We're hunting for roots of polynomials. Where might they be? There are a lot of numbers out there, and we'd hate to have to test them all. The whole "there's an infinite number of numbers" thing would make it especially hard.
We're in luck, though, because we have the Rational Root Theorem to help us out. It won't tell what the roots are directly, but it will narrow our choices down.
Sample Problem
List all of the possible rational roots of y = 2x3 − 7x2 − 46x – 21.
To use the Rational Root Theorem, first we find all of the factors of the first and last coefficients of the polynomial. In this case, the factors of 2 are 1, 2, -1, and -2. Remember that both positive and negative numbers count as factors, so we better count them too. Otherwise, we can't count on finding the right answer. For 21, we have ±1, ±3, ±7, and ±21.
The Rational Root Theorem says that the only possible rational roots are a ratio of one of the constant coefficient's factors divided by one of the leading coefficient's factors. That's a mouthful, but here's what it means. Take all of the factors of the last term, one at a time, and stick them on top of all the factors of the first term, one at a time. Those are the only rational roots possible.
For this polynomial, that means we have:
Is a number not on this list? Then it isn't a rational root of the polynomial. If it is on the list? Then it might be a root—but it might not be as well. For now, we'll have to plug in every value, positive and negative, into the Remainder Theorem to check if they are roots.
At least we know that we can't have more than 3 roots. Once we've found that many, stop looking. Thanks, Fundamental Theorem of Algebra.
Once we find our first root, we can use synthetic division to factor the polynomial, and the result might make it easier to find the rest of the roots. For instance, we'll just come out and tell you that 
is a root for this polynomial.
= -2 + 2 = 0
See? There's no remainder, which makes it a root. A quick round of synthetic division gets us:
So that's 2x2 – 8x – 42. This definitely is factorable. You can trust us on this.
2(x2 – 4x – 21)
= 2(x – 7)(x + 3)
Our final factorization of the whole polynomial is 
. Looking back at our list from the Rational Root Theorem, we see that all of the roots we found are listed. Because we factored for the other two roots, we can be sure that we found them all, without having to plug any of the rest in.
It could have been the case, though, that none of the possible roots are actually roots. Some polynomials, like y = x2 + 1, don't touch the x-axis at all, and so they don't have any real roots. There might only be complex roots. That means we end up doing a lot of number crunching for no payoff. We know, it's a bummer.
Another possibility is that there are roots, but they are irrational, like 
. The Rational Root Theorem can't find those either. We don't have an Irrational Root Theorem, or a Complex Root Theorem, so we'll have to make do with what we have.
Example 1
Find the possible rational roots of y = x5 + 3x2 − 10x – 24. |
Example 2
Find the possible rational roots of y = 3x7 − 12x3 + 52x2 − 96x + 9. |
.Example 3
Find all the roots of y = x4 − 5x2 + 4. |
Exercise 1
Find the possible rational roots of y = 3x5 + 2x4 – x3 – x2 + 6x + 16
Exercise 2
Find the possible rational roots of f(x) = x4 − 10x3 + 37x2 − 60x + 36
Exercise 3
Could the polynomial y = 5x3 − 10x2 + 17x + 28 have (x – 5) as a factor?
Exercise 4
Could the polynomial y = x4 + x3 + x2 – x – 2 have a zero between 0 and 1?
Exercise 5
Find all of the roots of y = 2x3 + 4x2 + 5x + 10.
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