1. 5-5 Theorems About Roots of Polynomial Equations
Solve equations by using the Rational Root Theorem and the Conjugate Root Theorem.
5-5 Theorems About Roots of Polynomial Equations

2. Rational Root Theorem Let 𝑃 𝑥 = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 +…+ 𝑎 1 𝑥+ 𝑎 0
Integer roots must be factors of 𝑎 0
Rational roots must have reduced form 𝑝 𝑞 where p is a factor of 𝑎 0 and q is a factor of 𝑎 𝑛 .
Ex: 21 𝑥 2 +29𝑥+10=0
Factors of constant: ±1, ±2, ±5, 𝑎𝑛𝑑±10
Factors of lead coefficient: ±1, ±3, ±7, 𝑎𝑛𝑑±21
All possible roots are ± 𝑝 𝑞
Roots are − 2 3 𝑎𝑛𝑑 − 5 7
𝑥 𝑥 =0

3. Finding a Rational Root
Find the rational roots of 2𝑥 3 − 𝑥 2 +2𝑥+5=0
Constant factors: ±1,±5
Lead coefficient factors: ±1,±2
Possible factors:±1,± 1 2 ,±5,± 5 2
Looking at a table of possible factors,
We see that -1 is the only rational root (the only one that causes P(x) to equal 0)

4. Practice List all possible rational roots of 3𝑥 3 + 7𝑥 2 +6𝑥−8=0
±1,± 1 3 ,±2,± 2 3 ,±4,± 4 3 ,±8,± 8 3

5. Using Rational Root Theorem
Find rational roots of 15𝑥 3 −32 𝑥 2 +3𝑥+2=0
Possible: ±1,± 1 3 ,± 1 5 ,± 1 15 ,±2,± 2 3 ,± 2 5 ,± 2 15
Test each possible rational root until you find one.
1 3 is a root
Factor using synthetic division.
− −9 15 −27 − −2 0
𝑥− 𝑥 2 −27𝑥−6 =0
Factor the rest to find other roots

6. Continued Factor 𝑥− 1 3 15 𝑥 2 −27𝑥−6 =0 𝑥− 1 3 (3) 5 𝑥 2 −9𝑥−2 =0
𝑥− 1 3 (3) 5 𝑥 2 −9𝑥−2 =0
𝑥− 1 3 (3) 5𝑥+1 (𝑥−2)=0
Zero-product property: rational roots are 1 3 ,− 1 5 ,2

7. Conjugate Root Theorem
If a complex number (a+bi) or an irrational number (a + b ) is a root of a polynomial equation with rational coefficients, then so is its conjugate.
Conjugate of 𝑎+𝑏𝑖 is 𝑎−𝑏𝑖
Conjugate of 𝑎+ 𝑏 is 𝑎− 𝑏

8. Identifying Roots
A quartic polynomial P(x) has rational coefficients. If and 1+𝑖 are roots of P(x) = 0, what are the two other roots?
Using the Conjugate Root Theorem, we know that the conjugates are also roots.
− 2 and 1−𝑖

9. Constructing a Polynomial
What is a third-degree polynomial function P(x) with rational coefficients such that 𝑃 𝑥 =0 has roots -4 and 2i ?
Since 2i is a root, so is – 2i.
𝑃 𝑥 =(𝑥+4)(𝑥−2𝑖)(𝑥+2𝑖)
Multiply factors: 𝑃 𝑥 = 𝑥+4 𝑥 2 +4
𝑃 𝑥 = 𝑥 3 +4 𝑥 2 +4𝑥+16

10. Assignment
p.316 #15,16, Odds 17-29,39