Download Section 4-4 The Rational Root Theorem Page 229

Section 44-4
The Rational Root Theorem
Page 229
Objectives:
1. Student should be able to identify all
possible rational roots of a polynomial
equation by using the Rational Root
Theorem.
2. Student should be able to determine the
number of positive and negative real
roots a polynomial function has.
Rational Root Theorem
• The rational root theorem, or Rational Root Test
can be used to do a quick check whether a
polynomial has rational roots.
• The theorem states: If a polynomial with
coefficients an, an-1...a0 : anxn+a1xn-1...a0=0
has rational roots p/q, where p and q are integers,
then p is a factor of ±a0 and q is a factor of ±an or
using "|" to mean is a factor of, p|a0 and q|an.
• According to the Rational Root Theorem, the roots
can be plus or minus, although p|a0 implies that p
can be plus or minus (because, for instance, both 2
and -2 are factors of 4, so writing ± serves simply
as a reminder of this.).
Book Example 1
List the possible rational roots of x3 - x2 - 10x - 8 = 0. Then
determine the rational roots.
According to the Rational Root Theorem, if is a root of the equation,
then p is a factor of 8 and q is a factor of 1.
possible values of p: ±1, ±2, ±4, ±8
possible values of q: ±1
possible rational roots, : ±1, ±2, ±4, ±8
You can use a graphing utility to narrow down
the possibilities. You know that all possible
rational roots fall in the domain -8 ≤ x ≤ 8.
So, set your x-axis viewing window at [-9, 9].
Graph the related function f(x) = x3 - x2 - 10x - 8.
A zero appears to occur at -2. Use synthetic
division to check that -2 is a zero.
Thus, x3 - x2 - 10x - 8 = (x + 2)(x2 - 3x - 4). Factoring x2 - 3x - 4 yields (x
- 4)(x + 1). The roots of x3 - x2 - 10x - 8 = 0 are -2, -1, and 4.
Example 1
• We should not ashamed to give trivial examples!
Consider: x2-3x+2=0
If this has rational roots p/q (which we know it has), then p is a factor
of 2 and q is a factor of 1.
So the roots are some combination of the factors in the fraction
below:
We must use all the numbers, that is we need to use both 1 and 2 to
make possible roots. Where all the roots are rational, we cannot use
just the "1" as a numerator. We need to use the 2 also. So 1 and 1
won't do as numerators, although 2 and 2 are possible (because the
1 is always automatically included in any integer.)
We test each of the roots in the original equation and, in this case,
we find x=1/1 or 2/1, which are 1 and 2, of course.
Example 1
Example 2
• x2+2x-15
If this has rational roots, then they are factors of:
• We must use all the numbers, so we cannot take 1 and 3
as possible numerators, because we must include 5
somewhere. So if 1 is one numerator, then 15 must be
the other (because there are 2 roots and we need to use
up all the numbers.)
By knowing or trying, we find -5 and 3 are the required
roots.
Example 3
• 2x2-3x+5
If this has rational roots, then they are factors of:
• We can try the combinations 1/2, 1/1, 5/1, 5/2 both plus and minus
(but Descartes Rule of Signs tells us all the roots are positive if they
are real, so we need to check positive roots only.), but we fail to find
any roots. The roots of this are approximately:
x0=0.75+1.3919i
x1=0.75 – 1.3919i
And these aren't rational numbers. So, as expected, when the roots
aren't rational, they do not conform with the rational root theorem.
Example 3
Example 4
• 21x2+11x-2
If this has rational roots, then they are factors of:
•
So, if there are rational roots, they are among: 1/1, 1/3,
1/7, 2/1, 2/3, 2/7
Because we need to use up all the factors, so for
example, if one root is |1/3|, then the other is |2/7|, etc..
In fact they are 1/7 and -2/3
Example 5
• 6x3-73x2-86x+273
If this has rational roots, then they are factors of:
•
Because the roots p/q are such that p and q have no
common factors, then ±3/3 isn't an option for a root. Also
we need to use up all the numbers, so if 3 is a
numerator, the other two numerators must be 7 and 13
or 1 and 91. Either way the numbers are used up.
In fact, 3/2, -7/3 and 13 satisfy the equation
Example 5
Example 6
• 18x3-57x2+53x-12
p|12=1.2.2.3
It is always useful to eliminate ±1, and these aren't roots.
Possible p's:
1.1.12 [1]
1.2.6 [2]
1.4.3 [3]
2.2.3 [4]
Possible q's:
1.1.18 [a]
1.3.6 [b]
1.2.9 [c]
2.3.3 [d]
q|18=1.2.3.3
In the p's, 3 can be the numerator of 1 or 2 only, so possible roots
are: ±3,±3/2.
Also in the p's 2 can be the numerator of 1 or 3 only
The roots are: 1/3, 4/3, 3/2.
Descartes Rule of Signs
• Suppose we have an equation:
f(x)= +x4-3x3-15x2+19x+30 [1]
I have put a + sign before the highest power above for clarification.
This equation has the signs: + - - ++
It has 2 sign changes, so the equation has 2 or (multiples of 2 less)
positive real roots. That is, it has 2 or 0 positive real roots.
We can check the number of negative real roots by noting:
f(-x)=+x4 -3x3-15x2-19x+30
The signs are:
+---+
That is, there are 2 sign changes, so the equation has 2 (or
multiples of 2 less) negative real roots. Or, 2 or 0 negative real roots.
Looking at the original equation [1], we can note the signs for the
negative case by reversing the sign of all the odd powers to get the
same answer.
Descartes Rule of Signs
• Consider:
x3-2x2-13x-10
The signs are:
+ - - - (the first plus comes from the x3term)
There is one sign change, so there is one real positive
root. When the indicated number of roots is an odd
number, we can be sure that there is one real root of that
kind, in this case, a positive reeal root.
For the negative roots, either reversing the signs of the
odd powers, or computing f(-x):
--+There are 2 sign changes, so the equation has 2 or 0
negative roots.
The roots are -1, -2, and 5
Descartes Rule of Signs
• x4-37x2+24x+180
The signs are:
+-++
So there 2 or 0 positive real roots
For the negative roots, the signs are:
+--+
There are 2 sign changes, so there are 2
or 0 negative real roots
The roots are: -2, -6, 3, 5
Descartes Rule of Signs
x5-4x4-3x3+28x2+60x+15
The signs for the positive root are:
+--+++
There are 2 sign changes, so the equation has 2 or 0
real positive roots.
For the negative roots:
--++-+
There are also 3 sign changes and therefore 3 or 1 real
negative roots.
The roots of this equation are:
Root0= 3.63998984 – 1.92654117i
Root1= 3.63998984 +1.92654117i
Root2= – 1.49501719 +0.90281543i
Root3= – 0.28994531
Root4= – 1.49501719 – 0.90281543i
There is one negative real root.
MANUFACTURING The volume of a box of pasta must be 120 cubic inches. The box
is
7 inches longer than it is wide and six times longer than it is tall. Find the dimensions
of the box.
The formula for the volume of a rectangular prism is V = lwh, where l is the length, w is the
width, and h is the height. From the given information, l = 6h and w = 6h - 7.
V
= lwh
120 = (6h)(6h - 7)(h)
120 = 36h3 - 42h2
0
= 36h3 - 42h2 - 120
0
= 6h3 - 7h2 - 20
Divide each side by 6.
According to Descartes’ Rule of Signs, there is
one positive real root and zero negative real
roots. Use the graphing utility to graph the
related function V(h) = 6h3 - 7h2 - 20.
Using the Rational Root Theorem, the possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20,
±1/2, ±1/3, ±1/6, ±2/3, ±4/3, ±5/2, ±5/3, ±5/6, ±10/3, and ±10/3. Use the Factor Theorem
until the one zero is found.
V(h) = 6h3 - 7h2 - 20
V(2) = 6(2)3 - 7(2)2 - 20
V(2) = 48 - 28 - 20
V(2) = 0
The height of the box of pasta is 2 inches, the length is 12 inches, and the width is 5 inches.
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