Understanding Polynomial Root Theorem and Complex Roots

When presented with a polynomial, it is possible to determine the number of roots by using the Fundamental Theorem of Algebra. However, this theorem does not give any detailed information about the roots — whether they are integer, rational, real, or imaginary. This lesson will explore methods to find these amazing inner workings of a root.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Polynomial
Complex Number
Synthetic Division
Leading Coefficient
Degree of a Polynomial

Explore

Polynomial Functions and Their Roots

Recalling the Fundamental Theorem of Algebra, a polynomial function of degree greater than

always has a root. These roots can be found using diverse methods like the quadratic formula for quadratic polynomial functions.

However, factoring a polynomial function of greater degrees can be tough, as there are not easy to use formulas to do so. The good news is that looking at the terms of a polynomial function can be useful to find more characteristics of a given function. Consider the following functions and their roots.

Polynomial Function	Roots

How can the roots of the polynomial be related to the constant term of a polynomial function?

Discussion

Rational Root Theorem

Consider a polynomial where every coefficient is an integer.

The following properties are true for the roots of the polynomial.

Every integer root is a factor of the constant
All rational roots must have a numerator that is a factor of and a denominator that is a factor of the leading coefficient

Proof

Rational Root Theorem

To prove this theorem, each of the properties will be considered individually. To start with the first property, consider a polynomial

with integer coefficients that has an integer root

This means that

Now

is subtracted from both sides of the equation.

SubEqn

FactorOut

Factor out

Since the coefficients of

are integers and

is an integer, the expression between the parentheses results in an integer. Also,

is the product of an integer and

which means that

is a factor of

This completes the proof of the first property.

For the second property, suppose that the polynomial has a rational root such that the fraction is written in its simplest form. Again, it is obtained that

This time the equation will be modified in a different way.

Factor out

Since the coefficients of

and

are all integers, the expression between parentheses results in an integer. Also, since

is written in its simplest form,

and

do not have a common factor, which means that

and

do not have a common factor either. Therefore,

is a factor of

The numerator of the root, is a factor of

If the modification is done differently, it is possible to come to the other property.

SubEqn

FactorOut

Factor out

With the same reasoning as before, it is possible to conclude that

the denominator of the root, is a factor of

finishing the proof.

As an example of how to apply this theorem, consider the following polynomial.

The integer roots of the equation must be factors of the value of

which in this case is

The factors of

are

and

Each of these values is substituted for

into the equation

to determine which is a root.

Equation	Result
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$\text{-}90 = 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\text{-}924 = 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$

As determined, the integer roots are and The rational roots have a numerator that is a factor of and a denominator that is a factor of In this case, is with factors of and The following table displays the possible fractions that can be formed with the factors. The columns of the table are the factors of and the rows are the factors of

It is important to understand that the integers do not need to be tested because the integer factors have already been found. Now, each fraction will be tested to see if they are roots of the equation.

Equation	Result
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$\dfrac{22}{27}= 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\text{-}\dfrac{100}{9}= 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\text{-}\dfrac{7280}{27}= 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\dfrac{235}{108} = 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$

The rational numbers

and

are roots of the equation. Since the given polynomial is of degree

by the Fundamental Theorem of Algebra, it has four solutions. Therefore, the roots of

are

and

Example

Finding the Integer Roots of a Polynomial Function

Izabella is tutoring some of her friends for an upcoming math exam.

To study, they decided to go over some exercises that the math professor assigned for homework in the past. One polynomial function, in particular, caught Izabella's attention.

Solve following exercises.

a Find the integer roots of the given polynomial function.

b Find the remaining roots of the function.

Hint

a Use the Rational Root Theorem.

b Use synthetic division.

Solution

a The integer roots of a polynomial function can be found using the Rational Root Theorem. This theorem indicates that the integer roots of a polynomial are factors of the constant term

which in this case is

Below are the factors of

listed.

To determine which of these is a root of the given polynomial function, each of the factors will be substituted into the equation

to check which results in a true statement. First, the positive factors will be substituted.

Equation	Solution
	$\text{-}4 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$30 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$140 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$936 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$18\, 300 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$

The value of is a root of the function. Since the polynomial function is of degree it is possible that one of the negative factors is also a root, so these values will be substituted next.

Equation	Solution
	$\text{-}30 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\text{-}40 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$156 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$1200 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$21\, 420 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$

As can be seen in the table, is also a root of Since every integer root has to be a factor of and every factor was tested, the only integer roots of the given function are and

b The Factor Theorem can be used to factor the given polynomial into a polynomials of a lesser degree. Since

is a root of

is a factor of

This means that multiplying a polynomial

results in

To find the polynomial

which is one degree less than

the polynomial

has to be divided by the binomial

This division can be performed by synthetic division. First, the coefficients are ordered and the leftmost coefficient goes down, below the horizontal line.

Bring down

Each time, the newest number under the horizontal line is multiplied by

then the product is aligned with the following number inside the division, and the newest terms inline are added. This procedure is shown below.

▼

Evaluating the division

The resulting polynomial of the division can be written using the quotient. It is important to note that the quotient is one degree less than the original polynomial.

Now the given function can be rewritten factoring the binomial

A similar thing can be done to rewrite

A root of the function

is also a root of

Therefore, the binomial

can be factored from

using synthetic division again.

▼

Evaluate the division

Bring down

With this result, the given function can be rewritten again.

Finally, to find the remaining two roots, the roots of the polynomial at the rightmost parentheses are needed. This time the Quadratic Formula will be used. First, the coefficients of the quadratic equation are identified.

Now these coefficients can be substituted into the Quadratic Formula.

SubstituteValues

Substitute values

▼

Simplify

CalcPow

Calculate power

Multiply

SubTerms

Subtract terms

There are two possible values for

to solve the polynomial. Since there is a square root of a negative number involved, the results are complex numbers.

Therefore, the remaining roots of the given polynomial function are the complex numbers and

Example

Finding the Rational Roots of a Polynomial Function

Izabella now wants to challenge her study group to try a problem all on their own. Kevin is ready for it!

He receives the following polynomial function.

Help Kevin find all the roots of the function.

Hint

Remember the Rational Root Theorem.

Solution

Considering the Rational Root Theorem, it is possible to find the integer and the rational roots. According to the theorem, the integer roots of the polynomial must be factors of the constant term of the polynomial, which is

Each of these factors is substituted into the equation

to determine which, if any, is a root of the function.

Substitution	Simplify
	$\text{-}14 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$4 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\text{-}2 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$10 \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$

No value resulted in a true statement. This means there is no integer root of the function. However, the Rational Root Theorem also allows to look for the rational roots of a polynomial. These roots can be written as follows.

In this expression,

the numerator, is a factor of the constant term

the denominator, is a factor of the leading coefficient, which in this case is

Previously the factors of

were presented. Below are the factors of

Since these factors must be denominators and the integer roots were already considered, the only factors to be considered are

and

which result in the following rational numbers.

Again, each of these numbers is substituted into the equation as before to verity which, if any, is a root of the function.

Substitution	Simplify
	$\tfrac{14}{3} \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$\tfrac{34}{3} \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$
	$0 = 0 \quad{\color{#009600}{\bm{\Large{\checkmark }}}}$
	$\text{-}\tfrac{14}{9} \neq 0 \quad{\color{#FF0000}{\bm{\Large{\times }}}}$

The value

is a root of the given function. This means that there are

roots remaining. Using synthetic division, the root

can be factored from the given function to obtain a quadratic equation. The first step of the synthetic division is to write the coefficients and the root as shown below.

Then, the

is moved down, below the horizontal line.

In synthetic division, the number is multiplied by the newest number below the horizontal line. It is then added to the following number above the line, and the sum is moved below the horizontal line. The division can then be done.

▼

Evaluate division

Using this result, it is possible to rewrite the given function. It is important to remember that the resulting polynomial is always one degree less than the original one.

The roots of the function are also roots of the polynomial inside the rightmost parentheses. Because of this, this polynomial has to be factored.

FactorOut

Factor out

FacDiffSquares

Now that the polynomial is factored, the given function can be rewritten again.

By the Factor Theorem, the roots of the function are

and

Discussion

The Irrational Roots of a Polynomial

The Rational Root Theorem helps determine the integer and rational roots of a polynomial. One might expect a similar theorem exists used to find irrational roots. Unfortunately, there is not. But wait, it is not all bad news. A theorem exists that does involve the irrational roots of a polynomial.

Rule

Irrational Conjugate Root Theorem

Let be a polynomial function whose coefficients are rational. If and are rational such that is an irrational number and is a root of then its irrational conjugate is also a root of

Proof

Let

be rational numbers and

be an irrational number. Suppose that the irrational number

is a root of the polynomial

with rational coefficients

Since the conjugate of zero is zero, the expression's conjugate on the left-hand side is also zero.

Next, using the properties of conjugation, the polynomial can be rewritten.

	Polynomial
	Conjugate of a sum is the sum of the conjugates
	Conjugate of a product is the product of the conjugates
	Conjugate of a power is the power of the conjugate
	Conjugate of a rational number is itself

This final expression is equal to

Notice that the middle expression is equal to the value of the polynomial at

Using this and the fact that the conjugate of

this equation can be rewritten as follows.

In conclusion, the irrational number

makes the polynomial

zero, and, therefore, it is also a root of

Example

Rational and Irrational Roots

Other than being really good at math, Izabella is also known for making excellent cakes. To motivate her friends, she decided to organize a little contest and give a cake to the winner.

The challenge is to write a polynomial function

of degree

that only has integer coefficients. Additionally, the function must have roots

and

Each of her friends wrote the following functions as their answers.

Who will win Izabella's tasty cake?

Hint

Remember the Irrational Conjugate Root Theorem

Solution

It is mentioned that the polynomial is of degree

Considering the Fundamental Theorem of Algebra, the given polynomial has three roots. Two of those roots are already given.

All the roots of the polynomial are needed to determine the correct polynomial. Since every coefficient of the polynomial is an integer and the solution

is an irrational number, by the Irrational Conjugate Root Theorem, the irrational conjugate of

has to be a root of the function.

Using the Factor Theorem, it is possible to write the polynomial by multiplying the binomials produced by the roots.

To expand this expression clearly, first the two rightmost binomials will be multiplied.

▼

Simplify

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	The conjugate of the polynomial evaluated at
	The conjugate of a sum is the sum of the conjugates.
	The conjugate of a product is the product of the conjugates.
	The conjugate of a power is the power of the conjugate.
	The conjugate of a real number is itself.
	The conjugate of is
	The polynomial evaluated at

Root	Binomial

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Catch-Up and Review

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Solution

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