Transcript Slide 1

The Rational Zero Theorem
The Rational Zero Theorem gives a list of possible rational zeros of a
polynomial function. Equivalently, the theorem gives all possible rational roots
of a polynomial equation. Not every number in the list will be a zero of the
function, but every rational zero of the polynomial function will appear
somewhere in the list.
The Rational Zero Theorem
p
If f(x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and
q
p
(where is reduced) is a rational zero, then p is a factor of the constant
q
term a0 and q is a factor of the leading coefficient an.
EXAMPLE: Using the Rational Zero Theorem
List all possible rational zeros of f(x) = 15x3 + 14x2 - 3x – 2.
Solution
The constant term is –2 and the leading coefficient is 15.
Factors of the constant term, - 2
Factors of the leading coefficient, 15
1,  2
=
1,  3,  5,  15
Possible rational zeros =
= 1,  2,
 13 ,  23 ,
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
 15 ,  52 ,
Divide 1
and 2
by 5.
1 ,  2
 15
15
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f(x) = 15x3 +
14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.
• Verify that x = -1 is a root in the previous
example…
Integral Root Theorem
• Suppose your leading coefficient in your
polynomial function is 1.
• What does this mean for you possible
rational roots?
• Ex: Find all the roots
of…
f ( x) = x + 8x + 16x + 5
3
2
Descartes' Rule of Signs
If f(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial with real
coefficients.
1. The number of positive real zeros of f is either equal to the number
of sign changes of f(x) or is less than that number by an even integer.
If there is only one variation in sign, there is exactly one positive real
zero.
2. The number of negative real zeros of f is either equal to the number
of sign changes of f(-x) or is less than that number by an even
integer. If f(-x) has only one variation in sign, then f has exactly one
negative real zero.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x) = x3 + 2x2 + 5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f(x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f(-x). We obtain this equation by replacing x
with -x in the given function.
f(x) = x3 + 2x2 + 5x + 4
Replace x with -x.
f(-x) = (-x)3 + 2(-x)2 + 5(-x) + 4
= -x3 + 2x2 - 5x + 4
This is the given polynomial function.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x) = x3 + 2x2 + 5x + 4.
Solution
Now count the sign changes.
f(-x) = -x3 + 2x2 - 5x + 4
1
2
3
There are three variations in sign.
# of negative real zeros of f is either equal to 3, or is less than this number by
an even integer.
This means that there are either 3 negative real zeros
or 3 - 2 = 1 negative real zero.
Use Descartes’ Rule of signs to find the number of possible
real roots… The, find all the roots.
f ( x) = x - 2x - 8x
3
2