In the real world, problems do not always easily fit into quadratic or even cubic equations. Financial models, population models, fluid activity, etc., all often require many degrees of the input variable in order to approximate the overall behavior. While it can be challenging to model some of these more complex interactions, the effort can be well worth it. Mathematical models of stocks are used constantly as a way to "look into the future" of finance and make the kinds of educated guesses that are behind some of the largest fortunes in the world.
What benefits can you think of to modeling the behavior of large populations? Can you think of other useful applications not mentioned here?
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PatrickJMT: Finding all Zeroes of a Polynomial Function
Guidance
There are three theorems and a rule that we will be referring to during this lesson in order to help make the discovery of the roots of polynomial functions easier. You should review them and be prepared to refer to them often during the practice problems.
The Remainder Theorem
If a polynomial \begin{align*}f(x)\end{align*} of degree @$\begin{align*}n>0\end{align*}@$ is divided by @$\begin{align*}xc\end{align*}@$ , then the remainder @$\begin{align*}R\end{align*}@$ is a constant and it is equal to the value of the polynomial when @$\begin{align*}c\end{align*}@$ is substituted for @$\begin{align*}x\end{align*}@$ . That is
@$$\begin{align*}f(c)=R\end{align*}@$$
The Factor Theorem
If @$\begin{align*}f(x)\end{align*}@$ is a polynomial of degree @$\begin{align*}n>0\end{align*}@$ and @$\begin{align*}f(c)=0\end{align*}@$ , then @$\begin{align*}xc\end{align*}@$ is a factor of the polynomial @$\begin{align*}f(x)\end{align*}@$ . Further, if @$\begin{align*}xc\end{align*}@$ is a factor, then @$\begin{align*}c\end{align*}@$ is a zero of @$\begin{align*}f\end{align*}@$ .
The Rational Zero Theorem
Given the polynomial
@$$\begin{align*}f(x)=a_{n}x^{n}+a_{n1}x^{n1}+\cdots+a_{1}x+a_{0}\end{align*}@$$
@$\begin{align*}a_{n}\ne 0\end{align*}@$ and @$\begin{align*}n\end{align*}@$ is a positive integer. If the coefficients are integers and @$\begin{align*}\frac{p}{q}\end{align*}@$ is a rational zero in lowest terms, then @$\begin{align*}p\end{align*}@$ is a divisor of @$\begin{align*}a_{0}\end{align*}@$ and @$\begin{align*}q\end{align*}@$ is a divisor of @$\begin{align*}a_{n}\end{align*}@$ .
Descartes' Rule of Signs
Given any polynomial, @$\begin{align*}p(x)\end{align*}@$ ,
 Write it with the terms in descending order, i.e. from the highest degree term to the lowest degree term.
 Count the number of sign changes of the terms in @$\begin{align*}p(x)\end{align*}@$ . Call the number of sign changes @$\begin{align*}n\end{align*}@$ .
 Then the number of positive roots of @$\begin{align*}p(x)\end{align*}@$ is less than or equal to @$\begin{align*}n\end{align*}@$ .
 Further, the possible number of positive roots is @$\begin{align*}n, n2, n4, \ldots\end{align*}@$
 To find the number of negative roots of @$\begin{align*}p(x)\end{align*}@$ , write @$\begin{align*}p(x)\end{align*}@$ in descending order as above (i.e. change the sign of all terms in @$\begin{align*}p(x)\end{align*}@$ with odd powers), and repeat the process above. Then the maximum number of negative roots is @$\begin{align*}n\end{align*}@$ .
Example A
Use synthetic division and the remainder and factor theorems to find the quotient @$\begin{align*}Q(x)\end{align*}@$ and the remainder @$\begin{align*}R\end{align*}@$ if @$\begin{align*}f(x)=2x^{3}3x^{2}+6\end{align*}@$ is divided by @$\begin{align*}x5\end{align*}@$ .
Solution:
@$$\begin{align*}& \ 5 \ \big ) \overline{2 3 \ \ \ 0 \ \ \ \ \ 6}\\ & \quad \ \ \underline{\downarrow \ 10 \ \ 35 \ \ 175}\\ & \quad \ \ 2 \ \ \ 7 \ \ 35 \ \ 181\end{align*}@$$
Hence
@$$\begin{align*}2x^{3}3x^{2}+6=(2x^{2}+7x+35)(x5)+181\end{align*}@$$
Notice that the remainder is 181 and it can also be obtained if we simply substituted @$\begin{align*}x=5\end{align*}@$ into @$\begin{align*}f(x)\end{align*}@$ ,
@$$\begin{align*}f(5) & = 2(5)^3  3(5)^2+6\\ & = 25075+6\\ & = 181\end{align*}@$$
Example B
Use the rational zero theorem and synthetic division to find all the possible rational zeros of the polynomial
@$$\begin{align*}f(x)=x^{3}2x^{2}x+2\end{align*}@$$
Solution:
From the rational zero theorem, @$\begin{align*}\frac{p}{q}\end{align*}@$ is a rational zero of the polynomial @$\begin{align*}f\end{align*}@$ . So @$\begin{align*}p\end{align*}@$ is a divisor of 2 and @$\begin{align*}q\end{align*}@$ is a divisor of 1. Hence, @$\begin{align*}p\end{align*}@$ can take the following values: 1, 1, 2, 2 and @$\begin{align*}q\end{align*}@$ can be either 1 or 1. Therefore, the possible values of @$\begin{align*}\frac{p}{q}\end{align*}@$ are
@$$\begin{align*}\frac{p}{q}:1,1,2,2\end{align*}@$$
So there are four possible zeros. Of these four, not more than three can be zeros of @$\begin{align*}f\end{align*}@$ because @$\begin{align*}f\end{align*}@$ is a polynomial with degree 3. To test which of the four possible candidates are zeros of @$\begin{align*}f\end{align*}@$ , we use the synthetic division. Recall from the remainder theorem if @$\begin{align*}f(c)=0\end{align*}@$ , then @$\begin{align*}c\end{align*}@$ is a zero of @$\begin{align*}f\end{align*}@$ . We have
@$$\begin{align*}& \ 2 \ \big ) \overline{1 \ 2 \ 1 \ \ \ \ 2\;}\\ & \quad \ \ \underline{\downarrow \ \ \ 2 \ \ \ \ \ 0 \ 2}\\ & \quad \ \ 1 \ \ \ \ 0 \ 1 \ \ \ \ 0\end{align*}@$$
Hence, 2 is a zero of @$\begin{align*}f\end{align*}@$ . Further, by the division algorithm,
@$$\begin{align*}f(x) & = (xc)Q(x)+R(x)\\ & = (x2)(x^21)+0\end{align*}@$$
The remaining zeros of @$\begin{align*}f\end{align*}@$ are simply the zeros of @$\begin{align*}Q(x)=x^{2}1\end{align*}@$ which is easier to manipulate,
@$$\begin{align*}Q(x) & = x^21\\ & = (x1)(x+1)\end{align*}@$$
and thus the remaining zeros are 1 and 1. Thus the rational zeros of @$\begin{align*}f\end{align*}@$ are 1, 1, and 2.
Example C
Graph the polynomial function @$\begin{align*}h(x)=2x^{3}9x^{2}+12x5\end{align*}@$ .
Solution:
Notice that the leading term is @$\begin{align*}2x^{3}\end{align*}@$ , where @$\begin{align*}n=3\end{align*}@$ odd and @$\begin{align*}a_{n}=2>0\end{align*}@$ . This tells us that the end behavior will take the shape of a power function with an odd exponent.
Here, as you can see, there is no straightforward way to find the zeros of @$\begin{align*}h(x)\end{align*}@$ . However, with the use of the factor theorem and the synthetic division, we can find the rational roots of @$\begin{align*}h(x)\end{align*}@$ .
First, we use the rational zero theorem and find that the possible rational zeros are
@$$\begin{align*}\frac{p}{q}:1,1,2,2,5,5,\frac{5}{2},\frac{5}{2}\end{align*}@$$
testing all these numbers by the synthetic division,
@$$\begin{align*}& \ 1 \ \big ) \overline{2 \ 9 \ \ 12 \ 5\;}\\ & \qquad \ \ \underline{\downarrow \ 2 \ \ \ 11 \ 23}\\ & \qquad \ \ 2 \ 11 \ 23 \ 28\end{align*}@$$
1 is not a root. Now let's test @$\begin{align*}x=1\end{align*}@$ .
@$$\begin{align*}& \ 1 \ \big ) \overline{2 \ 9 \ \ \ 12 \ 5\;}\\ & \quad \ \ \underline{\downarrow \ 2 \ 7 \ \ \ \ \ 5\;}\\ & \quad \ \ 2 \ 7 \ \ \ 5 \ 28\end{align*}@$$
we find that 1 is a zero of @$\begin{align*}h\end{align*}@$ and so we can rewrite @$\begin{align*}h(x)\end{align*}@$ ,
@$$\begin{align*}h(x)=(2x^{2}7x+5)(x1)\end{align*}@$$
Looking at quadratic part,
@$$\begin{align*}2x^{2}7x+5=(2x5)(x1)\end{align*}@$$
and so
@$$\begin{align*}h(x)=(2x5)(x1)^{2}\end{align*}@$$
Thus 1 and @$\begin{align*}\frac{5}{2}\end{align*}@$ are the @$\begin{align*}x\end{align*}@$ intercepts of @$\begin{align*}h(x)\end{align*}@$ . The @$\begin{align*}y\end{align*}@$ intercept is
@$$\begin{align*}h(0)=5\end{align*}@$$
Further, the synthetic division can be also used to form a table of values for the graph of @$\begin{align*}h(x)\end{align*}@$ :
@$$\begin{align*}& x && 1 && \ \ \ 0 && 1 && \ \ \ 2 && \frac{5}{2} && 3\\ & h(x) && 28 && 5 && 0 && 1 && 0 && 4\end{align*}@$$
We choose test points from each interval and find @$\begin{align*}g(x)\end{align*}@$ .
Interval  Test Value @$\begin{align*}x\end{align*}@$  @$\begin{align*}h(x)\end{align*}@$  Sign of @$\begin{align*}h(x)\end{align*}@$  Location of points on the graph 

@$\begin{align*}(\infty,1)\end{align*}@$  1  28    below the @$\begin{align*}x\end{align*}@$ axis 
@$\begin{align*}\left ( 1,\frac{5}{2} \right )\end{align*}@$  @$\begin{align*}\frac{3}{2}\end{align*}@$  @$\begin{align*}\frac{11}{4}\end{align*}@$    below the @$\begin{align*}x\end{align*}@$ axis 
@$\begin{align*}\left ( \frac{5}{2},\infty \right )\end{align*}@$  3  4  +  above the @$\begin{align*}x\end{align*}@$ axis 
From this information, the graph of @$\begin{align*}h(x)\end{align*}@$ is shown in the two graphs below. Notice that the second graph is a magnification of @$\begin{align*}h(x)\end{align*}@$ in the vicinity of the @$\begin{align*}x\end{align*}@$ axis.
Concept question wrapup: Were you able to identify some valuable realworld uses for modeling higherdegree polynomials? Here are a few possibilities:
There are many, many more. 

Guided Practice
1) Show that @$\begin{align*}x+3\end{align*}@$ is a factor of @$\begin{align*}g(x)=x^{4}+2x^{3}3x^{2}+4x+12\end{align*}@$ .
 Find the quotient @$\begin{align*}Q(x)\end{align*}@$ and express @$\begin{align*}f(x)\end{align*}@$ in factored form.
2) Use the 'rational zero' theorem and synthetic division to find all the possible rational zeros of the polynomial
@$$\begin{align*}f(x)=x^{3}2x^{2}5x+6\end{align*}@$$
3) Use Descartes' Rule of Signs to identify the possible number of positive and negative roots of
 @$\begin{align*}f(x)=2x^{3}+x^{2}3x^{5}+5x1\end{align*}@$ .
4) Find the root(s) of @$\begin{align*}f(x) = 4x^2  3x  7\end{align*}@$
5) Find the root(s) of @$\begin{align*}f(x) = x ^4 + 1\end{align*}@$
Answers
1) By the factor theorem, if @$\begin{align*}f(c)=0\end{align*}@$ , then @$\begin{align*}xc\end{align*}@$ is a factor of the polynomial. In other words, if the synthetic division produces a remainder equal to zero, then @$\begin{align*}c\end{align*}@$ is a factor of the polynomial.: Using the synthetic division with @$\begin{align*}c=3\end{align*}@$ ,
 @$$\begin{align*}& \ 3 \ \big ) \overline{1 \ \ \ 2 3 \ \ 4 \ \ \ 12\;\;}\\ & \qquad \quad \underline{\downarrow 3 \ \ \ 3 \ \ 0 12}\\ & \qquad \quad 1 1 \ \ \ 0 \ \ 4 \ \ \ \ 0\end{align*}@$$
 Hence, @$\begin{align*}g(3)=0\end{align*}@$ , and the quotient is
 @$$\begin{align*}Q(x)=x^{3}x^{2}+4\end{align*}@$$
 so that @$\begin{align*}g(x)\end{align*}@$ can be written as
 @$$\begin{align*}g(x) & = (x(3))(x^3x^2+4)\\ g(x) & = (x+3)(x^3x^2+4)\end{align*}@$$
2) Assume @$\begin{align*}\frac{p}{q}\end{align*}@$ is a rational zero of @$\begin{align*}f\end{align*}@$ . By the rational zero theorem, @$\begin{align*}p\end{align*}@$ is a divisor of 6 and @$\begin{align*}q\end{align*}@$ is a divisor of 1. Thus @$\begin{align*}p\end{align*}@$ and @$\begin{align*}q\end{align*}@$ can assume the following respective values
 @$$\begin{align*}p:1,1,2,2,3,3,6,6\end{align*}@$$

 and
 @$$\begin{align*}q:1,1\end{align*}@$$
 Therefore, the possible rational zeros will be
 @$$\begin{align*}\frac{p}{q}:1,1,2,2,3,3,6,6\end{align*}@$$
 Notice that with these choices for @$\begin{align*}p\end{align*}@$ and @$\begin{align*}q\end{align*}@$ there could be @$\begin{align*}8 \cdot 2=16\end{align*}@$ rational zeros. But, eight of them are duplicates. For example @$\begin{align*}\frac{1}{1}=\frac{1}{1}=1\end{align*}@$ . The next step is to test all these values by the synthetic division (we'll let you do this on your own for practice) and we finally find that
 @$$\begin{align*}1,2, \ \text{and} \ 3\end{align*}@$$
 are zeros of @$\begin{align*}f\end{align*}@$ . That is
 @$$\begin{align*}f(x) & = x^3  2x^2  5x +6\\ & = (x1)(x+2)(x3)\end{align*}@$$
3) First, rewrite @$\begin{align*}f(x)\end{align*}@$ in descending order
 @$$\begin{align*}f(x)=3x^{5}2x^{3}+x^{2}+5x1.\end{align*}@$$
 The number of sign changes of @$\begin{align*}f(x)\end{align*}@$ is 2, so the number of positive roots is either 2 or 0.
 For the negative roots, write
 @$$\begin{align*}f(x)=3x^{5}+2x^{3}+x^{2}5x1\end{align*}@$$
 The number of sign changes of @$\begin{align*}f(x)\end{align*}@$ is 2, so the maximum number of negative roots is 2.
 The graph of @$\begin{align*}f(x)\end{align*}@$ below shows that there is one negative root and two positive roots.
4) Solve by factoring and applying the zero product rule:
 @$\begin{align*}4x^2  3x  7 = 0\end{align*}@$
 @$\begin{align*} (4x  7)(x + 1) = 0\end{align*}@$
 @$\begin{align*}4x  7 = 0\end{align*}@$ or @$\begin{align*}x + 1 = 0\end{align*}@$
 @$\begin{align*}\therefore x = \frac{7}{4}\end{align*}@$ or @$\begin{align*} x = 1\end{align*}@$
 (each zero has a multiplicity of one)
5) This one is easy:
 @$\begin{align*}f(x) = x^4 + 1\end{align*}@$
 @$\begin{align*}x^4 = 1\end{align*}@$
 Since there are no real roots of even powers, this function has zero real solutions.
Explore More
Problems 1  3: Use a) long division and b) synthetic division to perform the divisions.
 Express each result in the form: @$\begin{align*}f(x)=D(x)\cdot Q(x)+R\end{align*}@$
 @$\begin{align*}5x^{5}3x^{4}+2x^{3}+x^{2}7x+3\end{align*}@$ by @$\begin{align*}x2\end{align*}@$
 @$\begin{align*}4x^{6}5x^{3}+3x^{2}+x+7\end{align*}@$ by @$\begin{align*}x1\end{align*}@$
 @$\begin{align*}2x^{3}5x^{2}+5x+11\end{align*}@$ by @$\begin{align*}x\frac{1}{2}\end{align*}@$
 Use synthetic division to find @$\begin{align*}Q(x)\end{align*}@$ and @$\begin{align*}f(c)\end{align*}@$ so that @$\begin{align*}f(x)=(xc)Q(x)+f(c)\end{align*}@$ if @$\begin{align*}f(x)=3x^{4}3x^{3}+3x^{2}+2x4\end{align*}@$ and @$\begin{align*}c=2\end{align*}@$
 If @$\begin{align*}f(x)=x^{3}+2x^{2}10x+10\end{align*}@$ , use synthetic division to determine the following: a) @$\begin{align*}f(1)\end{align*}@$ b) @$\begin{align*}f(3)\end{align*}@$ c) @$\begin{align*}f(0)\end{align*}@$ d) @$\begin{align*}f(4)\end{align*}@$ e) What are the factors of @$\begin{align*}f(x)\end{align*}@$ ?
 Find @$\begin{align*}k\end{align*}@$ so that @$\begin{align*}x2\end{align*}@$ is a factor of @$\begin{align*}f(x)=3x^{3}+4x^{2}+kx20\end{align*}@$
 Use synthetic division to determine all the zeros of the polynomials: a) @$\begin{align*}f(x)=3x^{3}7x^{2}+8x2\end{align*}@$ b) @$\begin{align*}g(x)=4x^{4}4x^{3}7x^{2}+4x+3\end{align*}@$
 Graph the polynomial function @$\begin{align*}f(x)=x^{3}2x^{2}5x+6\end{align*}@$ by using synthetic division to find the @$\begin{align*}x\end{align*}@$ intercepts and locate the @$\begin{align*}y\end{align*}@$ intercepts.
 Graph the polynomial function @$\begin{align*}h(x)=x^{3}3x^{2}+4\end{align*}@$ by using synthetic division to find the @$\begin{align*}x\end{align*}@$ intercepts and locate the @$\begin{align*}y\end{align*}@$ intercepts.
 Write a 3rd degree equation of a polynomial function with the zeroes: 0, 2, and 5.
 Write a 7th degree equation of a polynomial function with the zeroes: 0 (multiplicity 2), 2 (multiplicity 3), and 5 (multiplicity 2)
 Write a quadratic equation which has 4 (multiplicity 2) as the zero and opens downward.
 Write a 3rd degree polynomial function with the zeroes: 2, 2, and 6, passing through the point (3, 4)
 Let @$\begin{align*}f(x)=2x^{3}5x^{2}4x+3\end{align*}@$ and find the solutions: a) @$\begin{align*}f(x)=0\end{align*}@$ b) @$\begin{align*}f(2x)=0\end{align*}@$
 Graph and find the solution set of the inequality @$\begin{align*}x^{3}2x^{2}5x+6 \leq 0\end{align*}@$ .
 Use the graph of @$\begin{align*}f(x)=x (x  1)(x + 2)\end{align*}@$ to find the solution set of the inequality @$\begin{align*}x (x  1)(x + 2) > 0\end{align*}@$ .