6-5 theorems about roots of polynomial equations
TRANSCRIPT
![Page 1: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/1.jpg)
6-5 Theorems About Roots of Polynomial
Equations
![Page 2: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/2.jpg)
Objectives
The Rational Root Theorem
Irrational Root Theorem & Imaginary Root Theorem
![Page 3: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/3.jpg)
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
![Page 4: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/4.jpg)
(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
![Page 5: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/5.jpg)
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
![Page 6: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/6.jpg)
(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
![Page 7: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/7.jpg)
(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
![Page 8: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/8.jpg)
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
![Page 9: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/9.jpg)
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
![Page 10: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/10.jpg)
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
![Page 11: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/11.jpg)
(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
![Page 12: 6-5 Theorems About Roots of Polynomial Equations](https://reader036.vdocuments.us/reader036/viewer/2022081813/56649ee55503460f94bf3ca4/html5/thumbnails/12.jpg)
Homework
Pg 339 #1, 7, 13, 14, 15, 19, 20