# 5.5 Zeros of polynomial functions

 Page 1 / 14
In this section, you will:
• Evaluate a polynomial using the Remainder Theorem.
• Use the Factor Theorem to solve a polynomial equation.
• Use the Rational Zero Theorem to find rational zeros.
• Find zeros of a polynomial function.
• Use the Linear Factorization Theorem to find polynomials with given zeros.
• Use Descartes’ Rule of Signs.
• Solve real-world applications of polynomial equations

A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?

This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we will discuss a variety of tools for writing polynomial functions and solving polynomial equations.

## Evaluating a polynomial using the remainder theorem

In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem    . If the polynomial is divided by $\text{\hspace{0.17em}}x–k,\text{\hspace{0.17em}}$ the remainder may be found quickly by evaluating the polynomial function at $\text{\hspace{0.17em}}k,\text{\hspace{0.17em}}$ that is, $\text{\hspace{0.17em}}f\left(k\right)\text{\hspace{0.17em}}$ Let’s walk through the proof of the theorem.

Recall that the Division Algorithm    states that, given a polynomial dividend $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and a non-zero polynomial divisor $\text{\hspace{0.17em}}d\left(x\right)\text{\hspace{0.17em}}$ where the degree of $\text{\hspace{0.17em}}\text{\hspace{0.17em}}d\left(x\right)\text{\hspace{0.17em}}$ is less than or equal to the degree of $\text{\hspace{0.17em}}f\left(x\right)$ , there exist unique polynomials $\text{\hspace{0.17em}}q\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\left(x\right)\text{\hspace{0.17em}}$ such that

$\text{\hspace{0.17em}}f\left(x\right)=d\left(x\right)q\left(x\right)+r\left(x\right)$

If the divisor, $\text{\hspace{0.17em}}d\left(x\right),\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}x-k,\text{\hspace{0.17em}}$ this takes the form

$f\left(x\right)=\left(x-k\right)q\left(x\right)+r$

Since the divisor $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is linear, the remainder will be a constant, $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ And, if we evaluate this for $\text{\hspace{0.17em}}x=k,\text{\hspace{0.17em}}$ we have

$\begin{array}{ccc}\hfill f\left(k\right)& =& \left(k-k\right)q\left(k\right)+r\hfill \\ & =& 0\cdot q\left(k\right)+r\hfill \\ & =& r\hfill \end{array}$

In other words, $\text{\hspace{0.17em}}f\left(k\right)\text{\hspace{0.17em}}$ is the remainder obtained by dividing $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}x-k.\text{\hspace{0.17em}}$

## The remainder theorem

If a polynomial $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is divided by $\text{\hspace{0.17em}}x-k,\text{\hspace{0.17em}}$ then the remainder is the value $\text{\hspace{0.17em}}f\left(k\right).\text{\hspace{0.17em}}$

Given a polynomial function $\text{\hspace{0.17em}}f,$ evaluate $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=k\text{\hspace{0.17em}}$ using the Remainder Theorem.

1. Use synthetic division to divide the polynomial by $\text{\hspace{0.17em}}x-k.\text{\hspace{0.17em}}$
2. The remainder is the value $\text{\hspace{0.17em}}f\left(k\right).\text{\hspace{0.17em}}$

## Using the remainder theorem to evaluate a polynomial

Use the Remainder Theorem to evaluate $\text{\hspace{0.17em}}f\left(x\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x-7\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=2.\text{\hspace{0.17em}}$

To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by $\text{\hspace{0.17em}}x-2.\text{\hspace{0.17em}}$

The remainder is 25. Therefore, $\text{\hspace{0.17em}}f\left(2\right)=25.\text{\hspace{0.17em}}$

Use the Remainder Theorem to evaluate $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=-3.\text{\hspace{0.17em}}$

$\text{\hspace{0.17em}}f\left(-3\right)=-412\text{\hspace{0.17em}}$

## Using the factor theorem to solve a polynomial equation

The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm.

$f\left(x\right)=\left(x-k\right)q\left(x\right)+r$

If $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a zero, then the remainder $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}f\left(k\right)=0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)=\left(x-k\right)q\left(x\right)+0\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}f\left(x\right)=\left(x-k\right)q\left(x\right).\text{\hspace{0.17em}}$

Notice, written in this form, $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is a factor of $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ We can conclude if $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a zero of $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}x-k\text{\hspace{0.17em}}$ is a factor of $f\left(x\right).\text{\hspace{0.17em}}$

#### Questions & Answers

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar