3.4 (1) Complex and Rational Zeros of Polynomials
Next year consider presenting this as a CAS document, followed along by paper
This sheet contains notes as well as problems to work. Please work the problems,
using another sheet of paper if you need, and hand it in as a homework assignment.
Answers are on the second page, so you can check your work.
Most of this material we’ve seen before, so this should be largely review. If you have
questions, of course ask them
First, a little POD:
Find all zeros and their multiplicities for f (x) x 4 21x 2 100 .
Now, the lesson:
1. Factor f (x) x 4 81 over the set of real numbers.
2. Next, factor it into linear factors. How many will you have?
3. Factor f (x) x 3 4 x into linear factors. How many will you have?
What do you notice about the complex answers for each of the three polynomials you
have just factored?
Consider the following Theorem on Conjugate Pair Zeros of a Polynomial:
If a polynomial f(x) of degree n > 1 has real coefficients and if z a bi with
b 0 is a complex zero of f(x), then the conjugate z a bi is also a zero of
In other words, if -2i is a zero, +2i must also be a zero, and so on.
4. Find a polynomial of degree 4 with a leading coefficient of 1, and with zeros at
x 0,2,and 2 i . Give it with linear factors and then with real factors.
5. Find a polynomial with minimal degree that has real coefficients and zeros at
2 i and 3i . Give it with real coefficients.
Notice the degrees of these real factors—they are either linear or quadratic. This leads to
the following theorem:
Every polynomial with real coefficients and positive degree n can be expressed as
a product of linear and quadratic polynomials with real coefficients such that the
quadratic factors are irreducible over .
In other words, the real number factors of a polynomial will be either linear or
6. Express f (x) x 5 4x 3 x 2 4 as a product of linear and quadratic factors with
real coefficients and then as a product of linear factors.
Note that all the polynomials we’ve factored have had real coefficients. If we introduce
imaginary coefficients, the rules change. We still have the same number of roots, but the
imaginary solutions may not come in complex conjugate pairs, and we might not follow
the rule above.
POD: f (x) (x 2 25)(x 2 4) (x 5i)(x 5i)(x 2)(x 2)
So, zeros are x = 5i, 2 all with multiplicity of 1.
1. f (x) x 4 81 (x 2 9)(x 2 9) (x 3)(x 3)(x 2 9)
2. There should be four linear factors.
f (x) x 4 81 (x 2 9)(x 2 9) (x 3)(x 3)(x 3i)(x 3i)
3. There should be three linear factors.
f (x) x 3 4x x(x 2 4) x(x 2i)(x 2i)
4. Since complex solutions come in conjugate pairs, if 2 i is a solution, then
2 i must also be a solution. This fourth degree polynomial will have four
f (x) x(x 2)(x (2 i))(x (2 i))
These are linear factors, but obviously they are not all real. Multiply the
imaginary factors to determine real factors. You might fall back on your FOILing
skills to do this.
f (x) x(x 2)(x (2 i))(x (2 i)) x(x 2)(x 2 4x 5)
Keep in mind we can make that leading coefficient equal something besides 1.
5. Given these two imaginary solutions, we know there will be two more imaginary
solutions, for a grand total of four solutions. This polynomial is thus a fourth
f (x) (x (2 i))(x (2 i))(x (3i))(x (3i))
(x 2 4x 5)(x 2 9)
6. Factor by grouping.
f ( x) x 5 4 x 3 x 2 4 ( x 5 4 x 3 ) ( x 2 4)
x 3 ( x 2 4) ( x 2 4) ( x 3 1)( x 2 4)
( x 3 1)( x 2)( x 2)
(Remember how to factor a sum of cubes?)
f (x) (x 1)(x 2 x 1)(x 2)(x 2)
These are the factors with real coefficients. To break it down into linear factors,
use the quadratic formula.
For x 2 x 1 ,
(1) (1)2 4 11 1 1 4 1 3
2 1 2 2
1 3i 1 3
2 2 2
1 3 1 3
So, the factors of f(x) are (x 1) x i x i (x 2)(x 2) .
2 2 2 2
There are five linear factors, as there should be, and the complex factors come in