Learning Objectives
Introduction
Tutorial
Rational Zero (or Root) Theorem
If ,
where
are integer coefficients and the reduced fraction is
a rational zero, then p is a factor of the constant term and q is a factor of the leading coefficient .
Make sure that you include both the positive and negative factors.
Make sure that you include both the positive and negative factors.
Generally we don’t write a number over 1, I just did it to emphasize that the denominator comes for factors of q which are . It would have been ok to write out the 2nd line without writing out the 1st line.
Note, how when you reduce down the fractions, some of them are repeated.
Here is a final list of all the POSSIBLE rational zeros, each one written once and reduced:
The factors of the constant term 100 are .
The factors of the leading coefficient 1 are .
Writing the possible factors as we get:
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to
go to Tutorial
37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root form the list of . I would suggest to start with smaller easier numbers and then go from there.
I’m going to choose 2 to try:
Since the reminder came out -126, this means f(2) = -126, which means x = 2 is NOT a zero for this polynomial function.
That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board.
We need to choose another number that comes from that
same list of POSSIBLE
rational roots.
This time I’m going to choose -2:
Again, the reminder is not 0, so x = -2 is not a zero of this polynomial function.
This time let’s choose - 4:
At last, we found a number that has a remainder of 0. This means that x = - 4 is a zero or root of our polynomial function.
Since, x = - 4 is a zero, that means x + 4 is a factor of our polynomial function.
Rewriting f(x) as (x + 4)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
*Set 2nd factor = 0
*Set 3rd factor = 0
The factors of the constant term -16 are .
The factors of the leading coefficient 1 are .
Writing the possible factors as we get:
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to
go to Tutorial
37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root form the list of . I would suggest to start with smaller easier numbers and then go from there.
I’m going to choose -1 to try:
Since the reminder came out -30, this means f(-1) = -30, which means x = -1 is NOT a zero for this polynomial function.
That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board.
We need to choose another number that comes from that same list of POSSIBLE rational roots.
This time I’m going to choose 1:
At last, we found a number that has a remainder of 0. This means that x = 1 is a zero or root of our polynomial function.
Since, x = 1 is a zero, that means x - 1 is a factor of our polynomial function.
Rewriting f(x) as (x - 1)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Factor the difference
of squares
*Set 1st factor = 0
*Set 2nd factor = 0
*Set 3rd factor = 0
*Set 4th factor = 0
The number of POSITIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f(x) or is less than that number by an even number (until 1 or 0 is reached).
The number of NEGATIVE REAL
ZEROS of f is either equal to the number of sign changes of successive terms of f(-x)
or is less than that number by an even integer (until 1 or 0 is
reached).
This can help narrow down your possibilities when you do go on to find the zeros.
Possible number of positive real zeros:
The up arrows are showing where there are sign changes between successive terms, going left to right. The first arrow on the left shows a sign change from positive 3 to negative 5. The 2nd arrow shows a sign change from negative 5 to positive 2. The third arrow shows a sign change from positive 2 to negative 1. And the last arrow shows a sign change from negative 1 to positive 10.
There are 4 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0.
Since we have 4 sign changes with f(x), then there is a possibility of 4 or 4 - 2 = 2 or 4 - 4 = 0 positive real zeros.
Note how there are no sign changes between successive terms.
This means there are no negative real zeros.
Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This piece of information would be helpful when determining real zeros for this polynomial. However, for this problem we will stop here.
Possible number of positive real zeros:
The up arrow is showing where there is a sign change between successive terms, going left to right. This arrow shows a sign change from positive 2 to negative 7.
There is only 1 sign change between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 1, and then go down by even integers from that number until you get to 1 or 0.
If we went down by even integers from 1, we would be in the negative numbers, which is not a feasible answer, since we are looking for the possible number of positive real zeros. In other words, we can’t have a -1 of them.
Therefore, there is exactly 1 positive real zero.
The up arrows are showing where there are sign changes between successive terms, going left to right. The first arrow on the left shows a sign change from negative 2 to positive 7. The 2nd arrow shows a sign change from positive 7 to negative 8.
There are 2 sign changes between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 2, and then go down by even integers from that number until you get to 1 or 0.
Since we have 2 sign changes with f(-x), then there is a possibility of 2 or 2 - 2 = 0 negative real zeros.
The factors of the constant term -2 are .
The factors of the leading coefficient 3 are .
Writing the possible factors as we get:
Possible number of positive real zeros:
The up arrows are showing where there are sign changes between successive terms, going left to right.
There are 3 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 3, and then go down by even integers from that number until you get to 1 or 0.
Since we have 3 sign changes with f(x), then there is a possibility of 3 or 3 - 2 = 1 positive real zeros.
Note how there are no sign changes between successive terms.
This means there are no negative real zeros.
Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This will help us narrow things down in the next step.
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to
go to Tutorial
37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root form the list of . Above, we found that there are NO negative rational zeros, so we do not have to bother with trying any negative numbers. See how Descartes’ has helped us. I would suggest to start with smaller easier numbers and then go from there.
I’m going to choose 1 to try:
Since the reminder came out -2, this means f(1) = -2, which means x = 1 is NOT a zero for this polynomial function.
That doesn’t mean that we pack up our bags and quit. It’s back to the drawing board.
We need to choose another number that comes from that same list of POSSIBLE rational roots.
This time I’m going to choose 2:
At last, we found a number that has a remainder of 0. This means that x = 2 is a zero or root of our polynomial function.
Since, x = 2 is a zero, that means x - 2 is a factor of our polynomial function.
Rewriting f(x) as (x - 2)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
*Set 2nd factor = 0
*This is a quadratic that does not
factor
*Use the quadratic
formula
The factors of the constant term -18 are .
The factors of the leading coefficient 1 are .
Writing the possible factors as we get:
Possible number of positive real zeros:
The up arrow is showing where there is a sign change between successive terms, going left to right.
There is 1 sign change between successive terms, which means that is the highest possible number of positive real zeros.
Since we have 1 sign change with f(x), then there is exactly 1 positive real zero.
The up arrows are showing where there are sign changes between successive terms, going left to right.
There are 4 sign changes between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0.
Since we have 4 sign changes with f(x), then there are possibility of 4, 4 - 2 = 2 or 4 - 4 = 0 negative real zeros.
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root from the list of . Above, we found that there is exactly 1 positive rational zero. Since we know that there is 1 for sure, then we may want to go ahead and start with trying positive rational roots. I would suggest to start with smaller easier numbers and then go from there.
I’m going to choose 1 to try:
Bingo!!!! We found a number that has a remainder of 0. This means that x = 1 is a zero or root of our polynomial function.
Since, x = 1 is a zero, that means x - 1 is a factor of our polynomial function.
Rewriting f(x) as (x - 1)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
Recall, that in Descartes’s Rule of Signs we already
found that there
is exactly one positive real zero. It looks like we already found
that, so when we go trying again we can focus on finding a negative
real zero.
Note that we can still pick from the same list of numbers as we did above, since we are still looking at solving the same
overall problem. However when we set up the synthetic division,
we
will just look at the remaining factor, to help us factor that down
farther.
I’m going to choose -1 to try:
Bingo!!!! We found a number that has a remainder of 0. This means that x = -1 is a zero or root of our polynomial function.
Since, x = -1 is a zero, that means x +1 is a factor of our polynomial function.
Rewriting f(x) as (x - 1)(x + 1)(quotient) we get:
Recall, that in Descartes’ Rule of Signs we already
found that there
is exactly one positive real zero. It looks like we already found
that, so when we go trying again we can focus on finding a negative
real zero.
Note that we can still pick from the same list of numbers as we did above, since we are still looking at solving the same
overall problem. However when we set up the synthetic division,
we
will just look at the remaining factor, to help us factor that down
farther.
I’m going to choose -2 to try:
Bingo!!!! We found a number that has a remainder of 0. This means that x = -2 is a zero or root of our polynomial function.
Since, x = -2 is a zero, that means x + 2 is a factor of our polynomial function.
Rewriting f(x) as (x - 1)(x + 1)(x + 2)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
*Set 2nd factor = 0
*Set 3rd factor = 0
*Set 4th factor = 0
Practice Problems
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problem 1a: Use the Rational Zero Theorem to list all the possible rational zeros for the given polynomial function.
Practice Problem 2a: Find the possible number of positive and negative real zeros of the given polynomial function using Descartes’ Rule of Signs.
Practice Problems 3a - 3b: List all of the possible zeros, use Descartes’ Rule of Signs to possibly narrow it down, use synthetic division to test the possible zeros and find an actual zero, and use the actual zero to find all the zeros of the given polynomial function.
Need Extra Help on these Topics?
The following are webpages that can assist you in the topics that were covered on this page:
http://www.purplemath.com/modules/rtnlroot.htm
This webpage goes over the Rational Zero Theorem.
http://www.purplemath.com/modules/drofsign.htm
This webpage helps you with Descartes' Rule of Signs.
Last revised on March 15, 2012 by Kim Seward.
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