1
Algebra II – Chapter 6 Day #5
Topic: Theorems About Roots of Polynomial Equations
Today, we will focus our attention on theorems that will help us solve polynomial equations.
Standards/Goals:
N.CN.7.: I can solve quadratic equations with real coefficients that have complex solutions.
I can use the Rational Root Theorem to solve equations.
I can use the Conjugate Root Theorem to solve equations.
I can use the Descartes’ Rule of Signs to determine the number of roots of a polynomial equation.
I can use synthetic division to divide two polynomials.
We want to first look at the Rational Root Theorem
Rational Root TheoremThis theorem says that any rational zero of a function must be in the form of ,
where ‘p’ is a factor of the constant term and
‘q’ is a factor of the leading coefficient.
This theorem will NOT help us find irrational or imaginary zeros; to find these we must use the reduced polynomial and quadratic formula.
EXAMPLES: List all of the potential rational zeros of the following polynomials. Then use polynomial division and the quadratic formula to identify the actual zeros.
- f(x) =
- p(x) =
We also want to familiarize ourselves with the CONJUGATE ROOTS THEOREM.
Conjugate Roots TheoremThis theorem basically says that when we consider a function that is a polynomial with only real coefficients, if a + bi is a zero of ‘f’, then so is the complex number a – bi. In terms of the linear factors of ‘f’, this means that x – (a + bi) is a factor of f, then so is x – (a – bi).
EXAMPLES:
#1. A quartic polynomial P(x) has rational coefficients. If 2 – i and are roots of P(x) = 0, what are the other roots?
#2. A cubic polynomial P(x) has rational coefficients. If 2 + 3i and 2/3 are two roots of P(x) = 0, what is one additional root?
#3. What is a quartic polynomial function P(x) with rational coefficients so that P(x) = 0 has roots 1, 3, and 1 + 2i?
#4. A cubic polynomial P(x) with rational coefficients so that P(x) = 0 has roots -3 and i?
EXAMPLES: Use all available methods (in particular, the Conjugate Roots Theorem, if applicable) to factor the following polynomial equations completely, making use of the given zero).
#1. f(x) = ; -7i is a zero
#2. p(x) = ; 2i is a zero.
We will also use the Descartes’ Rule of Signs to determine how many roots there are in a polynomial.
Descartes’ Rule of SignsLet P(x) be a polynomial with real coefficients written in standard form.
The number of positive real roots of P(x) = 0 is either equal to the number of sign changes between consecutive coefficients of P(x) or is less than that by an even number.
The number of negative real roots of P(x) = 0 is either equal to the number of sign changes between consecutive coefficients of P(-x) or is less than that by an even number.
In both cases, count multiple roots according to their multiplicity.
EXAMPLES:
#1. What does Descartes’ Rule of Signs tell you about the real roots of ?
#2. What does Descartes Rule of Signs tell you about the real roots of ?
HOMEWORK – Chapter 6 Day #5
Use the Rational Root Theorem to list all possible rational roots for each equation.
1. x3 + 5x2 2x 15 = 0
2. 36x3 + 144x2x 4 = 0
Use the Rational Root Theorem to list all possible rational roots for each equation. Then find any actual rational roots.
3. 2x3 + 5x2 + 4x + 1 = 0
4. 12x4 + 14x3 5x2 14x 4 = 0
A polynomial function P(x) with rational coefficients has the given roots. Find two additional roots of P(x) = 0.
5. 2 + 3i and
6. and
7. 4i and 6 i
Write a polynomial function with rational coefficients so that P(x) = 0 has the given roots.
8. 4 and 6
9. 5 and 1
10. 3i and
11. 2 + iand
12. 5 and 3i
EXAMPLES: Use all available methods (in particular, the Conjugate Roots Theorem, if applicable) to factor the following polynomial equations completely, making use of the given zero).
13. ; -5i is a zero.
14.
What does Descartes’ Rule of Signs say about the number of positive real roots and negative real roots for each polynomial function?
15. P(x) = 3x3 + x2 8x 12
16. P(x) = 2x4x3 3x + 7
17. P(x) = 4x5x4x3 + 6x2 5