Learning Objectives
Introduction
In this tutorial we will be looking at several aspects dealing with zeros of polynomial functions. If you need a review on how to find zeros, the Rational Zero Theorem or Descartes' Rule of Signs, feel free to go to Tutorial 38: Zeros of Polynomial Functions, Part I. On this page we dive a little deeper into the concept of zeros. One thing that we will look at is finding the upper and lower bounds for roots of a polynomial function. This can help us narrow down the possibilities of rational zeros. Another concept on this page is the Intermediate Value Theorem. This can help narrow down the possibilities of real zeros, especially those that land between integer values. We will also work with non-real complex numbers. Did you know that if a non-real complex number is a zero of a polynomial function, that its conjugate is too? We will follow that up by using the Fundamental Theorem of Algebra and the Linear Factorization Theorem to find polynomial functions given zeros. Wow, it looks like we have our work cut out for us. I guess you better get started.
Tutorial
Note that two things must occur for c to be an upper bound. One is c > 0 or positive. The other is that all the coefficients of the quotient as well as the remainder are positive.
Note that two things must occur for c to be a lower bound. One is c < 0 or negative. The other is that successive coefficients of the quotient and the remainder have alternating signs.
Checking the
Lower Bound:
Lets apply synthetic division with - 4 and see if we get alternating signs:
Note how c = -4 < 0 AND the successive signs in the bottom row of our synthetic division alternate.
You know what that means?
- 4 is a lower bound for the real roots of this equation.
Note how c = 4 > 0 AND the all of the signs in the bottom row of our synthetic division are positive.
You know what that means?
4 is an upper bound for the real roots of this
equation.
Since - 4 is a lower bound and 4 is an upper bound for the real roots of the equation, then that means all real roots of the equation lie between - 4 and 4.
The Intermediate Value TheoremIf f(x) is a
polynomial function and f(a)
and f(b) have different signs, then there is at least
one
value, c,
between a and b such that f(c)
= 0.
This works because 0 separates the positives from the negatives. So to go from positive to negative or vice a versa you would have to hit a point in between that goes through 0.
Finding f(2):
Finding f(3):
Since there is a sign change between f(2)
= -2 and f(3) = 5, then according to the Intermediate
Value Theorem, there is at least one value between 2 and 3 that
is a zero of this polynomial function.
Checking functional values at intervals of one-tenth for a sign change:
Finding f(2.1):
Finding f(2.2):
Finding f(2.3):
Finding f(2.4):
Finding f(2.5):
Hey we have a sign change!!!!!
Now we want to find the zero to the nearest tenth. So is it going to be x = 2.4 or x = 2.5. We can not necessarily go by which functional value is closer to zero.
We will need to dig a little bit deeper and go by intervals of one-hundredths:
Finding f(2.41):
Finding f(2.42):
Finding f(2.43):
Finding f(2.44):
Finding f(2.45):
Since it would land slightly below 2.45, the nearest tenth would be 2.4.
The work here is not hard, it is just tedious.
Every polynomial function equation f(x) = 0 of degree one or higher has at least one complex root.
In other words, if a complex number with an imaginary part is a root of a polynomial function equation, then its conjugate is also a root of that same function.
Remember that a conjugate of a + bi is a - bi. So, if 2 + 3i is a known root of a polynomial function equation, then 2 - 3i is also.
If you need a review on complex numbers, feel free to go to Tutorial 12: Complex Numbers.
The Linear Factorization TheoremIf
where n > 1 and
then
where
are complex numbers
(possibly real and not necessarily
distinct)
Keep in mind that some factors may occur more than one time. For each time a linear factor appears it is counted as a linear factor. For example, if , the linear factor (x + 2) has a multiplicity of 3, which means that factor occurs three times. So technically there are 4 linear factors, one (x - 3) and three (x + 2)’s. This matches the degree of the polynomial function.
This is done in the same fashion as when you are given a real zero.
If you need a review on finding roots of a polynomial equation f(x) = 0 when given a root, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems.
Using synthetic division to find the quotient we get:
Yuck!!! Look at all of those complex numbers in the quotient. Don’t fear, when we put in our conjugate using that quotient those complex numbers will disappear and we will be left with a nice quotient with real number coefficients.
Check it out:
*The given complex zero
*The conjugate zero
*Setting the 3rd factor = 0
Factor as the product of factors that are irreducible over real numbers:
*Factor the difference of squares
Completely factored form involving complex non-real numbers:
Step 1: Use the given zeros and the Linear Factorization Theorem to write out all of the factors of the polynomial function.
Also keep in mind that the degree tells you how many linear factors over the complex numbers (possibly real and not necessarily distinct) that you will have.
The factors are written in the following way: if c is a zero than (x - c) is a factor of the polynomial function.
Step 2: Multiply all of the factors found in Step 1.
Note how we are only given two zeros. We need to come up with a third one. Do you have any ideas?
Oh yeah, if a non-real complex number is a zero than its conjugate is also a zero. Since 2 + 3i is a zero, that means 2 - 3i is also a zero.
Using the Linear Factorization Theorem we get:
*Multiply comp. factors
*Simplify (i squared
= -1)
*Multiply remaining factors
This will help us find in this problem.
*Solve for a sub n
Practice Problems
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problem 1a: Show that all real roots of the given equation lie between -3 and 4.
Practice Problem 2a: Show that the given polynomial has a real zero between the given integers. Use the Intermediate Value theorem to find an approximation for this zero to the nearest tenth.
Practice Problem 3a: Use the given root to find all of the roots of the given polynomial equation.
Practice Problem 4a: Factor the given polynomial function a) as the product of factors that are irreducible over rational numbers, b) as the product of factors that are irreducible over real numbers, and c) in completely factored form involving complex non-real numbers.
Practice Problem 5a: Find an nth degree polynomial function with the given conditions
Need Extra Help on these Topics?
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on March 15, 2012 by Kim Seward.
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