6-5 Theorems About Roots of Polynomial

Equations

Objectives

The Rational Root Theorem

Irrational Root Theorem & Imaginary Root Theorem

Find the rational roots of 3x3 – x2 – 15x + 5.

Step 1: List the possible rational roots.

The leading coefficient is 3. The constant term is 5. By the Rational

Root Theorem, the only possible rational roots of the equation have

the form .factors of 5factors of 3

The factors of 5 are ±5 and ±1 and ±5. The factors of 3 are

±3 and ±1. The only possible rational roots are ±5, ± , ±1, ± .53

13

Finding Rational Roots

(continued)

Step 2: Test each possible rational root.

1: 3(1)3 – (1)2 – 15(1) + 5 = –8 ≠ 0–1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 ≠ 0

5: 3(5)3 – (5)2 – 15(5) + 5 = 280 ≠ 0–5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 ≠ 0

: 3 3 – 2 – 15 + 5 = –8.8 ≠ 0

: 3 3 – 2 – 15 + 5 = –13.3 ≠ 0

5353

( )53

( )53

( )53

– ( )53

– ( )53

– ( )53

–

: 3 3 – 2 – 15 + 5 = 0 So is a root.

: 3 3 – 2 – 15 + 5 = 9.7 ≠ 013

( )13

– ( )13

– ( )13

–

13

( )13

( )13

( )13

13

–

The only rational root of 3x3 – x2 – 15x + 5 = 0 is .13

Continued

Find the roots of 5x3 – 24x2 + 41x – 20 = 0.

Step 1: List the possible rational roots.

The leading coefficient is 5. The constant term is 20. By the

Rational Root Theorem, the only possible roots of the equation

have the form .factors of – 20factors of 5

The factors of –20 are ±1 and ±20, ±2 and ±10, and ±4 and ±5.

The only factors of 5 are ±1 and ±5. The only possible rational roots

are ± , ± , ± , ±1, ±2, ±4, ±5, ±10, and ±20.15

25

45

Using the Rational Root Theorem

(continued)

Step 2: Test each possible rational root until you find a root.

Step 3: Use synthetic division with the root you found in Step 2to find the quotient.

5 –24 41 –20 4 –16 205 –20 25 0

5x2 – 20x + 25  Remainder

45

45

Test : 5 3 – 24 2 ± 41 – 20 = –12.72 ≠ 0

Test – : 5 3 – 24 2 ± 41 – 2 = –29.2 ≠ 0

Test : 5 3 – 24 2 ± 41 – 20 = –7.12 ≠ 0

Test – : 5 3 – 24 2 ± 41 – 20 = –40.56 ≠ 0

Test : 5 3 – 24 2 ± 41 – 20 = 0 So is a root.

( )15

15

15

25

25

45

( )25

( )45

( )15

( )25

( )45

( )45

( )25

( )15

(– )15

(– )15

(– )15

(– )25

(– )25

(– )25

Continued

(continued)

Step 4: Find the roots of 5x2 – 20x + 25 = 0.

5x2 – 20x + 25 = 0 5(x2 – 4x + 5) = 0 Factor out the GCF, 5. x2 – 4x + 5 = 0

x = Quadratic Formula = Substitute 1 for a, –4 for b,

and 5 for c.

–b ± b2 – 4ac2a

–(–4) ± (–4)2 – 4(1)(5)2(1)

= Use order of operations.

=    –1 = i.

= 2 ± i Simplify.

4 ± –42

4 ± 2i 2

The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i.45

Continued

By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate 2 + 5 is also a root.

A polynomial equation with rational coefficients has the roots

2 – 5 and 7 . Find two additional roots.

If 7 is a root, then its conjugate – 7 also is a root.

Finding Irrational Roots

A polynomial equation and real coefficients has the roots 2 + 9i with 7i. Find two additional roots.

By the Imaginary Root Theorem, if 2 + 9i is a root, then its complex conjugate 2 – 9i also is a root.

If 7i is a root, then its complex conjugate –7i also is a root.

Finding Imaginary Roots

Find a third degree polynomial with rational coefficients that has roots –2, and 2 – i.

Step 1: Find the other root using the Imaginary Root Theorem.

Since 2 – i is a root, then its complex conjugate 2 + i is a root.

Step 2: Write the factored form of the polynomial using the Factor Theorem.

(x + 2)(x – (2 – i))(x – (2 + i))

Writing a Polynomial Equation from Its Roots

(continued)

Step 3: Multiply the factors.

(x + 2)[x2 – x(2 – i) – x(2 + i) + (2 – i)(2 + i)] Multiply (x – (2 – i))(x – (2 + i)).

(x + 2)(x2 – 2x + ix – 2x – ix + 4 – i 2) Simplify. (x + 2)(x2 – 2x – 2x + 4 + 1) (x + 2)(x2 – 4x + 5) Multiply. x3 – 2x2 – 3x + 10

A third-degree polynomial equation with rational coefficients and roots –2 and 2 – i is x3 – 2x2 – 3x + 10 = 0.

Continued

Homework

Pg 339 #1, 7, 13, 14, 15, 19, 20