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

0

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

a x^{2} + b x + c = 0 ⇓ x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

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
$p (x) = x^{3} - 6 x^{2} + 11 x - 6$	$1,$ $2,$ $3$
$g (x) = x^{4} - 4 x^{3} - 19 x^{2} + 106 x - 120$	$- 5,$ $2,$ $3,$ $4$
$m (x) = x^{3} + 6 x^{2} - 55 x - 252$	$- 9,$ $- 4,$ $7$

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.

P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}

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

Every integer root is a factor of the constant $a_{0} .$
All rational roots must have a numerator that is a factor of $a_{0}$ and a denominator that is a factor of the leading coefficient $a_{n} .$

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

P (x)

with integer coefficients that has an integer root

x_{r} .

This means that

P (x_{r}) = 0 .

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} + a_{0} = 0

Now

a_{0}

is subtracted from both sides of the equation.

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} + a_{0} = 0

SubEqn

$LHS - a_{0} = RHS - a_{0}$

a_{n} x_{r}^{n} + a_{n - 1} x_{r}^{n - 1} + \dots + a_{1} x_{r} = - a_{0}

FactorOut

Factor out $x_{r}$

x_{r} (a_{n} x_{r}^{n - 1} + a_{n - 1} x_{r}^{n - 2} + \dots + a_{1}) = - a_{0}

Since the coefficients of

P (x)

are integers and

x_{r}

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

- a_{0}

is the product of an integer and

x_{r},

which means that

x_{r}

is a factor of

a_{0} .

This completes the proof of the first property.

For the second property, suppose that the polynomial $P (x)$ has a rational root $\frac{p}{q},$ such that the fraction is written in its simplest form. Again, it is obtained that $P (\frac{p}{q}) = 0 .$

$a_n \left(\dfrac{p}{q}\right)^n + a_{n-1}\left(\dfrac{p}{q}\right)^{n-1}+\cdots+a_1\left(\dfrac{p}{q}\right)+a_0 = 0$

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

a_{n} (\frac{p}{q})^{n} + a_{n - 1} (\frac{p}{q})^{n - 1} + \dots + a_{1} (\frac{p}{q}) + a_{0} = 0

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

a_{n} (\frac{p ^{n}}{q ^{n}}) + a_{n - 1} (\frac{p ^{n - 1}}{q ^{n - 1}}) + \dots + a_{1} (\frac{p}{q}) + a_{0} = 0

MultEqn

$LHS \cdot q^{n} = RHS \cdot q^{n}$

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) + a_{0} (q^{n}) = 0

SubEqn

$LHS - a_{0} (q^{n}) = RHS - a_{0} (q^{n})$

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) = - a_{0} (q^{n})

FactorOut

Factor out $p$

p (a_{n} p^{n - 1} + a_{n - 1} p^{n - 2} q + \dots + a_{1} q^{n - 1}) = - a_{0} q^{n}

Since the coefficients of

P (x),

p,

and

q

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

\frac{p}{q}

is written in its simplest form,

p

and

q

do not have a common factor, which means that

p

and

q^{n}

do not have a common factor either. Therefore,

p

is a factor of

a_{0} .

The numerator of the root, $p,$ is a factor of $a_{0} .$

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

a_{n} (p^{n}) + a_{n - 1} (p^{n - 1} q) + \dots + a_{1} (p q^{n - 1}) + a_{0} (q^{n}) = 0

SubEqn

$LHS - a_{n} p^{n} = RHS - a_{n} p^{n}$

a_{n - 1} p^{n - 1} q + \dots + a_{1} p q^{n - 1} + a_{0} q^{n} = - a_{n} p^{n}

FactorOut

Factor out $q$

q (a_{n - 1} p^{n - 1} + \dots + a_{1} p q^{n - 2} + a_{0} q^{n - 1}) = - a_{n} p^{n}

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

q,

the denominator of the root, is a factor of

a_{n},

finishing the proof.

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

p (x) = 6 x^{4} - 59 x^{3} + 94 x^{2} - 49 x + 8

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

a_{0},

which in this case is

8 .

The factors of

8

are

1,

2,

4,

and

8 .

Each of these values is substituted for

x

into the equation

p (x) = 0

to determine which is a root.

Equation	Result
$6 \cdot 1^{4} - 59 \cdot 1^{3} + 94 \cdot 1^{2} - 49 \cdot 1 + 8 = 0$	$0 = 0 ✓$
$6 \cdot 2^{4} - 59 \cdot 2^{3} + 94 \cdot 2^{2} - 49 \cdot 2 + 8 = 0$	$- 90 = 0 \times$
$6 \cdot 4^{4} - 59 \cdot 4^{3} + 94 \cdot 4^{2} - 49 \cdot 4 + 8 = 0$	$- 924 = 0 \times$
$6 \cdot 8^{4} - 59 \cdot 8^{3} + 94 \cdot 8^{2} - 49 \cdot 8 + 8 = 0$	$0 = 0 ✓$

As determined, the integer roots are $1$ and $8 .$ The rational roots have a numerator that is a factor of $a_{0}$ and a denominator that is a factor of $a_{n} .$ In this case, $a_{n}$ is $6$ with factors of $1,$ $2,$ $3,$ and $6 .$ The following table displays the possible fractions that can be formed with the factors. The columns of the table are the factors of $8,$ and the rows are the factors of $6 .$

	$1$	$2$	$4$	$8$
$2$	$\frac{1}{2}$	$\frac{2}{2} = 1$	$\frac{4}{2} = 2$	$\frac{8}{2} = 4$
$3$	$\frac{1}{3}$	$\frac{2}{3}$	$\frac{4}{3}$	$\frac{8}{3}$
$6$	$\frac{1}{6}$	$\frac{2}{6} = \frac{1}{3}$	$\frac{4}{6} = \frac{2}{3}$	$\frac{8}{6} = \frac{4}{3}$

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
$6 (\frac{1}{2})^{4} - 59 (\frac{1}{2})^{3} + 94 (\frac{1}{2})^{2} - 49 (\frac{1}{2}) + 8 = 0$	$0 = 0 ✓$
$6 (\frac{1}{3})^{4} - 59 (\frac{1}{3})^{3} + 94 (\frac{1}{3})^{2} - 49 (\frac{1}{3}) + 8 = 0$	$0 = 0 ✓$
$6 (\frac{2}{3})^{4} - 59 (\frac{2}{3})^{3} + 94 (\frac{2}{3})^{2} - 49 (\frac{2}{3}) + 8 = 0$	$\frac{2 2}{2 7} = 0 \times$
$6 (\frac{4}{3})^{4} - 59 (\frac{4}{3})^{3} + 94 (\frac{4}{3})^{2} - 49 (\frac{4}{3}) + 8 = 0$	$- \frac{1 0 0}{9} = 0 \times$
$6 (\frac{8}{3})^{4} - 59 (\frac{8}{3})^{3} + 94 (\frac{8}{3})^{2} - 49 (\frac{8}{3}) + 8 = 0$	$- \frac{7 2 8 0}{2 7} = 0 \times$
$6 (\frac{1}{6})^{4} - 59 (\frac{1}{6})^{3} + 94 (\frac{1}{6})^{2} - 49 (\frac{1}{6}) + 8 = 0$	$\frac{2 3 5}{1 0 8} = 0 \times$

The rational numbers

\frac{1}{2}

and

\frac{1}{3}

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

4,

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

p (x)

are

\frac{1}{2},

\frac{1}{3},

1

and

8 .

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.

p (x) = x^{4} - x^{3} - 6 x^{2} + 14 x - 12

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

a_{0},

which in this case is

- 12 .

Below are the factors of

- 12

listed.

Positive : Negative : 1, 2, 3, 4, 6, 12 - 1, - 2, - 3, - 4, - 6, - 12

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

p (x) = 0

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

Equation	Solution
$(1)^{4} - (1)^{3} - 6 (1)^{2} + 14 (1) - 12 = 0$	$- 4 \neq = 0 \times$
$(2)^{4} - (2)^{3} - 6 (2)^{2} + 14 (2) - 12 = 0$	$0 = 0 ✓$
$(3)^{4} - (3)^{3} - 6 (3)^{2} + 14 (3) - 12 = 0$	$30 \neq = 0 \times$
$(4)^{4} - (4)^{3} - 6 (4)^{2} + 14 (4) - 12 = 0$	$140 \neq = 0 \times$
$(6)^{4} - (6)^{3} - 6 (6)^{2} + 14 (6) - 12 = 0$	$936 \neq = 0 \times$
$(12)^{4} - (12)^{3} - 6 (12)^{2} + 14 (12) - 12 = 0$	$18300 \neq = 0 \times$

The value of $x = 2$ is a root of the function. Since the polynomial function is of degree $4,$ it is possible that one of the negative factors is also a root, so these values will be substituted next.

Equation	Solution
$(- 1)^{4} - (- 1)^{3} - 6 (- 1)^{2} + 14 (- 1) - 12 = 0$	$- 30 \neq = 0 \times$
$(- 2)^{4} - (- 2)^{3} - 6 (- 2)^{2} + 14 (- 2) - 12 = 0$	$- 40 \neq = 0 \times$
$(- 3)^{4} - (- 3)^{3} - 6 (- 3)^{2} + 14 (- 3) - 12 = 0$	$0 = 0 ✓$
$(- 4)^{4} - (- 4)^{3} - 6 (- 4)^{2} + 14 (- 4) - 12 = 0$	$156 \neq = 0 \times$
$(- 6)^{4} - (- 6)^{3} - 6 (- 6)^{2} + 14 (- 6) - 12 = 0$	$1200 \neq = 0 \times$
$(- 12)^{4} - (- 12)^{3} - 6 (- 12)^{2} + 14 (- 12) - 12 = 0$	$21420 \neq = 0 \times$

As can be seen in the table, $- 3$ is also a root of $p (x) .$ Since every integer root has to be a factor of $- 12$ and every factor was tested, the only integer roots of the given function are $- 3$ and $2 .$

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

2

is a root of

p (x),

(x - 2)

is a factor of

p (x) .

This means that multiplying a polynomial

q (x)

(x - 2)

results in

p (x) .

p (x) = (x - 2) q (x)

To find the polynomial

q (x),

which is one degree less than

p (x),

the polynomial

p (x)

has to be divided by the binomial

(x - 2) .

q (x) = \frac{x ^{4} - x ^{3} - 6 x ^{2} + 1 4 x - 1 2}{x - 2}

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

2 1 - 1 - 6 14 - 12

Bring down $1$

2 1 - 1 - 6 14 - 12 1 - 1 - 6 14 - 12

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

2,

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

2 1 - 1 - 6 14 - 12 1 - 1 - 6 14 - 12

Evaluating the division

$2 \cdot 1 = 2$

2 1 - 1 2 - 6 14 - 12 1 - 1 - 6 14 - 12

$- 1 + 2 = 1$

2 1 - 1 2 - 6 14 - 12 11 - 6 14 - 12

$2 \cdot 1 = 2$

2 1 - 1 2 - 6 2 14 - 12 11 - 6 14 - 12

$- 6 + 2 = - 4$

2 1 - 1 2 - 6 2 14 - 12 11 - 4 14 - 12

$2 \cdot - 4 = - 8$

2 1 - 1 2 - 6 2 14 - 8 - 12 11 - 4 14 - 12

$14 - 8 = 6$

2 1 - 1 2 - 6 2 14 - 8 - 12 11 - 4 - 6 - 12

$2 \cdot 6 = 12$

2 1 - 1 2 - 6 2 14 - 8 - 12 12 11 - 4 6 - 12

$- 12 + 12 = 0$

2 1 - 1 2 - 6 2 14 - 8 - 12 12 11 - 4 6 . 0

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.

q (x) = \frac{x ^{4} - x ^{3} - 6 x ^{2} + 1 4 x - 1 2}{x - 2} ⇓ q (x) = x^{3} + x^{2} - 4 x + 6

Now the given function can be rewritten factoring the binomial

(x - 2) .

p (x) = (x - 2) q (x) ⇓ p (x) = (x - 2) (x^{3} + x^{2} - 4 x + 6)

A similar thing can be done to rewrite

q (x) .

A root of the function

p (x)

is also a root of

q (x) .

Therefore, the binomial

(x + 3)

can be factored from

q (x)

using synthetic division again.

- 3 11 - 4 6

Evaluate the division

Bring down $1$

- 3 11 - 4 6 11 - 4 6

$- 3 \cdot 1 = - 3$

- 3 1 1 - 3 - 4 6 11 - 4 6

$1 - 3 = - 2$

- 3 1 1 - 3 - 4 6 1 - 2 - 4 6

$- 3 \cdot - 2 = 6$

- 3 1 1 - 3 - 4 6 6 1 - 2 - 4 6

$- 4 + 6 = 2$

- 3 1 1 - 3 - 4 6 6 1 - 2 26

$- 3 \cdot 2 = - 6$

- 3 1 1 - 3 - 4 6 6 - 6 1 - 2 26

$6 - 6 = 0$

- 3 1 1 - 3 - 4 6 6 - 6 1 - 2 20

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

p (x) = (x - 2) (x^{3} + x^{2} - 4 x + 6) ⇓ p (x) = (x - 2) (x + 3) (x^{2} - 2 x + 2)

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.

a x^{2} + b x + c = 0 ⇓ 1 x^{2} + (- 2) x + 2 = 0

Now these coefficients can be substituted into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 2 ) \pm ( 2 ) ^{2} - 4 ( 1 ) ( 2 )}{2 ( 1 )}

Simplify

CalcPow

Calculate power

x = \frac{- ( - 2 ) \pm 4 - 4 ( 1 ) ( 2 )}{2 ( 1 )}

Multiply

x = \frac{2 \pm 4 - 8}{2}

SubTerms

Subtract terms

x = \frac{2 \pm - 4}{2}

There are two possible values for

x

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

$x = \frac{2 + - 4}{2}$	$x = \frac{2 - - 4}{2}$
$x = \frac{2 + 2 i}{2}$	$x = \frac{2 - 2 i}{2}$
$x = 1 + i$	$x = 1 - i$

Therefore, the remaining roots of the given polynomial function are the complex numbers $1 + i$ and $1 - i .$

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.

g (x) = 3 x^{3} - x^{2} - 6 x + 2

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

2 .

Factors of 2: - 2, - 1, 1, 2

Each of these factors is substituted into the equation

g (x) = 0

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

Substitution	Simplify
$3 (- 2)^{3} - (- 2)^{2} - 6 (- 2) + 2 = 0$	$- 14 \neq = 0 \times$
$3 (- 1)^{3} - (- 1)^{2} - 6 (- 1) + 2 = 0$	$4 \neq = 0 \times$
$3 (1)^{3} - (1)^{2} - 6 (1) + 2 = 0$	$- 2 \neq = 0 \times$
$3 (2)^{3} - (2)^{2} - 6 (2) + 2 = 0$	$10 \neq = 0 \times$

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.

x = \frac{p}{q}

In this expression,

p,

the numerator, is a factor of the constant term

q,

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

3 .

Previously the factors of

2

were presented. Below are the factors of

3 .

Factors of 3: - 3, - 1, 1, 3

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

- 3

and

3,

which result in the following rational numbers.

- \frac{2}{3}, - \frac{1}{3}, \frac{1}{3}, \frac{2}{3}

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
$3 (- \frac{2}{3})^{3} - (- \frac{2}{3})^{2} - 6 (- \frac{2}{3}) + 2 = 0$	$\frac{1 4}{3} \neq = 0 \times$
$3 (- \frac{1}{3})^{3} - (- \frac{1}{3})^{2} - 6 (- \frac{1}{3}) + 2 = 0$	$\frac{3 4}{3} \neq = 0 \times$
$3 (\frac{1}{3})^{3} - (\frac{1}{3})^{2} - 6 (\frac{1}{3}) + 2 = 0$	$0 = 0 ✓$
$3 (\frac{2}{3})^{3} - (\frac{2}{3})^{2} - 6 (\frac{2}{3}) + 2 = 0$	$- \frac{1 4}{9} \neq = 0 \times$

The value

x = \frac{1}{3}

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

2

roots remaining. Using synthetic division, the root

\frac{1}{3}

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.

\frac{1}{3} 3 - 1 - 6 2

Then, the

3

is moved down, below the horizontal line.

\frac{1}{3} 3 - 1 - 6 2 3 - 1 - 6 2

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.

\frac{1}{3} 3 - 1 - 6 2 3 - 1 - 6 2

Evaluate division

$\frac{1}{3} \cdot 3 = 1$

\frac{1}{3} 3 - 1 1 - 6 2 3 - 1 - 6 2

$- 1 + 1 = 0$

\frac{1}{3} 3 - 1 1 - 6 2 30 - 6 2

$\frac{1}{3} \cdot 0 = 0$

\frac{1}{3} 3 - 1 1 - 6 0 2 30 - 6 2

$- 6 + 0 = - 6$

\frac{1}{3} 3 - 1 1 - 6 0 2 30 - 6 2

$\frac{1}{3} \cdot (- 6) = - 2$

\frac{1}{3} 3 - 1 1 - 6 0 2 - 2 30 - 6 2

$2 - 2 = 0$

\frac{1}{3} 3 - 1 1 - 6 0 2 - 2 30 - 6 0

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.

g (x) = 3 x^{3} - x^{2} - 6 x + 2 ⇓ g (x) = (x - \frac{1}{3}) (3 x^{2} - 6)

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

3 x^{2} - 6

FactorOut

Factor out $3$

3 (x^{2} - 2)

$a = (a)^{2}$

3 (x^{2} - (2)^{2})

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

3 (x - 2) (x + 2)

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

g (x) = (x - \frac{1}{3}) (3 x^{2} - 6) ⇓ g (x) = 3 (x - \frac{1}{3}) (x - 2) (x + 2)

By the Factor Theorem, the roots of the function are

\frac{1}{3},

- 2,

and

2 .

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 $P (x)$ be a polynomial function whose coefficients are rational. If $a$ and $b$ are rational such that $b$ is an irrational number and $a + b$ is a root of $P (x),$ then its irrational conjugate $a - b$ is also a root of $P (x) .$

$P (a + b) = 0 \Rightarrow P (a - b) = 0$

Proof

Let

a,

b

be rational numbers and

b

be an irrational number. Suppose that the irrational number

a + b

is a root of the polynomial

P (x)

with rational coefficients

c_{i} .

P (a + b) = i = 0 \sum n c_{i} (a + b)^{i} = 0

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

\overline{P (a + b)} = \overline{i = 0 \sum n c_{i} \cdot (a + b)^{i}} = 0

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

$\overline{i = 0 \sum n c_{i} \cdot (a + b)^{i}}$	Polynomial
$i = 0 \sum n \overline{c_{i} \cdot (a + b)^{i}}$	Conjugate of a sum is the sum of the conjugates
$i = 0 \sum n \overline{c_{i}} \overline{(a + b)^{i}}$	Conjugate of a product is the product of the conjugates
$i = 0 \sum n \overline{c_{i}} (\overline{a + b})^{i}$	Conjugate of a power is the power of the conjugate
$i = 0 \sum n c_{i} (\overline{a + b})^{i}$	Conjugate of a rational number is itself

This final expression is equal to

0 .

\overline{P (a + b)} = i = 0 \sum n c_{i} (\overline{a + b})^{i} = 0

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

\overline{a + b} .

Using this and the fact that the conjugate of

a + b

a - b,

this equation can be rewritten as follows.

i = 0 \sum n c_{i} (\overline{a + b})^{i} P (\overline{a + b}) P (a - b) = 0 ⇓ = 0 ⇓ = 0

In conclusion, the irrational number

a - b

makes the polynomial

p (x)

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

P (x) .

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

p (x)

of degree

3

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

- 5

and

1 - 3 .

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

Vincenzo : Zain : Ali : Kevin : p (x) = x^{3} + 2 x^{2} + 10 x - 12 p (x) = x^{3} - 4 x^{2} - 16 x - 14 p (x) = x^{3} + 3 x^{2} - 12 x - 10 p (x) = x^{3} - 2 x^{2} + 10 x + 13

Who will win Izabella's tasty cake?

Hint

Remember the Irrational Conjugate Root Theorem

Solution

It is mentioned that the polynomial is of degree

3 .

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

x_{1} x_{2} = - 5 = 1 - 3

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

x_{2} = 1 - 3

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

1 - 3

has to be a root of the function.

x_{3} = 1 + 3

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

p (x) (x + 5) (x - (1 - 3)) (x - (1 + 3))

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

(x - (1 - 3)) (x - (1 + 3))

Simplify

Distr

Distribute $(x - (1 - 3))$

x (x - (1 - 3)) - (1 + 3) (x - (1 - 3))

Distr

Distribute $x$

x^{2} - x (1 - 3) - (1 + 3) (x - (1 - 3))

Distr

Distribute $(1 + 3)$

x^{2} - x (1 - 3) - x (1 + 3) + (1 - 3) (1 + 3)

Distr

Distribute $x$

x^{2} - x + x 3 - x - x 3 + (1 - 3) (1 + 3)

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

x^{2} - x + x 3 - x - x 3 + 1 - 3

SubTerms

Subtract terms

x^{2} - 2 x - 2

Now, this result will be multiplied by the binomial

(x + 5) .

(x + 5) (x^{2} - 2 x - 2)

Simplify

Distr

Distribute $(x + 5)$

x^{2} (x + 5) - 2 x (x + 5) - 2 (x + 5)

Distr

Distribute $x^{2}$

x^{3} + 5 x^{2} - 2 x (x + 5) - 2 (x + 5)

Distr

Distribute $2 x$

x^{3} + 5 x^{2} - 2 x^{2} - 10 x - 2 (x + 5)

Distr

Distribute $2$

x^{3} + 5 x^{2} - 2 x^{2} - 10 x - 2 x - 10

SubTerms

Subtract terms

x^{3} + 3 x^{2} - 12 x - 10

With this result, it is possible to write the correct answer.

p (x) = x^{3} + 3 x^{2} - 12 x - 10

Therefore, Ali will be the winner. Additionally, it can be checked that the polynomials written by the other friends did not meet the conditions Izabella requested. Zain, in particular, feels sad to have not gotten their answer correct.

Example

All Irrational Roots

After Ali won the prized cake, Zain promised to study more efficiently. They thought this could be accomplished by putting on some shades and a fedora while studying at home. Nobody knows if it will work.

Zain begins this noble mission with a polynomial function

p (x)

with a degree of four and integer coefficients with the following roots.

x_{1} x_{2} = 2 - 5 = 7

Find the remaining roots of the equation

p (x) = 0 .

Hint

Consider the Irrational Conjugate Root Theorem.

Solution

It is given that the polynomial function

p (x)

has a degree of

4 .

Then, by the Fundamental Theorem of Algebra, the function has four roots, counting repeating roots as multiple roots. Two roots are already given.

x_{1} x_{2} = 2 - 5 = 7

Since it is known that the coefficients are integers, they can be used to find the other two roots. Recall that the Irrational Conjugate Root Theorem states that if an irrational number

a + b

is a root of a polynomial function with rational coefficients, then the irrational conjugate

a - b

is also a root. The given roots can be written as indicated.

x x_{1} x_{2} = a + b = 2 + (- 5) = 0 + 7

Considering the Irrational Conjugate Theorem, the remaining two roots are the irrational conjugates of the given roots.

x_{3} = 2 - (- 5) x_{4} = 0 - 7 \Rightarrow \Rightarrow x_{3} = 2 + 5 x_{4} = - 7

Zain could be on to something with rocking this new fashion while studying!

Discussion

The Complex Roots of a Polynomial

The Rational and Irrational Roots Theorems provide information about the integer, rational, and irrational roots of a polynomial. Therefore, in some way, the real roots of a polynomial are covered by these theorems. Next, a theorem involving the complex roots of a polynomial is presented.

Rule

Complex Conjugate Root Theorem

Let $P (x)$ be a polynomial function whose coefficients are real. If a complex number $a + b i$ is a root of $P (x),$ then the root's complex conjugate, $a - b i,$ is also a root of $P (x) .$

$P (a + b i) = 0 \Rightarrow P (a - b i) = 0$

Proof

Let

p (x)

be a polynomial with real coefficients

c_{k} .

P (x) = k = 0 \sum n c_{k} x^{k}

Using the properties of conjugation, the conjugate of the polynomial evaluated at

a + b i

can be rewritten as expressed in the following table.

$\overline{P (a + b i)} = \overline{k = 0 \sum n c_{k} (a + b i)^{k}}$	The conjugate of the polynomial evaluated at $a + b i$
$\overline{P (a + b i)} = k = 0 \sum n \overline{c_{k} (a + b i)^{k}}$	The conjugate of a sum is the sum of the conjugates.
$\overline{P (a + b i)} = k = 0 \sum n \overline{c_{k}} \overline{(a + b i)^{k}}$	The conjugate of a product is the product of the conjugates.
$\overline{P (a + b i)} = k = 0 \sum n \overline{c_{k}} (\overline{a + b i})^{k}$	The conjugate of a power is the power of the conjugate.
$\overline{P (a + b i)} = k = 0 \sum n c_{k} (\overline{a + b i})^{k}$	The conjugate of a real number is itself.
$\overline{P (a + b i)} = k = 0 \sum n c_{k} (a - b i)^{k}$	The conjugate of $a + b i$ is $a - b i .$
$\overline{P (a + b i)} = P (a - b i)$	The polynomial evaluated at $a - b i$

It is given that

a + b i

is a root of the polynomial

p (x) .

P (a + b i) = 0

The conjugate of the real number

0

0 .

Therefore, the conjugate of the polynomial evaluated at

a + b i

0 .

\overline{P (a + b i)} = 0

This equation, along with the last equality in the table, shows that

a - b i,

the conjugate of

a + b i,

is also a root of the polynomial.

{\overline{P (a + b i)} = P (a - b i) \overline{P (a + b i)} = 0 ⇓ P (a - b i) = 0

Example

Integer and Complex Roots

As Zain progresses with their studies, they build up the courage to ask Izabella for extra help with polynomials that have complex roots.

They were given the numbers

x_{1} = 4

and

x_{2} = 3 + i

which are roots of the following equation. Consider that

b,

c,

and

d

are integers.

x^{3} + b x^{2} + c x + d = 0

Zain needs to answer the following questions.

a What are the remaining roots of the equation?

b Find the values of

b,

c,

and

d .

Hint

a Remember the Complex Conjugate Root Theorem.

b Multiply the binomials produced by the roots of the equation.

Solution

a By the Fundamental Theorem of Algebra, since the given polynomial equation is of degree

3,

the equation has three roots. Two were already given.

x_{1} x_{2} = 4 = 3 + i

It can be noted that the root

x_{2}

is a complex root. Since all the coefficients are integers, the conjugate of

x_{2}

is also a root of the equation, by the Complex Conjugate Root Theorem.

x_{2} = 3 + i \Rightarrow x_{3} = 3 - i

Since this is the third root, it is the only root remaining aside from the given two.

b The given polynomial equation is the multiplication of the binomials formed with each of its solutions.

x^{3} + b x^{2} + c x + d = (x - 4) (x - (3 + i)) (x - (3 - i))

This is true because of the Factor Theorem. Expanding the right-hand side expression will give the values of

b,

c,

and

d .

To do this multiplication, the binomials with complex numbers will be multiplied first.

(x - (3 + i)) (x - (3 - i))

Simplify

AssociativePropAdd

\AssociativePropAdd

((x - 3) - i)) ((x - 3) + i)

$(a + b i) (a - b i) = a^{2} + b^{2}$

(x - 3)^{2} + 1

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

x^{2} - 6 x + 9 + 1

AddTerms

Add terms

x^{2} - 6 x + 10

Now the resulting polynomial is multiplied by the third binomial.

(x^{2} - 6 x + 10) (x - 4)

Simplify

Distr

Distribute $(x - 4)$

x^{2} (x - 4) - 6 x (x - 4) + 10 (x - 4)

Distr

Distribute $x^{2}$

x^{3} - 4 x^{2} - 6 x (x - 4) + 10 (x - 4)

Distr

Distribute $6 x$

x^{3} - 4 x^{2} - 6 x^{2} + 24 x + 10 (x - 4)

Distr

Distribute $10$

x^{3} - 4 x^{2} - 6 x^{2} + 24 x + 10 x - 40

AddSubTerms

Add and subtract terms

x^{3} - 10 x^{2} + 34 x - 40

Looking at the equation, the desired values can be found.

b c d = - 10 = - 34 = - 40

Example

Only Complex Roots

After studying polynomial functions with complex roots, Zain feels confident enough to finish an assignment all on their own.

The final exercise of the assignment displays a polynomial with real coefficients and it has the following roots.

x_{1} = 5 + 3 i x_{2} = 2 - 6 i

Along with Zain, answer the following questions.

a What is the least degree of the polynomial?

b Assuming the polynomial's degree is the least possible one, find the remaining roots.

Hint

a What does the Fundamental Theorem of Algebra say?

b Remember the Complex Conjugate Root Theorem.

Solution

a By the Complex Conjugate Root Theorem, since the polynomial's coefficients are real, the conjugates of any complex root must also be a root. Two complex roots were given.

x_{1} = 5 + 3 i x_{2} = 2 - 6 i

Therefore, the polynomial has at least another two different roots. Then, by the Fundamental Theorem of Algebra, the least degree of the polynomial is

4 .

b As mentioned in Part A, the conjugates of the given roots are also roots of the polynomial by the Complex Conjugate Root Theorem.

5 + 3 i \Rightarrow 5 - 3 i 2 - 6 i \Rightarrow 2 + 6 i

Since the least degree of the polynomial is

4,

these are the remaining roots of the polynomial.

Example

Irrational and Complex Roots

Feeling so proud of Zain, Izabella feels really confident that all of her friends will do well when dealing with their math assignments. To complete this study session, she proposes a problem that has a polynomial with both irrational and complex roots.

Consider a polynomial function

g (x)

of degree

4

with integer coefficients has the following roots.

x_{1} x_{2} = 7 - 3 i = 4 + 5

Help answering the following.

a What are the remaining roots of the equation

g (x) = 0 ?

b Write the expression of

g (x)

with a leading coefficient of

1 .

Hint

a Recall the Complex Conjugate and the Irrational Conjugate Root Theorems

b Multiply the binomials produced by the roots of the function.

Solution

a Considering the Fundamental Theorem of Algebra, it is known that a polynomial of degree

4

has

4

solutions. Therefore, there are two remaining solutions. The given solutions can be used to find the remaining solutions. The solution

x_{1}

is a complex number.

x_{1} = 7 - 3 i

Also, it is stated that the polynomial has integer coefficients, which are real coefficients. If a complex number is a root of a polynomial with real coefficients, the Complex Conjugate Root Theorem states that the complex conjugate of such root must also be a solution.

x_{1} = 7 - 3 i \Rightarrow x_{3} = 7 + 3 i

On the other hand, the other solution is a real number. Since

5

is a prime number, its square root is an irrational number. Therefore, the given root is an irrational number.

x_{2} = 4 + 5

This time the Irrational Conjugate Root Theorem is useful. This theorem says that if an irrational number is a root of a polynomial with rational coefficients, the corresponding irrational conjugate is also a root of the polynomial.

x_{2} = 4 + 5 \Rightarrow x_{4} = 4 - 5

Consequently, the two remaining roots are the following.

x_{3} x_{4} = 7 + 3 i = 4 - 5

b The expression of the polynomial

g (x)

can be obtained by multiplying the binomials generated by each root.

Root	Binomial
$7 + 3 i$	$x - (7 + 3 i)$
$7 - 3 i$	$x - (7 - 3 i)$
$4 + 5$	$x - (4 + 5)$
$4 - 5$	$x - (4 - 5)$

Multiplying four binomials can be messy. To make the multiplication a little easier, the binomials will be multiplied in pairs. First, the binomials of complex roots will be multiplied.

(x - (7 + 3 i)) (x - (7 - 3 i))

Simplify

AssociativePropAdd

\AssociativePropAdd

((x - 7) - 3 i) ((x - 7) + 3 i)

$(a + b i) (a - b i) = a^{2} + b^{2}$

(x - 7)^{2} + 9

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

x^{2} - 14 x + 49 + 9

AddTerms

Add terms

x^{2} - 14 x + 58

Next, the binomials of the irrational roots will be multiplied.

(x - (4 + 5)) (x - (4 - 5))

Simplify

AssociativePropAdd

\AssociativePropAdd

((x - 4) - 5) ((x - 4) + 5)

ExpandDiffSquares

$(a + b) (a - b) = a^{2} - b^{2}$

(x - 4)^{2} - (5)^{2}

CalcPow

Calculate power

(x - 4)^{2} - 5

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

x^{2} - 8 x + 16 - 5

SubTerms

Subtract terms

x^{2} - 8 x + 11

The resulting quadratic polynomials can then be multiplied together.

(x^{2} - 14 x + 58) (x^{2} - 8 x + 11)

Simplify

Distr

Distribute $(x^{2} - 8 x + 11)$

x^{2} (x^{2} - 8 x + 11) - 14 x (x^{2} - 8 x + 11) + 58 (x^{2} - 8 x + 11)

Distr

Distribute $x^{2}$

x^{4} - 8 x^{3} + 11 x^{2} - 14 x (x^{2} - 8 x + 11) + 58 (x^{2} - 8 x + 11)

Distr

Distribute $14 x$

x^{4} - 8 x^{3} + 11 x^{2} - 14 x^{3} + 112 x^{2} - 154 x + 58 (x^{2} - 8 x + 11)

Distr

Distribute $58$

x^{4} - 8 x^{3} + 11 x^{2} - 14 x^{3} + 112 x^{2} - 154 x + 58 x^{2} - 464 x + 638

SubTerms

Subtract terms

x^{4} - 22 x^{3} + 11 x^{2} + 112 x^{2} + 58 x^{2} - 618 x + 638

AddTerms

Add terms

x^{4} - 22 x^{3} + 181 x^{2} - 618 x + 638

Finally, it is possible to write the expression for the function

g (x) .

Closure

About the Roots

Previously, it was explained how the coefficients of a polynomial function can give information about its roots, including the irrational and complex roots. This can be especially helpful considering that the graph of a function also gives some information. For example, consider a polynomial function

p (x) .

p (x) = (x - 2)^{2} (x - 5) ⇓ p (x) = x^{3} - 9 x^{2} + 24 x - 20

By the Factor Theorem, the roots of the polynomial are

2

and

5,

with a multiplicity of

2

and

1,

respectively. Now consider the graph of

p (x) .

Looking at the graph, it can be noted that the graph does not cross the

x -

axis when touching the root with multiplicity of

2,

but the graph crosses the axis on the next root, with multiplicity of

1 .

Now consider a different function. It will be written factored because the coefficients are big.

p_{2} (x) = (x - 2)^{3} (x - 5)^{4}

The roots of the polynomial are

2

and

5

again, but this time the roots have a multiplicity of

3

and

4

respectively. Consider the graph of

p_{2} (x) .

This time, the function's graph crosses the $x -$ axis on root $2$ and it does not on root $5 .$ From this, it is possible to draw two conclusions.

The graph of a function crosses the $x -$ axis on a root if the root's multiplicity is odd.
The graph of a function does not cross the $x -$ axis on a root if the root's multiplicity is even.

But what happens if the root is a complex number? Think about the following function.

p_{3} (x) = (x - 2)^{2} (x - 5) (x - 2 i) (x + 2 i)

The graph of this function is presented next.

Note that the function's roots of

2

and

5

are visible on the graph, while on the other hand, the complex roots do not shape the graph at all. If everything is considered, both the function's coefficients and its graph can be used to obtain information and help make better conclusions.

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Catch-Up and Review

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