Solving a Polynomial without its Graph

The Rational Roots Test says that the possible zeroes are:

It is easy to check if x = −1 or x = 1 is a zero: I plug them in. I will be multiplying the coefficients by −1 or by +1, according to the x-value and the exponent, which is simple enough:

I didn't get a value of zero for either value, so neither x = −1 nor x = 1 is a zero.

But, because of the sign change between f(−1) = −120 and f(1) = 14, there must be a zero in between x = −1 and x = 1.

However, I'd rather avoid fractions if at all possible, so I'm going to try the next biggest non-fractional possible zeroes, being x = −5 and x = 5.

First I'll try x = −5:

I didn't get a zero remainder, so x = −5 is not a zero of this polynomial.

However, f(−5) = 4100 is positive and f(−1) = −120 is negative, so there must be a zero in between x = −5 and x = −1. (Dang; I'm gonna have to work with fractions.)

And since f(−1) = −120 is closer to zero than is f(−5) = 4100, I'm gonna guess that the zero is closer to x = −1 than to x = −5.

Also, I can see that there is no need to test zeroes below x = −5, because I divided by a negative, and the bottom row has alternating signs.

Okay; let me see if x = 5 is a zero:

I didn't get a zero remainder, so x = 5 is not a zero of this polynomial.

Unfortunately, f(5) = 1650 and f(1) = 14 are both positive, so I can't tell if there is a zero in between x = 1 and x = 5.

Descartes' Rule of Signs says that f(x) = 6x⁴ − 13x³ − 34x² + 80x − 25 has three or one positive zeroes (so we are assured of at least one positive zero) and one negative zero. So there are either two real-number zeroes (so the other solutions of this fourth-degree polynomial are complex) or four real-number zeroes. Since there is only one negative root and possibly as many as three positive roots, it looks to me as though my odds of finding a rational zero are better if I try the positive side.

Since f(−1) = −120 is negative and f(1) = 14 is positive, then there must be a zero between x = −1 and x = 1. I'll try :

Nope; it's not a zero, either.

But the remainder says that , which is a positive number. So I know, from the sign change, that there must be a zero between x = −1 and x = ½. I'll try :

Dang! This isn't a root either. (This stoopid thing is taking forever....)

Since the remainder says that (in particular, that it's negative), then the zero must be between and .

However, there are no rational candidates between and , so the zero on this interval must be irrational. I'll have to look elsewhere for a rational zero.

As mentioned above, I have better odds of finding positive zeroes than negative ones (from Descartes' Rule of Signs), so I will try to find a zero between x = 1 and x = 5. I'll try :

Still not a zero. (Argh!)

But this remainder says that , which is negative, and I already knew that f(1) = 14 was positive. This means that now I know that there is a zero between x = 1 and . So (because maybe the *sixth* time is the charm) I'll try :

Finally! I've *finally* found a zero. Yay!

I still have to find at least one more rational zero, so that the polynomial will be reduced to a quadratic, but now I only have to find the zeroes of the result of the last division; namely, now I'm working with the cubic polynomial 6x³ − 3x² − 39x + 15. That is, now all I'm trying to solve is 6x³ − 3x² − 39x + 15 = 0.

One thing I notice, which can make life easier on myself, is to divide through by 3, since all the coefficients of this cubic are multiples of 3. This leaves me with the cubic equation 2x³ − x² − 13x + 5 = 0.

And, helpfully, this also reduces the number of possible rational zeroes. My list now consists only of:

I have already eliminated −5, 5, −1, 1, , and from my list of possible zeroes. Combining these earlier deletions with my new shorter list leaves me with only and to try.

This polynomial has been messy so far, and I know that there's a negative root *somewhere*, so I'll try the negative possibility; namely, :

And there — at last! — is my other rational zero.

Now I'm left with just a simple quadratic, 2x² − 6x + 2, whose zeroes I can find with the Quadratic Formula:

(I don't actually need the decimal approximations for this exercise.)

For my hand-in answer, I'll only include the divisions that worked, plus my zeroes, which are: