Question 1155272: List all rational zeros that are possible according to the Rational Zero Theorem
The following function is given: f(X)=x3-5x2-4x+20
I am assuming that I look for the leading coefficient first
Then possible rational zeros Factors of the constant terms
Factors of the leading coefficient
I am going to assume that the constant term is 20.
The leading Coefficient would be 1, 5,-5,2-2,4.-4,20, -20
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source):
You can put this solution on YOUR website!
the Rational Zero Theorem tells us that if  is a zero of  , then  is a factor of  and  is a factor of  .
The factors of are ± and
the factors of are ± ,±  , ± , ± ,± ,±.
The possible values for are:
± , ± , ± , ± ,± ,±
These are the  rational zeros for the function.
We can determine which of the possible zeros are   by substituting these values for  in .
do it and you will see that are:
Answer by ikleyn(52873)  (Show Source):
You can put this solution on YOUR website! .
In your list, you missed -1, 10 and - 10.
Notice, that in your problem, the leading coefficient is 1 (one, ONE), and it makes the solution much EASIER.
The better way is to present your list in this form
+/-1, +/-2, +/-4, +/-5, +/-10, +/-20.
When you present in this form, you are (almost) guaranteed from missing the terms.
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Notices and guidelines
In this problem, the leading coefficient is 1 (at x^3).
The constant term is 20.
You must list all the numbers +/-  .
IGNORE THE POST BY @MathLover1, since she presented TOTALLY WRONG solution.
Ignore it for your safety (!) (!) (!)
Dear @MathLover1, in my view, it is ABSOLUTELY INAPPROPRIATE for a tutor to consult students/visitors
in a level as you did it in this case (and in many other cases I observed).
I know that you are able to consult satisfactory/successfully on many simple level problems.
Do it in those areas/topics, where you are confident, and do not go outside, where you have no enough knowledge/expertise.
One more advise: CHECK yourself, using both the formal checks and the common sense.
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