Pre-Calculus
Section 4.4
* Rational Root Theorem
* Integral Root Theorem
* Decartes’s Rule of Signs
Lesson Overview 4-4A
Lesson Overview 4-4B
Only one rational root: 1
the other 2 are imaginary
1
What?
In other words
We are doing more than guessing at the solutions
to a function. We can narrow them down to a finite
set of possibilities.
Also notice that the coefficients must all be
integers.
Lesson Overview 4-4A
Another example List the possible rational roots of
6x3 + 11x2 – 3x – 2 = 0. Then determine the
rational roots. ( with out graphing or using
a program on your calculator!)
Possible values of p:  1, 2
Possible values of q:  1, 2,  3, 6
Possible rational roots, p :
q
1, 2,
1 1 1  2
,
,
,
6
3
3
2
The graph will narrow it down more ( yes you can look at the graph on
your calculator now).
Lesson Overview 4-4A
Another example List the possible rational roots of
6x3 + 11x2 – 3x – 2 = 0. Then determine the
rational roots. ( with out graphing or using
a program on your calculator!
Possible rational roots,
p
: 1, 2,  1  1 ,  1 ,  2
,
q
6
3
3
2
Since the -2 is evident from the graph.
Divide it out. Then either factor the
depressed quadratic or use the
quadratic formula
Or use synthetic division on each
possible root to see which work and
which don’t.
Roots are
-2, - 1/3, and 1/2
2
Integral Root Theorem:
Let xn + an-1xn-1 + … + a1x + a0 = 0 represent a
polynomial equation that has a leading coefficient of 1,
integral coefficients, and a0  0. Any rational roots of
this equation must be an integral factor of a0.
So x3 + 6x2 – 13x – 6 = 0
consider only the factors of -6: 1, 2, 3, 6
2 is evident from the graph
so divide it out
Then use quadratic formula to find the
other 2 roots
Rational roots: 2,  4  13
Descartes’ Rule of Signs
Suppose P(x) is a polynomial whose terms are in
descending order. 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 signs of the
coefficients of P(-x), or less than this number by an even
number.
the above theorem shows up on the next slide as well
Descartes Rule of Signs
Suppose P(x) is a polynomial whose terms are in descending order. 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 signs of the coefficients of P(-x), or less than this
number by an even number.
Find the number of possible positive real zeros and the # of possible
negative real zeros for f(x) = 2x5 + 3x4  6x3 + 6x2 – 8x + 3. Then determine
the rational zeros.
f(x) = 2x5 + 3x4  6x3 + 6x2 – 8x + 3
has 4 sign changes
f(-x) = -2x5 + 3x4 + 6x3 + 6x2 + 8x + 3
has one sign change
So there is 1 negative real zero and 4 positive real zeros
or
or
1 negative and 2 positive real zeros
1 negative and NO positive real zeros
Continued on the next slide
3
Find the number of possible positive real zeros and the # of
possible negative real zeros for f(x) = 2x5 + 3x4  6x3 + 6x2 – 8x + 3.
Then determine the rational zeros.
f(x) = 2x5 + 3x4  6x3 + 6x2 – 8x + 3
has 4, 2, or 0 positive zeros and 1 negative zero
Possible values of p:  1, 3
Possible values of q:  1, 2
Possible rational roots, p/q :  1, 3,  1/2, 3/2
Test possible zeros using synthetic division ( use the graph as a hint as well)
on the board….
Solutions are:
-3, ½ and 1 but there are suppose to be 5 zeros???
Continued to the next slide
f(x) = 2x5 + 3x4  6x3 + 6x2 – 8x + 3.
We know it has 4 positive and 1 negative zero,
or only 2 positives and 1 negative OR no positive and 1 negative
And the negative one is namely, -3
and 2 of the positives are ½ and 1
so what about the other 2 zeros?
Use synthetic division to divide them out to see the depressed polynomial
-3
2
3 -6
-6
6 -8
3
9
-9 9 -3
2 -3 3
-3 1 0
1 2 -3 3
2
-3 1
2
-1
2 -1
-1
2
-1 0
1/2
2
2
-1
2
1
0
-1
1
0
2
0
f(x) = (x + 3)(x - 1)(x – ½)(2x2 + 2)
f(x) = 2(x + 3)(x - 1)( x – ½)(x2 + 1)
4