Download Finding Rational Solutions

Math 102
4.3 "Polynomial Equations"
Objectives:
* Know the concept of multiplicity of roots.
* Solve polynomial equations using the rational root theorem and the factor theorem.
* Use Descartes’rule of signs to determine the possibilities for the nature of the solutions of a polynomial equation.
* Given the solutions and an indicated degree, write a polynomial equation using integral coe¢ cients with those solutions.
Preliminaries:
We need to recall the remainder theorem and the factor theorem
Remainder Theorem:
kIf a polynomial f (x) is divided by x
c, then the remainder is equal to f (c) :k
Factor Theorem:
kA polynomial f (x) has a factor x
c if and only if f (c) = 0:k
De…nition:
"Multiplicity"
If the factor x c occurs k times in the complete factorization of the polynomial P (x) ,
then c is called a root of P (x) = 0 with multiplicity k:
Example 1: (Finding the multiplicity of polynomials)
State the degree of each polynomial equation. Find all of the real roots of each equation, state the multiplicity when it is
greater than one.
a) x2
b) x5
10x + 25 = 0
9x3 = 0
c) (x
2
3
4) (x + 5) = 0
Finding Rational Solutions
Rational Root theorem.
If f (x) = an xn + an 1 xn
1
+ an 2 xn 2 +
+ a1 x + a0 is a polynomial function with integral coe¢ cients
p
(an 6= 0 and a0 6= 0) and
(in lowest term) is a rational zero of f (x) ; then p is a factor of the constant
q
term a0 and q is a factor of the leading coe¢ cient an :
Page: 1
Notes by Bibiana Lopez
College Algebra by Kaufmann and Schwitters
4.3
Example 2: (Finding rational solutions)
Use the rational root theorem and the factor theorem to help solve each equation.
a) x3
2x2
c) 2x4 + 3x3
11x + 12 = 0
11x2
9x + 15 = 0
b) x3
x2 + x
d) x4 + x3
3x2
1=0
17x
30 = 0
Note: If a polynomial equation with real coe¢ cients has any nonreal complex solutions, they must occur in conjugate pairs.
Page: 2
Notes by Bibiana Lopez
College Algebra by Kaufmann and Schwitters
Descartes’Rule of Signs:
Let an xn + an 1 xn 1 + an
2x
n 2
+
4.3
+ a1 x + a0 = 0 be a polynomial equation with real coe¢ cients.
1: The number of positive real solutions of the given equation is either equal to the number of variations in sign
of the polynomial or less than that number of variations by a positive even number.
2: The number of negative real solutions of the given equation is either equal to the number of variations in sign
n
of the polynomial an ( x) + an
n 1
1
( x)
+ an
n 2
2
( x)
+
+ a1 ( x) + a0 or less than that number of
variations by a positive even integer.
Example 3: (Using the Descartes’rule of signs)
Use Descartes’ rule of signs to determine the possibilities for the nature of the solutions for each of the equations. Do not
solve the equations.
a) 3x3
2x2 + 6x + 5 = 0
b) 2x6 + 3x4
2x2
1=0
Example 4: (Building polynomials with prescribed zeros)
Find a polynomial equation with integral coe¢ cients that has the given numbers as solutions and indicated degree.
a) 2; 4;
c)
2; 1
3; degree 3
4i; degree 3
b) 1 of multiplicity 5
d) 1
Page: 3
i; 2i; degree 4
Notes by Bibiana Lopez