\({{x}^{3}}+6{{x}^{2}}4x3\) 
\(\begin{align}\frac{{\pm 1,\,\,\,\pm 3}}{{\pm 1}}&=\,\,1,\,\,1,\,\,3,\,\,3\\\\&=\pm \,\,1,\,\,\pm \,\,3\end{align}\) 
\(3{{x}^{2}}+8x2\)  \(\displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 1,\,\,\,\pm 3}}\,\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2}}{{\pm 3}}\\\\&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,\frac{1}{3},\,\,\pm \,\,\frac{2}{3}\end{align}\) 
\(2{{x}^{4}}27{{x}^{3}}+4{{x}^{2}}+8\)  \(\require{cancel} \displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\pm 1,\,\,\,\pm 2}}\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 4,\,\,\pm 8}}{{\,\,\,\pm 2}}\,\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,4,\,\,\pm \,\,8,\,\,\pm \,\,\frac{1}{2},\,\,\cancel{{\pm \,\,1}},\cancel{{\pm \,\,2}},\cancel{{\pm \,\,4}}\end{align}\) 
\(12{{x}^{3}}{{x}^{2}}+2x6\)  \(\displaystyle \begin{align}\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 1,\,\,\,\pm 2,\,\,\pm 3,\,\,\,\pm 4,\,\,\pm 6,\,\,\,\pm 12}}\,\,&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 1}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 2}}\\&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 3}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 4}}\\&=\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\pm 6}},\,\,\,\frac{{\pm 1,\,\,\,\pm 2,\,\,\,\pm 3,\,\,\pm 6}}{{\,\,\,\pm 12}}\\\,\,&=\,\,\pm \,\,1,\,\,\pm \,\,2,\,\,\pm \,\,3,\,\,\pm \,\,6,\,\,\pm \,\,\frac{1}{2},\,\,\pm \,\,\frac{3}{2},\\\,\,\,\,\,\,\,&\pm \,\,\frac{1}{3},\,\,\pm \,\,\frac{2}{3},\,\,\pm \,\,\frac{1}{4},\,\,\pm \,\,\frac{3}{4},\,\,\pm \,\,\frac{1}{6}\,\,,\,\,\pm \,\,\frac{1}{{12}}\end{align}\) 
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
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 \({{x}^{2}}9\), which it is), then when you divide 3 into \({{x}^{2}}9\) (like with synthetic division), you get a remainder of 0. Also, \(f\left( 3 \right)=0\) for \(f\left( x \right)={{x}^{2}}9\). Also, if 3 if a root of \({{x}^{2}}9\), then \((x3)\) is a factor.
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 \((xa)\). 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 \((x3)\).
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 \(\left( {x3} \right)\) 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 \(x=3\):
\(\begin{align}{{x}^{5}}15{{x}^{3}}10{{x}^{2}}+kx+72&=0\\{{\left( 3 \right)}^{5}}15{{\left( 3 \right)}^{3}}10{{\left( 3 \right)}^{2}}+k\left( 3 \right)+72&=0\\24340590+3k+72&=0\\3k&=180\\k&=60\end{align}\) 
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 {\, Now let’s solve for \(k\) to make the remainder 0: \(\displaystyle \begin{align}72+3\left( {k84} \right)&=0\\72+3k252&=0\\3k180&=0\\k&=\,\,60\end{align}\) Therefore, the 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 \(\left( {xa} \right)\). 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 {\, 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 \(P\left( x \right)\) 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 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 \(P\left( x \right)\) (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:
Math 
Notes 
Using DesCartes’ rule of signs, find the number of positive and negative real roots for the following polynomial:
\(P\left( x \right)={{x}^{4}}+{{x}^{3}}3{{x}^{2}}x+2\) 
To find the positive roots, we use the polynomial “as is” and look for the terms to turn from “\(+\) to \(\)“ or “\(\) to \(+\)”. Note that the sign of a term is in front of it.
\(P\left( x \right)\,\,=\,\,+\,{{x}^{4}}\color{red}{+}{{x}^{3}}\color{red}{}3{{x}^{2}}\color{lime}{}x\color{lime}{+}2\)
We have 2 changes of signs for \(P\left( x \right)\), so there might be 2 positive roots, or there might be 0 positive roots. From earlier, we saw “1” was a root with multiplicity 2; this counts as 2 positive roots of 1.
Now let’s find the number of negative roots: \(P\left( {x} \right)\,\,=\,\,\color{red}{+}\,{{x}^{4}}\color{red}{}{{x}^{3}}\color{lime}{}3{{x}^{2}}\color{lime}{+}x+2\)
We also have 2 changes of signs for \(P\left( {x} \right)\), so there might be 2 negative roots, or there might be 0 negative roots. From earlier, we saw that both “–1” and “–2” were roots; these are our 2 negative roots. 
Using DesCartes’ rule of signs, find the number of positive and negative real roots for the following polynomial:
\(P\left( x \right)={{x}^{4}}5{{x}^{2}}36\)

To find the positive roots, we use the polynomial “as is” and look for sign changes. Note that you can skip over terms that aren’t there (like \({{x}^{3}}\)), since the sign changes will be the same:
\(P\left( x \right)=\color{red}{+}{{x}^{4}}\color{red}{}5{{x}^{2}}36\)
We have 1 change of signs for \(P\left( x \right)\), so there might be 1 positive root. From earlier we saw that “3” is a root; this is the positive root.
Now let’s find the number of negative roots (polynomial stays the same): \(P\left( {x} \right)=\color{red}{+}{{x}^{4}}\color{red}{}5{{x}^{2}}36\)
We also see 1 change of signs for \(P\left( {x} \right)\), so there might be 1 negative root. From earlier we saw that “–3” is a root; this is the negative root. 
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:
 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.
 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.
 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?)!
 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 nonreal 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 \(3i\) is also a root (Conjugates Zeros Theorem). You will see this from the Quadratic Formula. 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.
 (Optional) Use the DesCartes’ Rule of Signs 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.
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 Zeroes Theorem):
Polynomial Roots Problem  Solution 
\({{x}^{4}}+4{{x}^{3}}5{{x}^{2}}18x+18\)
\(1+\sqrt{7}\) is a root 
By the Conjugate Zeros Theorem, we know that since \(1+\sqrt{7}\) is a root, \(1\sqrt{7}\) is also a root. We could perform synthetic division with these roots, like we do in the next problem, but I’d probably find the factor for these roots, and then use long division:
\(\begin{array}{c}\left( {x\left( {1+\sqrt{7}} \right)} \right)\left( {x\left( {1\sqrt{7}} \right)} \right)\\=\left( {x+1\sqrt{7}} \right)\left( {x+1+\sqrt{7}} \right)={{x}^{2}}+2x6\end{array}\) (Note that there’s another (easier) way to find a factored form for a polynomial, given an irrational root, and thus its conjugate. Using the example above: \(1\sqrt{7}\) is a root, so let \(x=1\sqrt{7}\) or \(x=1+\sqrt{7}\) (both get same result). Then we have: \(\begin{array}{c}x=1\sqrt{7};\,\,\,\,x1=\left( {\sqrt{7}} \right);\,\,\,\,{{\left( {x1} \right)}^{2}}={{\left( {\sqrt{7}} \right)}^{2}}\\{{x}^{2}}2x+1=7;\,\,\,\,{{x}^{2}}2x6=0\end{array}\) We get the same root as above: \({{x}^{2}}2x6\).) Let’s do long division with this root: Now we can factor our quotient: \(\displaystyle {{x}^{2}}+2x3=\left( {x+3} \right)\left( {x1} \right)\). The zeroes are \(1+\sqrt{7},\,\,1\sqrt{7},\,\,3\), and \(1\). 
Find the other zeros for the following function, given \(5i\) is a root:
\({{x}^{3}}15{{x}^{2}}+76x130\) 
By the Complex Conjugate Complex Zeros Theorem, we know that since \(5i\) is a root, \(5+i\) is also a root. Instead of using long division as in the last problem, let’s try performing synthetic division using first \(5+i\), and then, with what’s leftover, divide by \(5+i\). This is a little tricky, since we have to FOIL out some of the complex numbers when we multiply, for example, \(\left( {10i} \right)\left( {5i} \right)=50+10i5i+{{i}^{2}}=50+5i1=51+5i\).
Perform synthetic division the same way though, keeping the reals separate from the imaginaries when adding: Note that we see that both \(5i\) and \(5+i\) go in without a remainder (which they should!) and we are left with \(x5\) from the “1 –5”. The zeros are \(5i,\,\,\,5+i\) and 5. Just to check, we can put the original polynomial in the calculator and see that there is, in fact, a zero at \(x=5\). 
Two roots of the polynomial are \(i\) and 1. Find the 3^{rd} and 4th roots.^{ }
\({{x}^{4}}+3{{x}^{3}}3{{x}^{2}}+3x4\)

There are several ways to do this problem, but let’s try this: By the Complex Conjugate Complex Zeros Theorem, we know that since \(i\) is a root, \(i\) is also a root, so that’s the 3^{rd} root. Let’s perform synthetic division with the real root 1:
Now let’s factor what we end up with: \({{x}^{3}}+4{{x}^{2}}+x+4={{x}^{2}}\left( {x+4} \right)+1\left( {x+4} \right)=\left( {{{x}^{2}}+1} \right)\left( {x+4} \right)\). Aha! There’s our 4^{th} root: \(x=4\). The 3^{rd} and 4^{th} roots are \(i\) and \(4\).

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 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).
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 righthand 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 righthand 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 noninclusive (\(<\) 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 noninclusive 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:
Polynomial Inequality: Solving Graphically  Polynomial Inequality: Solving Algebraically  
\(\left( {x+1} \right)\left( {x+4} \right)\left( {x3} \right)\le 0\)
The polynomial is already factored, so just make the leading coefficient positive by dividing (or multiplying) by –1 on both sides (have to change inequality sign): \(\left( {x+1} \right)\left( {x+4} \right)\left( {x3} \right)\ge 0\)
End Behavior (of second inequality above): Leading Coefficient: Positive Degree: 3 (odd) \(\displaystyle \begin{array}{c}x\to \infty \text{, }\,y\to \infty \\x\to \infty \text{, }y\to \infty \end{array}\)
We want above (including) the \(x\)axis, because of the \(\ge \). The solution is \([4,1]\cup \left[ {3,\,\infty } \right)\). 
\(\left( {x+1} \right)\left( {x+4} \right)\left( {x3} \right)\le 0\)
The polynomial is already factored, so just make the leading coefficient positive by dividing by –1 on both sides (have to change inequality sign):
\(\left( {x+1} \right)\left( {x+4} \right)\left( {x3} \right)\ge 0\)
Draw a sign chart with critical values (where factors equal 0) –4, –1, and 3. Use closed circles for the critical values since we have a \(\ge \), so the critical values are inclusive.
Then check each interval with a sample value in the last inequality above and see if we get a positive or negative value.
For example, we can try 0 for the interval between –1 and 3: \(\left( {0+1} \right)\left( {0+4} \right)\left( {03} \right)=12\), which is negative:
We want the positive intervals, including the critical values, because of the \(\ge \).
The solution is \([4,1]\cup \left[ {3,\,\infty } \right)\). 
Here’s one more example:
Polynomial Inequality: Solving Graphically  Polynomial Inequality: Solving Algebraically  
\({{x}^{4}}<9{{x}^{2}}\) \(\begin{array}{c}{{x}^{4}}9{{x}^{2}}<0\\{{x}^{2}}\left( {{{x}^{2}}9} \right)<0\\{{x}^{2}}\left( {x3} \right)\left( {x+3} \right)<0\end{array}\)
End Behavior: Leading Coefficient: Positive Degree: 4 (even) \(\begin{array}{c}x\to \infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}\)
We want below (not including) the \(x\)axis. The solution is \(\left( {3,0} \right)\cup \left( {0,3} \right)\), since we can’t include 0, because of the \(<\). 
\({{x}^{4}}<9{{x}^{2}}\) \(\begin{array}{c}{{x}^{4}}9{{x}^{2}}<0\\{{x}^{2}}\left( {{{x}^{2}}9} \right)<0\\{{x}^{2}}\left( {x3} \right)\left( {x+3} \right)<0\end{array}\)
Draw a sign chart with critical values –3, 0, and 3. Use open circles for the critical values since we have a \(<\) and not a \(\le \) sign.
Then check each interval with a sample value and see if we get a positive or negative value. (For example, we can try 1 for the interval between 0 and 3: \({{\left( 1 \right)}^{2}}\left( {13} \right)\left( {1+3} \right)=8\), which is negative):
We have two minus’s in a row, since we have a bounce at \(x=0\). But we can’t include 0 since we have a \(<\) sign and not a \(\le \) sign.
We want the negative intervals, not including the critical values.
The solution is \(\left( {3,0} \right)\cup \left( {0,3} \right)\), since we have to “jump over” the 0, because of the \(<\) sign. 
Here’s one more where we can ignore a factor that can never be 0:
Polynomial Inequality  Notes 
\(\displaystyle \begin{array}{c}\color{#800000}{{{{x}^{4}}+3{{x}^{2}}\,\,\,\ge \,\,\,4}}\\\\{{x}^{4}}3{{x}^{2}}4\le 0\\\left( {{{x}^{2}}4} \right)\left( {{{x}^{2}}+1} \right)\,\,\,\le 0\\\left( {x2} \right)\left( {x+2} \right)\left( {{{x}^{2}}+1} \right)\,\,\,\le 0\end{array}\)
The problem calls for \(\le 0\), so we look for the minus sign(s), and our answers are inclusive (hard brackets).
The answer is \(\left[ { 2,2} \right]\). 
The first thing we need to do is to get everything on the left side, and 0 on the right side; note that it’s best to keep the \({{x}^{4}}\) positive, but we have to be really careful with our inequality sign.
Even though the polynomial has degree 4, we can factor by a difference of squares (and do it again!). The \({{x}^{2}}+1\) can never be 0, so we can ignore that factor.
After factoring, draw a sign chart, with critical values –2 and 2. Now check each interval with random points to see if the polynomial is positive or negative. We put the signs over the interval. Always try easy numbers, especially 0, if it’s not a boundary point!
Let’s try –2 for the leftmost interval: \(\left( {32} \right)\left( {3+2} \right)\left( {{{{\left( {3} \right)}}^{2}}+1} \right)=\left( {5} \right)\left( {1} \right)\left( {10} \right)=\text{ positive (}+\text{)}\). Now 0 for the next interval: \(\left( {02} \right)\left( {0+2} \right)\left( {{{{\left( 0 \right)}}^{2}}+1} \right)=\left( {2} \right)\left( 2 \right)\left( 1 \right)=\text{ negative (}\text{)}\). Now 3 for the next interval: \(\left( {32} \right)\left( {3+2} \right)\left( {{{{\left( 3 \right)}}^{2}}+1} \right)=\left( 1 \right)\left( 5 \right)\left( {10} \right)=\text{ positive (}+\text{)}\)
We want \(\le \) from the factored inequality, so we look for the – (negative) sign intervals, so the interval is \(\left[ { 2,2} \right]\).

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:
Polynomial Application Math  Notes 
a) From above, volume of the box in Factored Form is:
\(V\left( x \right)=\left( {2x+5} \right)\left( {2x} \right)\left( {2x+3} \right)\)
Let’s multiply out to get Standard Form: \(\begin{align}V\left( x \right)&=\left( {2x+5} \right)\left( {2x} \right)\left( {2x+3} \right)\\&=\left( {2x+5} \right)\left( {4{{x}^{2}}+6x} \right)\\&=8{{x}^{3}}+12{{x}^{2}}+20{{x}^{2}}+30x\\V\left( x \right)&=8{{x}^{3}}+32{{x}^{2}}+30x\end{align}\)
The volume of the cutout section is: \(\begin{align}V\left( x \right)&=\left( {x+1} \right)\left( {2x} \right)\left( {x+3} \right)\\&=\left( {x+1} \right)\left( {2{{x}^{2}}+6x} \right)\\V\left( x \right)&=2{{x}^{3}}+8{{x}^{2}}+6x\end{align}\) 
Since volume is \(\text{length }\times \text{ width }\times \text{ height}\), we can just multiply the three terms together to get the volume of the box.
Then we can multiply the length, width, and height of the cutout. Notice that the cutout goes to the back of the box, so it looks like this: 
b) To get the volume of the box remaining, just subtract the two volumes:
\(\begin{align}V\left( x \right)&=8{{x}^{3}}+32{{x}^{2}}+30x \left( {2{{x}^{3}}+8{{x}^{2}}+6x} \right)\\&=6{{x}^{3}}+24{{x}^{2}}+24x\end{align}\) 
We need to subtract two polynomials to get the volume of the box without the cutout section.
When we do the subtraction, we have to be careful to push through the negative sign into all the terms of the second volume.

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:
Polynomial Application Math  Notes 
Volume of the new box in Factored Form is:
\(V\left( x \right)=\left( {x+5} \right)\left( {x+4} \right)\left( {x+3} \right)\)
Let’s multiply out to get Standard Form and set to 120 (twice the original volume). \(\begin{array}{c}\left( {x+5} \right)\left( {x+4} \right)\left( {x+3} \right)=120\\\left( {{{x}^{2}}+9x+20} \right)\left( {x+3} \right)=120\end{array}\)
Using vertical multiplication (see right), we have: \(\begin{array}{l}{{x}^{3}}+12{{x}^{2}}+47x+60=120,\,\,\,\,\text{or}\\{{x}^{3}}+12{{x}^{2}}+47x60=0\end{array}\)
I got lucky and my first attempt at synthetic division worked: \begin{array}{l}\left. {\underline {\, We end up with \({{x}^{2}}+13x+60\), which doesn’t have real roots; 1 is the only real root. We could have also put the righthand side and lefthand sides into a graphing calculator, and used the “Intersect” function to find the real root. 
Again, the volume is \(\text{length }\times \text{ width }\times \text{ height}\), so the new volume is \(\displaystyle \left( {x+5} \right)\left( {x+4} \right)\left( {x+3} \right)\), and the new box will look like this:
The old volume is \(\text{5 }\times \text{ 4 }\times \text{ 3}\) inches, or 60 inches. We have to set the new volume to twice this amount, or 120 inches.
We used vertical multiplication for the polynomials: \(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+9x+20\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times \,\,\,\,\,x\,\,+3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,3{{x}^{2}}+27x+60\\\underline{{{{x}^{3}}+\,\,\,9{{x}^{2}}+20x\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\{{x}^{3}}+12{{x}^{2}}+47x+60\end{array}\)
We would need to add 1 inch to double the volume of the box. 
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 and then in standard form.
(b) What would be a reasonable domain for the polynomial? (Hint: Each side of the threedimensional box has to have a length of at least 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 threedimensional 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 \(\left( {302x} \right)\), the new width of the box will be \(\left( {152x} \right)\), and the height up of the box will “\(x\)” (since the outside pieces are folded up). The volume is length \(x\) width times height, so the volume of the box is the polynomial \(V\left( x \right)=\left( {302x} \right)\left( {152x} \right)\left( x \right)\).
Now let’s answer the questions with a little help from a graphing calculator:
Polynomial Application Math  Notes 
a) From above, volume in Factored Form is:
\(V\left( x \right)=\left( {302x} \right)\left( {152x} \right)\left( x \right)\) . Let’s multiply out to get Standard Form: \(\begin{align}V\left( x \right)&=\left( {302x} \right)\left( {152x} \right)\left( x \right)\\&=\left( {302x} \right)\left( {15x2{{x}^{2}}} \right)\\&=450x60{{x}^{2}}30{{x}^{2}}+4{{x}^{3}}\\V\left( x \right)&=4{{x}^{3}}90{{x}^{2}}+450x\end{align}\) 
Now that we have the volume in Factored Form, we need to multiply out the polynomial to get it in Standard Form.
Multiply the \(x\) through one of the other factors, and then use FOIL or “pushing through” to get the Standard Form. 
b) To get the reasonable domain for \(x\) (the cutout), we have to make sure that the length, width, and height all have to be greater than 0. All 3 conditions must be met:
\(\begin{array}{c}302x>0;\,\,\,\,\,\,x<15\\152x>0;\,\,\,\,\,\,\,x<7.5\\x>0\end{array}\) Domain: \(\left( {0,\,7.5} \right)\) 
We typically look for a subtraction, since we could potentially get a negative number. Remember that’s it’s usually the smaller of the numbers that work, since the larger one will make one of the expressions negative.
Our domain has to satisfy all equations; therefore, a reasonable domain is \(\left( {0,\,7.5} \right)\). 
c) Let’s use our graphing calculator to graph the polynomial and find the highest point. Note that the value of \(x\) at the highest point is 3.17 inches.
d) The volume is \(y\) part of the maximum, which is 649.52 inches. e) The dimensions of the open donut box with the largest volume is \(\left( {302x} \right)\) by \(\left( {152x} \right)\) by (\(x\)), which equals \(\left( {302\left( {2.17} \right)} \right)\) by \(\left( {152\left( {2.17} \right)} \right)\) by \(\left( {152\left( {2.17} \right)} \right)\), which equals 23.66 inches by 8.66 inches by 3.17 inches. 
We can put the polynomial in the graphing calculator using either the standard or factored form. In fact, you can even put in both forms and make sure if the same curve comes up to test your conversion from factored to standard form!
Remember that the \(x\) represents the height of the box (the cut out side length), and the \(y\) represents the volume of the box
To get the best window, I use ZOOM 6, ZOOM 0, then ZOOM 3 enter a few times. You can also hit WINDOW and play around with the Xmin, Xmax, Ymin and Ymax values. Since we know the domain is between 0 and 7.5, that helps with the Xmin and Xmax values.
Then I used 2^{nd} TRACE (CALC), 4 (maximum), moved the cursor to the left of the top after “Left Bound?”, moved the cursor to the right of the top after “Right Bound?”, and then hit ENTER twice to get the maximum point. 
Cost Revenue Profit Problem
The price \(p\) that a makeup company can charge for a certain kit is \(p=404{{x}^{2}}\), 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 \(\left( 404{{x}^{2}} \right)\left( x \right)\). The cost to make \(x\) thousand kits is \(15x\). So the total profit of is \(P\left( x \right)=\left( 404{{x}^{2}} \right)\left( x \right)15x=40x4{{x}^{3}}15x=4{{x}^{3}}+25x\).
(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 \(x=1.5\), 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 \(P\left( x \right)=4{{x}^{3}}+25x24\) (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 \(P\left( x \right)=4{{x}^{3}}+25x24\). We could find the other roots by using a graphing calculator, but let’s do it without:
Math 
Notes 
Let’s make sure that 1.5 is really a root of \(P\left( x \right)=4{{x}^{3}}+25x24\). Then we’ll find the other roots of this equation to find other number of kits to make with the same profit:
\begin{array}{l}\left. {\underline {\,
Use Quadratic Formula to find other roots:
\(\displaystyle \begin{align}\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}&=\frac{{6\pm \sqrt{{364\left( {4} \right)\left( {16} \right)}}}}{{8}}\\&=\frac{{6\pm \sqrt{{292}}}}{{8}}\approx 2.886,\,\,1.386\end{align}\) 
First use synthetic division to verify that 1.5 is a root to \(P\left( x \right)=4{{x}^{3}}+25x24\) – it is!
Since the remaining term is not factorable, use the Quadratic Formula to find another root. (We could have also factored out a “–2” first, but don’t have to.)
Note that the negative number –2.886 doesn’t make sense (you can’t make a negative number of kits), but the 1.386 would work (even though it’s not exact). The company could sell 1.386 thousand or 1,386 kits and still make the same profit as when it makes 1500 kits. 
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 righthand corner and click on “Examples” to drill down by topic.
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On to Exponential Functions – you are ready!