# 3.6 Zeros of polynomial functions  (Page 4/14)

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## Using the fundamental theorem of algebra

Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations.

Suppose $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is a polynomial function of degree four, and $\text{\hspace{0.17em}}f\left(x\right)=0.\text{\hspace{0.17em}}$ The Fundamental Theorem of Algebra states that there is at least one complex solution, call it $\text{\hspace{0.17em}}{c}_{1}.\text{\hspace{0.17em}}$ By the Factor Theorem, we can write $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ as a product of $\text{\hspace{0.17em}}x-{c}_{\text{1}}\text{\hspace{0.17em}}$ and a polynomial quotient. Since $\text{\hspace{0.17em}}x-{c}_{\text{1}}\text{\hspace{0.17em}}$ is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it $\text{\hspace{0.17em}}{c}_{\text{2}}.\text{\hspace{0.17em}}$ So we can write the polynomial quotient as a product of $\text{\hspace{0.17em}}x-{c}_{\text{2}}\text{\hspace{0.17em}}$ and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$

## The Fundamental Theorem of Algebra  States that, if f(x) Is a polynomial of degree n>0 , then f(x) Has at least one complex zero.

We can use this theorem to argue that, if $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is a polynomial of degree $\text{\hspace{0.17em}}n>0,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is a non-zero real number, then $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ has exactly $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ linear factors

$f\left(x\right)=a\left(x-{c}_{1}\right)\left(x-{c}_{2}\right)...\left(x-{c}_{n}\right)$

where $\text{\hspace{0.17em}}{c}_{1},{c}_{2},...,{c}_{n}\text{\hspace{0.17em}}$ are complex numbers. Therefore, $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ roots if we allow for multiplicities.

Does every polynomial have at least one imaginary zero?

No. A complex number is not necessarily imaginary. Real numbers are also complex numbers.

## Finding the zeros of a polynomial function with complex zeros

Find the zeros of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{3}+9{x}^{2}+x+3.\text{\hspace{0.17em}}$

The Rational Zero Theorem tells us that if $\text{\hspace{0.17em}}\frac{p}{q}\text{\hspace{0.17em}}$ is a zero of $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ is a factor of 3 and $\text{\hspace{0.17em}}q\text{\hspace{0.17em}}$ is a factor of 3.

The factors of 3 are $±1\text{\hspace{0.17em}}$ and $±3.\text{\hspace{0.17em}}$ The possible values for $\text{\hspace{0.17em}}\frac{p}{q},\text{\hspace{0.17em}}$ and therefore the possible rational zeros for the function, are We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with –3.

Dividing by $\text{\hspace{0.17em}}\left(x+3\right)\text{\hspace{0.17em}}$ gives a remainder of 0, so –3 is a zero of the function. The polynomial can be written as

$\left(x+3\right)\left(3{x}^{2}+1\right)$

We can then set the quadratic equal to 0 and solve to find the other zeros of the function.

The zeros of $f\left(x\right)$ are –3 and $\text{\hspace{0.17em}}±\frac{i\sqrt{3}}{3}.$

Find the zeros of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{3}+5{x}^{2}-11x+4.$

The zeros are

## Using the linear factorization theorem to find polynomials with given zeros

A vital implication of the Fundamental Theorem of Algebra    , as we stated above, is that a polynomial function of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ will have $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ factors. The Linear Factorization Theorem    tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form $\text{\hspace{0.17em}}\left(x-c\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a complex number.

what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
How would you find if a radical function is one to one?