Study Guide

# Polynomials - Roots of Higher-Degree Polynomials

## Roots of Higher-Degree Polynomials

The roots of a polynomial are important. They tell us what the factors are, where the graphs touch the x-axis, and how much money we could save on our car insurance. In this section, we're going to find all kinds of ways to find the roots of big, nasty looking polynomials. Don't worry, they're all friendly, no matter how intimidating they may look.

• ### Fundamental Theorem of Algebra

What could be more fundamental than the Fundamental Theorem of Algebra? A lot of things, actually. This theorem isn't actually fundamental to algebra at all—why else haven't we heard of it yet? It's very important for polynomials though.

The total number of roots in a polynomial—the number of real roots plus the number of complex roots—will add up to the degree of the polynomial.

Have a degree of 5? Then you have 5 roots. It's as simple as that. The only complication is that complex roots always come in conjugate pairs. That's the ± we always see hanging around complex numbers.

• ### Rational Root Theorem

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.

• ### Restricting the Roots

This section is not about wrestling a tree to the ground, tying its roots to a chair, or telling them that they're grounded and can't go to prom. Sorry if you got your hopes up.

What we're really talking about is the fact that there are usually a lot of possible roots for a polynomial. We saw how crazy things get with the Rational Root Theorem. We'd love it if there were a way to cut down on how much work we have to do when searching for roots.

One way we can do that is by using something called Descartes' Rule of Signs. The rule is not, "No one can put up any signs but Descartes." He was all about sharing stuff with people, including his rule for finding the number of positive and negative real roots.

### Sample Problem

How many positive and how many negative real roots does f(x) = -x6 + x5x4+ x3x2 – 5x – 2 have?

Descartes' Rule of Signs wants us to find out how many times the coefficients of a polynomial change signs. Let's write out the signs of each term, without the numbers or variables getting in the way.

– + – + – – –

What we do now is count how many times the signs change sign—that is, go from positive to negative, or negative to positive. The signs change four times, from negative, to positive, to negative, to positive, to negative again. Then they stay negative.  This tells us that there are at most 4 positive real roots (that is, 4 roots greater than 0).

Yes, that is weirdly specific. No, we don't know why it works. Yes, it does actually work.

But wait, it gets even more weirdly specific. It can have "at most" 4 positive real roots. It might also have 2 or 0 positive real roots, and we can't narrow that down any further. In general, the number of positive real roots is (# of sign changes) – (2n). Like we said, this is getting even weirder and even more specific.

The reason the number of possible real roots decreases by 2 is because any possible complex roots will come in pairs. They burst on in and push the real roots around, taking their spot on the couch and forcing them to leave.

We can also have real number roots that are negative. We're still going to look at changing signs, but this time for f(-x). That is, we replace x with -x throughout the whole function. Are we in the Twilight Zone? Is that why things have gotten so weird?

f(-x) = -(-x)6 + (-x)5 – (-x)4+(-x)3 – (-x)2 – 5(-x)– 2

= -x6x5x4x3x2 + 5x – 2

Now let's check out the signs: We have 2 sign changes here, meaning we have at most 2, and maybe 0, negative real roots. Like before, we have to account for the possibility of complex roots crashing the party.

How does all of this weirdness, with Descartes' Rule of Signs, help us when finding roots of a polynomial? In our example, if we happen to find one negative root using the Rational Root Theorem, then we know that there is at least one more.

And sometimes, either f(x) or f(-x) will not change sign at all, and we can get rid of half of our possible answers. We don't want to throw the baby out with the bathwater, but sometimes there is no baby.

### Bounds on Roots

Polynomials are pretty unimaginative. They don't have a lot of loops, swoops, and swirls on their graphs. They tend to do their own thing, going in one direction, until they reach the middle of the graph. Then they'll bob up and down a few times, and then resume heading off into some infinity.

That means that the zeros of a polynomial tend to be clustered in one spot, and that there are no more zeros outside that area. Here's a technique for finding the upper and lower bounds on the real roots using synthetic division.

### Sample Problem

Are x = 4 and x = -4 bounds on the roots of y = x4 – 16?

We start off by checking the results of dividing the function by xa. Our a is 4, which is greater than 0. That means we can check for an upper bound on the roots. We do that by checking the signs of the quotient. Yes, more sign checking. That never gets old. /sarcasm

Anyway, the signs are all positive, which is good. That means we've found an upper bound. There are no roots for this function that are larger than 4. That means we don't have to check +8 or +16. We like to hear that. /not sarcasm

Now we do the same thing to check -4, with just a few changes. Our a is negative, so we can only check for lower bounds. Those are the rules, even if they aren't written on a sign. This time, we want to see if the signs of the quotient alternate. That is, do they go (–) (+) (–) (+) or the other way around, without missing a beat? If they do, then we have a lower bound. And they do, because we can choose how we treat zeros. We have a lower bound. We know that -8 and -16 aren't roots. And knowing is half the battle. The other half is actually fighting, so we never get that far.

### Summary

When we looking for the roots of a polynomial, start off with Descartes. See how many positive and negative real roots there are. Then use the Rational Root Theorem to find potential real roots. Dive right into using synthetic division to test them. Even when we fail to find a root, we can check to see if we've stumbled up a bound.

One thing to remember is, when checking the bounds, to treat a zero as either positive or negative, depending on what will give us the most information given the circumstances. Turn those zeros into root-finding heroes.

Up until now, we've used the Remainder Theorem to find roots, and then used synthetic division afterwards to find the quotient and find more factors. Now that we can check the bounds on the roots, we suggest going straight to synthetic division. If we happen to stumble upon an upper or lower bound, we can save a lot of time, and the RT won't do that for us.

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