1. 6-5 Theorems About Roots of Polynomial Equations

2. Objectives
The Rational Root Theorem Irrational Root Theorem & Imaginary Root Theorem

3. Finding Rational Roots
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 5
factors 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, ± . 5 3 1

4. Continued ( ) ( ) ( ) ( ) (continued)
Step 2: Test each possible rational root.
5: 3(5)3 – (5)2 – 15(5) + 5 = 280 ≠ 0
–5: 3(–5)3 – (–5)2 – 15(–5) + 5 = –320 ≠ 0
: – – = –8.8 ≠ 0
: – – = –13.3 ≠ 0 5 3
( ) –
( )
1: 3(1)3 – (1)2 – 15(1) + 5 = –8 ≠ 0
–1: 3(–1)3 – (–1)2 – 15(–1) + 5 = 16 ≠ 0
: – – = 0 So is a root.
: – – = 9.7 ≠ 0 1 3
( ) –
( )
The only rational root of 3x3 – x2 – 15x + 5 = 0 is . 1 3

5. Using the Rational Root Theorem
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 – 20
factors 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. 1 5 2 4

6. Continued ( ) (– ) (continued)
Step 2: Test each possible rational root until you find a root. 4 5
Test : – ± – 20 = –12.72 ≠ 0
Test – : – ± – 2 = –29.2 ≠ 0
Test : – ± – 20 = –7.12 ≠ 0
Test – : – ± – 20 = –40.56 ≠ 0
Test : – ± – 20 = 0 So is a root.
( ) 1 2
(– )
5 –24 41 –20
4 –
5 –
5x2 – 20x   Remainder 4 5
Step 3: Use synthetic division with the
root you found in Step 2
to find the quotient.

7. Continued (continued) Step 4: Find the roots of 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 – 4ac 2a
–(–4) ± (–4)2 – 4(1)(5)
2(1)
= Use order of operations.
=    –1 = i.
= 2 ± i Simplify.
4 ± –4 2
4 ± 2i
The roots of 5x3 – 24x2 + 41x – 20 = 0 are , 2 + i, and 2 – i. 4 5

8. Finding Irrational Roots
A polynomial equation with rational coefficients has the roots 2 – 5 and Find two additional roots.
By the Irrational Root Theorem, if 2 – 5 is a root, then its conjugate
is also a root.
If is a root, then its conjugate – also is a root.

9. Finding Imaginary 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.

10. Writing a Polynomial Equation from Its 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))

11. Continued (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 )
(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.

12. Homework
Pg 339 #1, 7, 13, 14, 15, 19, 20