Bryan P.

asked • 05/18/17

f(x) = 2x3 + 5x2 – 39x + 18

How many possible rational roots are there for the polynomial function?
a. 6
b. 10
c. 14
d. 18

Arturo O.

Since the polynomial f(x) given in the problem statement is of degree 3, it cannot have more than 3 distinct roots of any kind.  There has to be something wrong with the wording of the question, or with the list of answers.
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05/18/17

3 Answers By Expert Tutors

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Carol H. answered • 05/18/17

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Using the Rational Root Theorem 

p: ±1, ±2, ±3, ±6, ±9, ±18

q: ±1, ±2

p/q: ±1, ±1/2, ±2, ±3, ±3/2, ±6, ±9/2, ±9, ±18

nine possible roots

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Ravi V.

Don't know if I should add to the confusion, but the question is poorly asked. As Arturo said, you can't have more than 3 rational roots, however, Ration Root Theorem says that to find those three (or less), you only have to search 18 possible numbers and that is probably what the questioner is asking and therefore use Andrew's answer.
 
How to come up with those 18 ?
 
First obtain all the factors or 18 (constant in the polynomial): They are 1,2,3, 6, 9 and 18
Then obtain all the factors of 2 (coefficient corresponding to the highest degree term x^3): They are 1 and 2
 
Then look at all numbers of the P/Q where P is a factor of 18 and Q is a factor of 2
 
So you have +/- (1/1, 2/1, 3/1, 6/1, 9/1, 18/1, 1/2, 2/2, 3/2, 6/2, 9/2, 18/2)
That is 24 possibilities before removing duplicates.
Now remove the duplicates and you are left with 18 potential candidates which is your answer.
However do note as Arturo mentioned at most only 3 of them can be valid. In fact -6, 1/2 and 3 are the actual roots for the polynomial in question.
 
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05/18/17

Arturo O.

Thank you Ravi,
 
That really clarifies the wording of the problem.  So the theorem gives 18 candidates of which only 3 can be distinct because of the degree of the polynomial.
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05/19/17

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