The Real Zeros of a Polynomial Function

### Objectives

#### Complex Zeros:

###### 3. Find the Complex Zeros of a Polynomial

Preparing for this section:

A zero is a number that solves the equation F(x)=0. It is sometimes referred to as roots or solutions.
A rational zero is a rational number, which is a number that can be written as a fraction of two integers
A irrational zero is a number that is not rational, so it has an infinitely non-repeating decimal.

###### The Rational Zeros Theorem

The rational zeros theorem helps you find the rational zeros of a polynomial function. Once you find some of the rational zeros of a function (even just one) the other zeros can usually be found through traditional factoring methods.

An easy way to find the zeros is by using the p/q system. The P represents the factor of the constant in a function. Q represents the factors of the leading coefficient in the function.
So, if we have the function f(x)= 4x^3+3x^2+6x+8, the first thing we do is determine what the leading coefficient and the constant are in the function. 8 is constant, therefore 1, 2, 4, and 8 are the factors that can go in for p. 4 is the leading coefficient, so 1, 2, and 4 can go in as q.
P: +/- 1, +/- 2, +/- 4, +/- 8
Q: +/- 1, +/- 2, +/- 4
It can be positive or negative, so be sure to write +/- before the number

The next step is to set up the p/q possibilities in order from smallest to greatest.
P/Q: +/- ¼, +/- ½ , +/- 1, +/- 2, +/- 4, +/- 8
Once you have these possible zeros, you then use synthetic division to see which ones are true. This next step can be found here.

Practice problems:

Find all possible zeros of the functions.
F(x)= 4x4-5x3+7x2+3x-1
F(x)= 2x2-8x+9
P(x)= x5-x4+4x3+8x2-32x+48

###### Find the Real Zeros of a Polynomial Function

Let's say you are given a polynomial and a few zeros and are asked to find the remaining zeros and their multiplicities. The polynomial is x4-10x3+37x2-60x+36, and the zeros you are given are x=2, and x=3.
The first step? Identify what some factors will be. X=2, so you know that (x-2) will be in the final factored form of the polynomial, as well as (x-3).
Next, you must use synthetic division.

2
. 1 -10 37 -60 36
. ^ +2 -16 +42
. 1 -8 21 -18 0

As you know, when you have a remainder of zero you are doing it correctly. This is telling you that x is in fact a factor, therefore the next step is to add in the x’s and exponents to what you have gotten in synthetic division: x3-8x2+21x-18
You then will add the factor (x-2) to this function.
(x-2) (x3-8x2+21x-18)
Now that we have one factor done, we must go back to the x=3. We then must plug 3 into the synthetic division.

3
. 1 -8 21 -18
. ^ +3 -15 +18
. 1 -5 6 0

The three values at the end of synthetic division will be set like this: x2-5x+6
(x-3) (x-2) (x2-5x+6) Next, you can factor the main function
(x-3) (x-2) (x-2) (x-3)
This will turn into: (x-3)2 and (x-2)2
The function (x-3)^2 equals 3 once you solve it, and (x-2)^2 equals 2. Each zeros multiplicity is 2.

##### Complex Zeros:

What is a complex zero you may ask?
A complex zero is a real or imaginary number, which when plugged into a function, makes the function come out to zero. Some functions have no zeros that are real numbers, and some have both real and imaginary, so the only way to express them is by using complex numbers.

###### 1. Use the Conjugate Pairs Theorem

In, p=a+bi, a is the real number, and the b (which is a real number) is scaling i, the imaginary number.
If you have the function a+bi, its conjugate will be a-bi. This only applies when all the coefficients are real numbers.

###### 2. Find a Polynomial Function with Specified Zeros

Let’s say you are given the zeros: -3, -1+i, 3-2i and are asked to write a polynomial function in factorable form with them. This is not too hard.
Let’s take it one zero at a time.
-3 will turn into (x+3) because that is the factorable form of it.
For -1+i, it will be (x-(-1+i)) and (x-(-1-i)). The x- is put in front and there are two terms with two different signs after the 1 because complex roots always happen in conjugate pairs.
For the last zero, 3-2i, it goes to (x-(3-2i)) and (x-(3i+2i))
The final answer should be:
(x+3) (x-(-1+i)) (x-(-1-i)) (x-(3-2i)) (x-(3+2i))

If a problem asks for a multiplicity with a zero, ie: the Zero 6 with multiplicity of 3, then it would turn into (x-6)3

Practice problems:

1. Zeros: -4, -2+3i, -i

2. Zeros: 3, -2i, 1+i

3. Zeros: 2 (multiplicity of 4), 4i, -8i

###### 3. Find the Complex Zeros of a Polynomial

Consider the function F(x)= x2-2x-8
In order to find the zeros of this function, you factor it into (x-4) (x+2), and after setting each term equal to zero, you know that the zeros of this function are x=4 and x=2
Now we’ll go onto a harder problem. Let’s say we have the function x3-4x2+25x-100.
The first step is to factor by grouping. Out of the first two terms, we can take out a x2 → x2 (x-4), and in the second we can take out 25→ 25 (x-4)
Next, we can separate the terms, making one (x-4) and the other (x2+25)
Then we will set each term equal to zero.
For the first term, x-4=0, we will get 4 as the zero.
For the second term, x2+25=0, we get x2=-25. We must take the square root of both sides, we get +/- the square root of 25. Because the square root of -1 is i, the answer will be +/- 5i.
The zero's for this function are -5, 4, 5,