SOLVING POLYNOMIAL EQUATIONS

Unit Overview

This unit is about solving polynomial equations by using variable substitution and by using the

rational root theorem. The unit concludes with verifying and proving polynomial identities.

Solving Polynomial Equations by Using Variable Substitution and

Factoring

One of the most important theorems in mathematics is the Fundamental Theorem of Algebra.

The polynomial, 4 29 14 0x x , is a polynomial of degree 4 so it should have 4 roots. The roots

may be real numbers, complex numbers, and/or the same number. These roots will be found in the

following example.

If a polynomial has a degree of more than 2, variable substitution can be used find the roots.

Example #1: Solve for x: 4 29 14 0x x

Step #1: Let u represent 2x and substitute u for

2x .

2 2 2 4 2 2

2

( ) 9( ) 14 0 express as ( )

9 14 0

x x x x

u u

Fundamental Theorem of Algebra

If P(x) is a polynomial of degree n, where n > 0, then P(x)

has exactly n roots including multiple and complex roots.

Step #2: Factor and solve.

( 7)( 2) 0

7 0 2 0

7 2

u u

u u

u u

Step #3: Remember that u represents 2x and replace u with 2x .

2 27 2

7 2

x x

x x

Example #2: Solve for z: 5 328 27 0z z z

5 328 27 0z z z factor out a GCF of z

4 2( 28 27) 0z z z substitute u for 2z

2( 28 27) 0z u u factor

z(u – 27)(u – 1) = 0 solve for u

z = 0 u – 27 = 0 u – 1 = 0

u = 27 u = 1 replace u with 2z

z = 0 2 27z 2 1z solve for z

z = 0, z = 27 , z = 1 1

Therefore, the roots of the polynomial 5 328 27 0z z z are 0, 1, –1, 27 , and

27 .

Finding the Zeros (04:08)

Stop! Go to Questions #1-5 about this section, then return to continue on to the next

section.

Rational Root Theorem

Example #1: List all the possible rational roots of the function

P(x) = 2x3 – 11x2 + 12x + 9.

According to the Rational Root Theorem, p

q is a root of 3 22 911 12x x x .

If p is a factor of the constant term, 9, and q is a factor of the leading coefficient, 2.

Step #1: Make an organized list of all factors of 9 and 2.

factors of 9: ±1, ±3, ±9

factors of 2: ±1, ±2

Step #2: Form all quotients that have factors of 9 in the numerator and factors of 2 in

the denominator.

These numbers represent all the possible rational zeros of the function.

Factors of 9 over 1

1

1 ,

3

1 ,

9

1

Factors of 9 over 2

1

2 ,

3

2 ,

9

2

Rational Root Theorem

Let P be a polynomial function with integer coefficients in standard form.

If p

q (in lowest terms) is a root of P(x) = 0, then

p is a factor of the constant term of P and

q is a factor of the leading coefficient of P

Let's practice determining the possible rational zeros of the polynomial, P(x) = 2x3 – x2 + 2x + 5.

Name the constant term. P(x) = 2x3 – x2 + 2x + 5

“Click here” to check the answer.

5

List the factors of the constant term. P(x) = 2x3 – x2 + 2x + 5

“Click here” to check the answer.

±1, ±5

Name the leading coefficient. P(x) = 2x3 – x2 + 2x + 5

“Click here” to check the answer.

2

List the factors of the leading coefficient. P(x) = 2x3 – x2 + 2x + 5

“Click here” to check the answer.

±1, ±2

List the possible rational roots. P(x) = 2x3 – x2 + 2x + 5

“Click here” to check the answer.

1 51, 5, ,

2 2

The Zero Property (06:50)

Stop! Go to Questions #6-7 about this section, then return to continue on to the next

section.

Solving Polynomial Equations Using the Rational Root Theorem

Note: The following examples are shown using a graphing calculator. The activities can be also

be done using an online graphing program. Click here to navigate to the online grapher.

Example #1: Find all of the rational roots of 3 28 10 11 2 0x x x .

According to the Rational Root Theorem, p

q is a root of 3 28 10 11 2 0x x x , if p is

a factor of the constant term, 2, and q is a factor of the leading coefficient, 8.

Step #1: Make an organized list of all factors of 2 and 8.

factors of 2: 1 , 2

factors of 8: 1 , 2 , 4 , 8

Step #2: Form all quotients that have factors of 2 in the numerator and factors of 8 in

the denominator.

*Some rational numbers are repeated, such as 1

4 and

2

8 .

Step #3: Graph the function on your graphing calculator. Use the button that is

marked XTθn for the variable x. Also, to raise a number to the third power, use the

button that is located above the key. The standard viewing screen is the

figure on the left. ZOOMIN ENTER is the figure on the right.

One root appears to be –2. Test whether P(–2) = 0 by using synthetic division.

1

1 ,

1

2 ,

1

4 ,

1

8

2

1 ,

2

2 ,

2

4 ,

2

8

–2 8 10 –11 2

–16 12 –2 remainder

8 –6 1 0

Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the

polynomial.

Step #4: From the graph there appears to be two real zeros between 0 and 1. Zoom in

( ZOOM 2) and use the trace feature to check. Press TRACE and move the

cursor around with the arrow keys until the y-value on your screen gets very

very close to 0. ZOOM , 2: Zoom In, ENTER if necessary to locate this point.

It looks like there are two real roots, one at 1

4 and the other at

1

2. Test whether

10

4P

and 1

02

P

by using synthetic division.

1

4 8 10 –11 2

1

2 8 10 –11 2

2 3 –2 4 7 –2

8 12 –8 0 8 14 –4 0

Since 1

04

P

and 1

02

P

, 1

4 and

1

2 are also roots of the polynomial.

Therefore, there are three roots –2, 1

4, and

1

2.

*When a polynomial function P(x) touches but does not cross the x-axis at (r, 0), then P(x) = 0

has a double root.

For example: P(x) = (x – 3)(x – 3)(x+1) or 3 2( ) 5 3 9P x x x x touches the x-axis at

(3, 0) but does not cross the x-axis there, as shown in the graphs below.

Therefore, 3 is a double root of P(x) = 0.

Example #2: Find all of the rational roots of 3 23 2 12 8 0x x x .

According to the Rational Root Theorem, p

q is a root of 3 23 2 12 8 0x x x ,

if p is a factor of the constant term, 8, and q is a factor of the leading coefficient, 3.

Step #1: Make an organized list of all factors of 8 and 3.

factors of 8: 1 , 2 , 4 , 8

factors of 3: 1 , 3

Step #2: Form all quotients that have factors of 8 in the numerator and factors of 3 in

the denominator.

1 1,

1 3

2 2,

1 3

4 4,

1 3

8 8,

1 3

Step #3: Graph the function on your graphing calculator. Use the button that is

marked XTθn for the variable x. Also, to raise a number to the third power, use the

button that is located above the key. The standard viewing screen is the

figure below on the left. ZOOMIN ENTER is the figure on the right.

There appears to be two real roots at –2 and +2. Test either P(–2) =0 or P(2)=0 by

using synthetic division. We’ll try P(–2)=0.

–2 3 –2 –12 8

–6 16 –8 remainder

3 –8 4 0

Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the

polynomial.

Now we’ll try P(2) = 0.

2 3 –2 –12 8

6 8 –8 remainder

3 4 – 4 0

Since the remainder is 0, this means that P(2) = 0 is true and that 2 is a root of the

polynomial.

Step #4: From the graph there appears to be a real zero between 0 and 1. Zoom in

( ZOOM 2) and use the trace feature to check. Press TRACE and move the

cursor around with the arrow keys until the y-value on your screen gets very

very close to 0. ZOOM , 2: Zoom In, ENTER if necessary to locate this point.

It looks like there is one real root, 2

3. Test whether

20

3P

and by using synthetic

division.

2

3 3 –2 –12 8

2 0 –8 remainder

3 0 –12 0

Since 2

03

P

, 2

3 is a root of the polynomial.

Therefore there are 3 roots –2, 2, and 2

3.

Some polynomials do not have all rational roots. We can still find all the zeros of a function

using some of the strategies already learned.

Example #3: Find all of the roots of 3 24 6 4 0x x x .

According to the Rational Root Theorem, p

q is a root of 3 24 6 4 0x x x ,

if p is a factor of the constant term, –4, and q is a factor of the leading coefficient, 1.

Step #1: Make an organized list of all factors of 4 and 1.

factors of –4: 1 , 2 , 4

factors of 1: 1

Step #2: Form all quotients that have factors of 4 in the numerator and factors of 1 in

the denominator.

Step #3: Graph the function on a graphing calculator. Use the button that is marked

XTθn for the variable x. Also, to raise a number to the third power, use the

button that is located above the key. The standard viewing screen is the figure

below on the left. ZOOMIN ENTER is the figure on the right.

1

1

2

1

4

1

The graph crosses the x-axis once. Therefore there is only one real root. This root

appears to be 2. Test whether P(2) = 0 by using synthetic division.

2 1 –4 +6 –4

2 –4 4 remainder

1 -2 2 0

Since the remainder is 0, this means that P(2) = 0 is true and that 2 is a root of the

polynomial.

The resulting numbers 1, –2 and 2 are the coefficients of the quotient. The quotient

will start with an exponent that is one less than the dividend. 21 2 2x x

This quotient is called a depressed polynomial. Since the depressed polynomial is

quadratic, use the Quadratic Formula to find the other two roots.

2 2 2 0x x a = 1 b = –2 c = 2

2

2

4Substitute 1, 2, 2

2

( 2) ( 2) 4(1)(2)

2(1)

2 4 8

2

2 4

2

2 ( 1)4

2

2 2 ( 1)

2

2 2

2

1

1

1

1 or 1

b b acx a b c

a

x

x

x

x

x

ix

ix

x i

x i x i

Therefore the roots of the equation are 2, 1 + i and 1 – i.

Stop! Go to Questions #8-12 about this section, then return to continue on to the next

section.

Proving Polynomial Identities

A polynomial identity is a statement of equality between two polynomials that is true for all

values of the variable(s) for which the expression is defined.

Example #1: Check the polynomial identity 2 2 2( ) 2a b a ab b by substituting a = 4

and b = 2.

2 2 2

2

(4 2) (4 +2(4)(2) + 2 ) Substitute 4 and 2 on both sides of the equation.

6 (16 16 4) Simplify

36 = 36

a b

Example #2: Prove the polynomial identity 2 2 2( ) 2a b a ab b by using algebraic

operations.

2

2 2

2 2

( )

( )( ) Definition of Squared

( ) ( ) Distributive Property

Distributive Property

2 Combine like terms

a b

a b a b

a a b b a b

a ab ab b

a ab b

Example #3: Check the polynomial identity 3 3 2 2( )( )a b a b a ab b by

substituting a = 4 and b = 2.

3 3 2 24 2 (4 2)(4 +4(2) + 2 ) Substitute 4 and 2 on both sides of the equation.

64 8 2(16 8 4) Simplify

56 = 2(28)

56 = 56

a b

Example #4: Prove the polynomial identity 3 3 2 2( )( )a b a b a ab b using algebraic

expressions.

2 2

2 2 2 2

3 2 2 2 2 3

3 3

( )( )

( ) ( ) Distributive Property

Distributive Property

Collect like terms

x y x xy y

x x xy y y x xy y

x x y xy x y xy y

x y

Stop! Go to Questions #13-33 to complete this unit.