2.12: Real Zeros of Polynomials
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
The Factor Theorem
If
The Rational Zero Theorem
Given the polynomial
Descartes Rule of Signs
Given any polynomial,
 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
p(x) . Call the number of sign changesn .  Then the number of positive roots of
p(x) is less than or equal ton .  Further, the possible number of positive roots is
n,n−2,n−4,…  To find the number of negative roots of
p(x) , writep(−x) in descending order as above (i.e. change the sign of all terms inp(x) with odd powers), and repeat the process above. Then the maximum number of negative roots isn .
Example A
Use synthetic division and the remainder and factor theorems to find the quotient
Solution
Hence
Notice that the remainder is 181 and it can also be obtained if we simply substituted
Example B
Use the rational zero theorem and synthetic division to find all the possible rational zeros of the polynomial
Solution
From the rational zero theorem,
So there are four possible zeros. Of these four, not more than three can be zeros of
Hence, 2 is a zero of
The remaining zeros of
and thus the remaining zeros are 1 and 1. Thus the rational zeros of
Example C
Graph the polynomial function
Solution
Notice that the leading term is
Here, as you can see, there is no straightforward way to find the zeros of
First, we use the rational zero theorem and find that the possible rational zeros are
testing all these numbers by the synthetic division,
1 is not a root. Now let's test
we find that 1 is a zero of
Looking at quadratic part,
and so
Thus 1 and
Further, the synthetic division can be also used to form a table of values for the graph of
We choose test points from each interval and find
Interval 
Test Value 

Sign of 
Location of points on the graph 


1  28   
below the 



 
below the 

3  4  + 
above the 
From this information, the graph of
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
Questions:
1) Show that

Find the quotient
Q(x) and expressf(x) in factored form.
2) Use the 'rational zero' theorem and synthetic division to find all the possible rational zeros of the polynomial
3) Use Descartes Rule of Signs to identify the possible number of positive and negative roots of

f(x)=−2x3+x2−3x5+5x−1 .
4) Find the root(s) of
5) Find the root(s) of
Solutions
1) By the factor theorem, if

−3 )1 2−3 4 12¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯↓−3 3 0−12−−−−−−−−−−−−1−1 0 4 0 
Hence,
g(−3)=0 , and the quotient is 
Q(x)=x3−x2+4 
so that
g(x) can be written as 
g(x)g(x)=(x−(−3))(x3−x2+4)=(x+3)(x3−x2+4)
2) Assume

p:1,−1,2,−2,3,−3,6,−6 
 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.
Practice
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*}.
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Descartes' Rule of Signs
Descartes' rule of signs is a technique for determining the number of positive and negative real roots of a polynomial.factor theorem
The factor theorem states that if is a polynomial of degree and , then is a factor of the polynomial .factorization theorem
The factorization theorem states that If , where , and is a positive integer, then where the numbers are complex numbers.Multiplicity
The multiplicity of a term describes the number of times the given term acts as a zero of the given function.Polynomial
A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.Rational Zero Theorem
The rational zero theorem states that for a polynomial, , where are integers, the rational roots can be determined from the factors of and . More specifically, if is a factor of and is a factor of , then all the rational factors will have the form .Remainder Theorem
The remainder theorem states that if , then is the remainder when dividing by .Roots
The roots of a function are the values of x that make y equal to zero.Synthetic Division
Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.Zeroes
The zeroes of a function are the values of that cause to be equal to zero.Zeros
The zeros of a function are the values of that cause to be equal to zero.Image Attributions
Here you will learn about using polynomial long division and synthetic division to find roots, and how to apply synthetic division to find the rational zeros of an unfactored polynomial (of degree >2).