### Rational Root Theorem PPT 2013

```The Real
Zeros of a
Polynomial
Function
REMAINDER THEOREM
Let f be a polynomial function. If f (x) is divided
by x – c, then the remainder is f (c).
Let’s look at an example to see how this theorem is useful.
3
2
f x  2x  3x  2x 1
So the remainder we get in synthetic division is the same
as the answer we’d get if we put -2 in the function.
The root of
x + 2 = 0 is
x = -2
Find f(-2)
-2
2
-3 2 -1
-4 14 -32
2 -7 16 -33
using synthetic
division let’s
divide by x + 2
the remainder
f  2  2 2  3 2  2 2 1  33
3
2
FACTOR THEOREM
Let f be a polynomial function. Then x – c is a
factor of f (x) if and only if f (c) = 0
If and only if means this will be true either way:
1. If f(c) = 0, then x - c is a factor of f(x) Try synthetic
2. If x - c is a factor of f(x) then f(c) = 0. division and see
if the remainder
3
2
Is x  3 a factorof  4 x  5x  8 ? is 0
Opposite -3
-4 5 0
8
NO it’s not a
sign goes
12 -51 153 factor. In fact,
here
-4 17 -51 161 f(-3) = 161
We could have
computed f(-3) at
3
2
f  3  4  3  5  3  8  161
first to determine
this.
Not = 0 so not a factor
 
 
 
Our goal in this section is to learn how we can factor higher
degree polynomials. For example we want to factor:
4
3
2
f x  x  x  3x  x  2
We could randomly try some factors and use synthetic
division and know by the factor theorem that if the
remainder is 0 then we have a factor. We might be trying
things all day and not hit a factor so in this section we’ll
learn some techniques to help us narrow down the things to
try.
The first of these is called
Descartes Rule of Signs named
after a French mathematician that
worked in the 1600’s.
Rene Descartes 1596 - 1650
Descartes’ Rule of Signs
Let f denote a polynomial function written in standard
form.
The number of positive real zeros of f either equals the
number of sign changes of f (x) or else equals that
number less an even integer.
The number of negative real zeros of f either equals the
number of sign changes of f (-x) or else equals that
number less an even integer.
1
2
starts Pos.
changes Neg.
changes Pos.
f x  x  x  3x  x  2
4
3
2
There are 2 sign changes so this means there could be
2 or 0 positive real zeros to the polynomial.
Descartes’Rule of Signs
Let f denote a polynomial function written in standard
form.
The number of positive real zeros of f either equals the
number of sign changes of f (x) or else equals that
number less an even integer.
The number of negative real zeros of f either equals the
number of sign changes of f (-x) or else equals that
number less an even integer.
  xPos. changes
f xstarts
x Neg.
3x changes
 x Pos.
2
4
31
2
2
f x   x 4  x3  32 x   x  2
f  x  x  x  3x  x  2
4
3
2
There
simplify
are f(-x)
2 sign changes so this means there could be
2 or 0 negative real zeros to the polynomial.
Use Descartes’ Rule of Signs to determine how many
positive and how many negative real zeros the
polynomial may have.
Counting multiplicities and complex
(imaginary) zeros, the total number
1
starts Neg.
changes Pos. of zeros will be the same as the
degree of the polynomial.
f x  3x  4x  2
5
4
There is one sign change so there is one positive real
zero.
starts Pos.
Never changes
f  x  3 x  4 x  2  3x  4x  2
5
4
5
4
There are no negative real zeros.
Descartes rule says one positive and no negative real
zeros so there must be 4 complex zeros for a total of 5.
Back to our original polynomial we want to factor:
f x  1x  x  3x  x  2
4
3
2
We’d need to try a lot of positive or negative numbers until we
found one that had 0 remainder. To help we have:
The Rational Zeros Theorem
Let f be a polynomialfunctionof degree1 or higher of theform
f  x   an x n  an 1 x n 1    a1 x  a0 , an  0, a0  0
p
, in lowest terms, is a
q
rationalzero of f , then p must be a factorof a0 , and q must be a factorof an .
where each coefficient is an integer. If
What this tells us is that we can get a list of the POSSIBLE
rational zeros that might work by taking factors of the constant
divided by factors of the leading coefficient.
Factors of the constant
 1, 2
1
Both positives
and negatives
would work for
factors
 1, 2
So a list of possible things to try would be
any number from the top divided by any
from the bottom with a + or - on it. In this
case that just leaves us with  1 or  2
1
f x  x  x  3x  x  2
1
1
1
x 1x
4
3
2
1
1
2
-3
2
-1
-1
-1
-2
3
2
-2
0

 2x  x  2
2
Since 1 is a zero, we can write the
factor x - 1, and use the quotient to
write the polynomial factored.
Let’s try 1
YES! It is a
zero since the
remainder is 0
We found a
positive real zero
so Descartes Rule
tells us there is
another one
We could try 2, the other positive possible.
IMPORTANT: Just because 1 worked doesn’t
mean it won’t work again since it could have
Let’s try 1 again,
a multiplicity.
but we try it on the
factored version
3
2
for the remaining
factor (once you have
 1, 2
1


f x  x 1 x  2x  x  2
1
1
2
1
3
1
-1
3
2
-2
2
0
it partly factored use
that to keep going--don't start over with
the original).
YES! the remainder is 0
x  3x  2  x  2x  1
2
Once you can get it down to 3 numbers here, you can put
the variables back in and factor or use the quadratic
formula, we are done with trial and error.
Let’s take our polynomial then and write all of the
factors we found:
There ended up
4
3
2
being two positive
real zeros, 1 and 1
2
and two negative
real zeros, -2, and -1.
f x  x  x  3x  x  2
 x 1 x  2x 1
In this factored form we can find intercepts and left
and right hand behavior and graph the polynomial
Left &
right
hand
behavior
Plot intercepts
Touches at 1
crosses at -1
and -2.
“Rough” graph
Let’s try another one from start to finish using the
theorems and rules to help us.
f x  2x  13x  29x  27x  9
4
3
2
Using the rational zeros
1, 3, 9 factors of constant
theorem let's find factors
of the constant over
1,
2
coefficient
coefficient to know what
numbers to try.
So possible rational zeros are
all possible combinations of
numbers on top with numbers
on bottom:

1
3
9
 1,  ,  3,  ,  9, 
2
2
2
starts Pos. Stays positive
f x  2x  13x  29x  27x  9
4
3
Let’s see if Descartes Rule
helps us narrow down the
choices.
2
1
3
9

 1, 
 ,
 3, 
 ,
 9, 

2
2
2
No sign changes in f(x) so no positive real zeros---we just
ruled out half the choices to try so that helps!
1
2
3
4
starts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos.
f  x  2x 13x  29x  27x  9
4
3
2
4 sign changes so 4 or 2 or 0 negative real zeros.
f x  2x  13x  29x  27x  9
4
3
2
1
3
9
Let’s try -1
 1,  ,  3,  ,  9, 
2
2
2
2 13 29 27
9 Yes! We found a zero.
-2 -11 -18 -9 Let’s work with reduced
polynomial then.
2 11 18
9
0
-1
Let’s try -1 again
-1
2
2
11
-2
9
18
-9
9
9
-9
0
2x  9x  9  2x  3x  3
2
Yes! We found another
one. We are done with
trial and error since we
can put variables back in
and solve the remaining
So remaining zeros found by setting these factors = 0
are -3/2 and -3. Notice these were in our list of choices.
f x  2x  13x  29x  27x  9
4
3
2
So our polynomial factored is:
f x  x  1 2x  3x  3
2
Intermediate
Value
Theorem
Let’s solve the equation:
2 x  3x  3x  5  0
3
2
To do that let’s consider the function
f x  2x  3x  3x  5
3
2
If we find the zeros of the function, we would be solving the
equation above since we want to know where the function = 0
f
By Descartes Rule:
There is one sign change
in f(x) so there is one
positive real zero.
 x  2x
3
 3x  3x  5
2
There are 2 sign changes
in f(-x) so there are 2 or 0
negative real zeros.
Using the rational zeros theorem, the possible
rational zeros are:
1
1,
5

 1,  5,  , 
1, 2
2
5
2
f x  2x  3x  3x  5
1 5
 1,  5,  , 
2 2
Let’s try 1
Let’s try 5
3
1
2
5
2
2
2
-3 -3
2 -1
-1 -4
-3 -3
10 35
7 32
-5
-4
-9
-5
160
155
2
1 is not a zero
and f(1) = -9
5 is not a zero
and f(5) = 155
On the next screen we’ll plot these points and the y
intercept on the graph and think about what we can tell
1 5
 1,  5,  , 
2 2
f(5) = 155
f(0) = -5
f x  2x  3x  3x  5
3
2
155 is a lot higher than this but that
gives us an idea it’s up high
To join these points in a
smooth, continuous curve, you
would have to cross the x axis
somewhere between 1 and 5.
This is the Intermediate Value
Theorem in action. We can see
that since Descartes Rule told
us there was 1 positive real
zero, that is must be between 1
f(1) = -9 and 5 so you wouldn’t try 1/2,
Intermediate Value Theorem
Let f denote a polynomial function. If a < b and if f(a) and f(b)
are of opposite sign, then there is at least one zero of f
between a and b.
f(5) = 155
In our illustration,
a = 1 and b =5
In our
illustration,
f(a) = -9 and
f(b) = 155
which
are opposite
signs
f(1) = -9
So if we find function
values for 2 different
x’s and one is positive
and the other
negative, there must
be a zero of the
function between
these two x values
We use this theorem to approximate zeros when they
are irrational numbers. The function below has a zero
between -1 and 0. We’ll use the Intermediate Value
Theorem to approximate the zero to one decimal place.
f x   x  8x  x  2
4
3
2
First let’s verify that there is a zero between -1 and 0. If we
find f(-1) and f(0) and they are of opposite signs, we’ll know
there is a zero between them by the Intermediate Value
Theorem.
-1
1
8 -1 0 2
-1 -7 8 -8
1 7 -8 8 -6
So f(-1) = -6 and f(0) = 2
These are opposite
signs.
f x   x  8x  x  2
f(-1) = -6 neg
Does the sign change
occur between f(-1)
and f(-0.5) or between
f(-0.7) = -0.9939
f(-0.5) and f(0)?
f(-0.6) = 0.0416
4
3
2
The graph must cross the x-axis
somewhere between -1 and 0
Let’s try half way
Sign
between at x = - 0.5
change
f(-0.5) = 0.8125 pos
-1
So let’s try something
-0.5
1 8 -1-1and0- 0.5.2
between
Let’s -0.5
try --3.75
0.7. 2.375
Do this
-1.1875
with synthetic division
1 7.5 substitution.
-4.75 2.375 0.8125
or direct
f(0) = 2 pos
Notice that the sign change is
between - 0.7 and - 0.6
- 0.6 is the closest to
zero so this is the
zero approximated to
one decimal place
```