How Many Rational Zeros Does the Polynomial f(x) = x³ + 4x² - 4x - 16 Have?

[Greywolfe1982]

Homework Statement

Consider
f(x) = x^3 + 4x^2 − 4x − 16.
a) How many possible rational zeros does f(x) have?
1. None
2. Two
3. Ten
4. Three
5. Five

The Attempt at a Solution

My first response was answer 4. - Three zeroes. I
a) solved for the zeroes on paper
b) graphed it on a graphing calculator
c) realized that a third degree polynomial can have three zeroes
d) answered the next question, asking what the zeroes were, correctly.

Apparently I'm misunderstanding the question though, because answer 4 is not correct. So just what is the question asking? I thought it was simply "how many zeroes does it have", but I guess I'm missing something.

 

[n!kofeyn]

The question asks: How many possible rational zeros does f(x) have? I think it is a bad question though, because the polynomial has a possibly of having 3 zeros, and if I'm not mistaken, all of these zeros could potentially be rational.

 

[Greywolfe1982]

All of the zeroes are rational - I'm going off memory, but I believe they were -4, -2, and 2. Regardless, they were all integers. That's why I'm so confused.

 

[n!kofeyn]

All of the zeroes are rational - I'm going off memory, but I believe they were -4, -2, and 2. Regardless, they were all integers. That's why I'm so confused.

Then it appears that the answer is incorrect.

 

[Tedjn]

It is possible that the question is asking how many different values are possible for a rational root from the rational roots theorem. I agree that it is a poorly worded question.

 

[Greywolfe1982]

So the answer is 5...correct?

 

[Tedjn]

Actually, I remembered that rational roots allows both positive and negative roots, so I would say 10... but who knows how this question is to be interpreted. Five is the closest to any sort of right answer if three isn't correct.

 

[Mark44]

The question might be asking how many rational candidates are there for zeroes in this cubic. If that's the correct interpretation, then the answer is ten. The candidates are $\pm$ 1, $\pm$ 2, $\pm$ 4, $\pm$ 8, and $\pm$ 16.

 

[Greywolfe1982]

Thought about it some more and remembered it's plus or minus, came on here and read what I was thinking.

Thanks guys

 

[HallsofIvy]

As n!kofeyn said, that is a badly worded question. Obviously, a cubic equation cannot have more than 3 zeros and so not more than three possible rational zeros.

However, there there are 10 rational numbers that, by the rational number theorem, are possible zeros.

 

[Count Iblis]

You can always substitute x = y + p and then obtain other possible rational solutions. The intersection of all these sets of solutions is the correct answer, so I think the correct answer is 3 if there are 3 rational solutions. In fact, you can save a lot of time by shifting your variable. E.g., if you see if x = 1 is a solution you find that:

f(1) = -15

So, x = 1 is not a zero. But this result is nevertheless useful, becaus it is clear that the polynomial g(y) defined as:

g(y) = f(1+y)

has 1 as the coefficient of y^3 and the constant term is

g(0) = f(1) = -15

So, if we apply the Rational Roots Theorem to g(y), we find that the zeroes of g(y) can be:

y = ±1, ±3, ±5,±15

Since g(y) = f(1+y), this means that we need to add 1 to these numbers to obtain the possible rational zeroes of f(x):


x = 0, 2, -2, 4, -4, 6, -14, 16

If you strike out those candidates that are not on the original list, you are left with:


x = 2, -2, 4, -4, 16

If you play the same game with x = -1, you find:

f(-1) = -9

therefore putting y = -1 + x gives the candidates:

y = ±1, ±3, ±9

So,

x = -2, 0, -4, 2, -10, 8

Striking out those candidates that are not on the previous list, leaves us with:

x = -2, -4, 2

Which are the three correct solutions.