The Real Zeros of a Polynomial Function

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The Real Zeros of a Polynomial Function
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Lets look at an example to see how this theorem
is useful.
So the remainder we get in synthetic division is
the same as the answer wed get if we put -2 in
the function.
using synthetic division lets divide by x 2
The root of x 2 0 is x -2
-2 2 -3 2 -1
-4 14 -32
2 -7 16 -33
the remainder
Find f(-2)
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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) 2. If x - c is a factor of f(x) then f(c)
0.
Try synthetic division and see if the remainder
is 0
Opposite sign goes here
NO its not a factor. In fact, f(-3) 161
We could have computed f(-3) at first to
determine this.
Not 0 so not a factor
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Our goal in this section is to learn how we can
factor higher degree polynomials. For example we
want to factor
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 well 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 1600s.
Rene Descartes 1596 - 1650
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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.
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starts Pos. changes Neg. changes Pos.
There are 2 sign changes so this means there
could be 2 or 0 positive real zeros to the
polynomial.
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DescartesRule 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.
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2
starts Pos. changes Neg. changes Pos.
simplify f(-x)
There are 2 sign changes so this means there
could be 2 or 0 negative real zeros to the
polynomial.
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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 of zeros will be the same
as the degree of the polynomial.
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starts Neg. changes Pos.
There is one sign change so there is one positive
real zero.
starts Pos. Never changes
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. Well learn more complex zeros in
Section 4.7.
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Back to our original polynomial we want to factor
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Wed 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
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.
Both positives and negatives would work for
factors
?
1, 2
Factors of the constant
1
Factors of the leading coefficient
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?
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, 2
1
Lets 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
Since 1 is a zero, we can write the factor x - 1,
and use the quotient to write the polynomial
factored.
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We could try 2, the other positive possible.
IMPORTANT Just because 1 worked doesnt mean it
wont work again since it could have a
multiplicity.
?
1, 2
1
Lets try 1 again, but we try it on the factored
version for the remaining factor (once you have
it partly factored use that to keep going---don't
start over with the original).
YES! the remainder is 0
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.
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Lets take our polynomial then and write all of
the factors we found
There ended up being two positive real zeros, 1
and 1 and two negative real zeros, -2, and -1.
In this factored form we can find intercepts and
left and right hand behavior and graph the
polynomial
Plot intercepts
Left right hand behavior
Touches at 1 crosses at -1 and -2.
Rough graph
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Lets try another one from start to finish using
the theorems and rules to help us.
Using the rational zeros theorem let's find
factors of the constant over factors of the
leading coefficient to know what numbers to try.
?
1, 3, 9
factors of constant
1, 2
factors of leading coefficient
So possible rational zeros are all possible
combinations of numbers on top with numbers on
bottom
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starts Pos. Stays positive
Lets see if Descartes Rule helps us narrow down
the choices.
No sign changes in f(x) so no positive real
zeros---we just ruled out half the choices to try
so that helps!
4 sign changes so 4 or 2 or 0 negative real zeros.
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Lets try -1
Yes! We found a zero. Lets work with reduced
polynomial then.
Yes! We found a zero. Lets work with reduced
polynomial then.
Lets try -1 again
Yes! We found another one. We are done with
trial and error since we can put variables back
in and solve the remaining quadratic equation.
So remaining zeros found by setting these factors
0 are -3/2 and -3. Notice these were in our
list of choices.
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So our polynomial factored is
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Intermediate Value Theorem
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Lets solve the equation
To do that lets consider the function
If we find the zeros of the function, we would be
solving the equation above since we want to know
where the function 0
By Descartes Rule There is one sign change in
f(x) so there is one positive real zero.
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, 5
?
1, 2
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1 is not a zero and f(1) -9
Lets try 1
5 is not a zero and f(5) 155
Lets try 5
On the next screen well plot these points and
the y intercept on the graph and think about what
we can tell about this graph and its zeros.
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155 is a lot higher than this but that gives us
an idea its up high
f(5) 155
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 and
5 so you wouldnt try 1/2, but you'd try 5/2
instead.
f(0) -5
f(1) -9
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Intermediate Value TheoremLet f denote a
polynomial function. If a lt 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
So if we find function values for 2 different xs
and one is positive and the other negative, there
must be a zero of the function between these two
x values
In our illustration, f(a) -9 and f(b) 155
which are opposite signs
f(1) -9
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We use this theorem to approximate zeros when
they are irrational numbers. The function below
has a zero between -1 and 0. Well use the
Intermediate Value Theorem to approximate the
zero to one decimal place.
First lets verify that there is a zero between
-1 and 0. If we find f(-1) and f(0) and they are
of opposite signs, well know there is a zero
between them by the Intermediate Value Theorem.
So f(-1) -6 and f(0) 2These are opposite
signs.
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f(-1) -6
The graph must cross the x-axis somewhere between
-1 and 0
f(-0.7) -0.9939
Lets try half way between at x - 0.5
Sign change
f(-0.6) 0.0416
f(-0.5) 0.8125
So lets try something between -1 and - 0.5.
Lets try - 0.7. Do this with synthetic division
or direct substitution.
- 0.6 is the closest to zero so this is the zero
approximated to one decimal place
f(0) 2
Notice that the sign change is between - 0.7 and
- 0.6