Chapter 4: Polynomial and Rational
Functions
Section 4-4: The Rational Root
Theorem
The Rational Root Theorem
•
Sometimes when we have large numbers as coefficients it is difficult to
test zeros because there are so many possibilities. The Rational Root
Theorem helps us narrow the possible zeros.
•
•
Let a0 x n + a1 xn-1 +a2 xn-2 + …. an x0 = 0 represent a polynomial p
equation of degree n with integral coefficients. If a rational number q ,
where p and q have no common factors, is a root of the equation, then
p is a factor of a n and q is a factor of a0 .
Example #1
•
List the possible rational roots of 3x -13x +2x +8=0
Then determine the rational roots.
Remember inspecting the graph on your graphing calculator is the
quickest way to narrow down the possible rational roots. When a root
is found, the depressed polynomial can then be used to complete the
problem. However, not all roots will necessarily be rational.
Solution: According to the Rational Root Theorem if p/q is a root of the
equation, then p is a factor of 8 and q is a factor of 3.
Possible values of p: ±1, ±2, ±4, ±8
Possible values of q: ±1, ±3
Possible rational roots, p/q :
1
2
± 1, ± , ± 2, ± , ± 4,
3
3
4
8
± , ± 8, ±
3
3
Continuation
• So now we know that these are the possible
roots. We also know that there are three
roots because the degree is 3. Now we use
synthethic division to see which three are the
roots of our possible list. Remember if it is a
root the remainder will be 0.
Continuation
• Possible roots:
1
2
± 1, ± , ± 2, ± , ± 4, r
3
3
1
4
8
± , ± 8, ±
3
3
-1
•
•
•
•
•
Table of synthetic division
So from this table we have
Two of the roots 1, 4. Thus
We know the other root
Must be either ± 1 ,± 2 ,± 4 ,± 8
3
3
3
3
2
-2
4
-4
8
-8
3
3
3
3
3
3
3
3
3
-13 2
-10 -8
-16 18
-7 -12
-19 40
-1 -2
-25 102
11 90
-27 X
8
0
-10
-16
-72
0
X
X
X
Test the fractions now
So the roots of the
2
−
polynomial are 1, 4, 3
r
3
1
3
−
1
3
2
3
−
2
3
4
3
4
3
−
8
3
−
8
3
-13
3 -12
3 -14
3 -11
3 -15
3
3
3
3
2
8
-2
x
x
12
x
x
x
0
x
x
x
x
Integral Root Theorem
n−1
n
a
x
x
• Let
+ 1 + …+ an−1x + a n represent a
polynomial that has a leading coefficient is 1
• ( a0 ), integral coefficients, and an ≠ 0 . Any
rational roots of the equation must be integral
factors of
• This makes sense because the possible roots
are p/q . Since the only factors of the leading
coefficient 1 are 1 then the possible roots will
be factors of an = a
1
n
Example
• Find the roots of x3 + 6x2 -13x-6=0 Since the leading
coefficient is 1, the possible rational roots
r
1 6 -13
will be factors of 6.
1 1 7 -6
• Possible roots: ±1, ±2,±3 ,±6 Now we -1 1 5 -18
know there are three roots because the
2 1 8 3
polynomial is degree 3. But not all
-2 1 4 -21
necessarily will be rational.
3 1 9 14
• Synthetic Division:
We have one rational root so the other
-3 1 3 -22
two must be complex.
6 1 12 x
We know the depressed polynomial
-6 1 0 -13
is x2 + 8x + 3
-6
-12
12
0
x
x
x
x
x
Quadratic formula
• Using the quadratic formula, we have
− 8 ± 64 − 4(−3)
2
− 8 ± 72
=
2
− 8 ± 2 13
=
2
= −4 ± 13
x=
• So the roots of the polynomial
x3 + 6 x 2 − 13 x − 6 = 0
are 2, - 4 ± 13
Descartes’ Rule of Signs
• Another tool you can use to narrow the possible roots is Descartes’
Rule of Signs. This can be used to determine the possible number of
positive real zeros of a polynomial.
• Suppose P (x) is a polynomial whose terms are arranged in
descending powers of the variable. Then the number of positive real
zeros of P (x) is the same as the number of changes in sign of the
coefficients of the terms or is less than this by an even number. The
number of negative real zeros of P (x) is the same as the number of
changes in sign of the coefficients of the term of P (-x), or less
than this number by an even number.
Example
•
Find the number of possible positive real zeros and the number of possible
•
•
negative real zeros for P(x)= 2x4 - x 3 -2x2 +5x +1
To determine the number of possible positive real zeros, count the sign changes
for the coefficients.
2
-1
-2
5
1
•
•
•
•
•
•
•
•
•
•
yes
no
yes
no
There are two changes. So there are two or zero positive real zeros. (same
number 2 or less by even 0)
To determine the number of possible negative real roots, find P (-x) and count the
number of sign changes.
2
1
-2
-5
1
P(-x)= 2(-x)4 –(-x)3 -2(-x)2+5(-x)+1
no
yes
no
yes
= 2x4 + x3 -2x 2 -5x+1
There are two changes. So there are two or zero possible negative real zeros.
Now let’s find the zeros!
•
•
We know that we have a degree of 4 so there are 4 possible rational
roots.
We know that the possible rational roots using the rational root theorem
are ± 1, ± 1 .
2
•
So we use synthetic division and we find that none of the possible roots
give us a remainder of zero so we have no positive real roots and no
negative real roots. All of them are imaginary.
•
What is helpful with the Descartes Rules of Sign is that it narrows down
for you the number of possibilities you need to test.
Multiplicity of Zeros
• Sometimes one number represents two or
more zeros of the function.
• For example (x-1)(x-1)=0, the root 1 is
shown twice. We say 1 has a multiplicity of 2.
HW#28
• Section 4-4
• Pp 233-235
• #11-19 odd, 29-31 all