I'll try to help get you started, but I don't want to do the entire assignment for you. The
Fundamental Theorem of Algebra says that every polynomial of degree n (meaning the highest power of x is n) has exactly n complex roots, if you count repeated ones separately. What in the world is this saying, and why is it important? It's important because this is not at all true for real roots, only complex roots. The best we can say over the
real numbers is that a polynomial of degree n has at most n real roots. You can see this isn't as helpful as the Fundamental Theorem, which guarantees
exactly n complex roots.
To get a sense of what's going on, let's look a few familiar examples -- some quadratics, when the degree is 2.
- The quadratic x2-1=0 has degree 2. How many real roots does it have? It has
two real roots. We can find them by factoring the left-hand side; we find (x+1)(x-1)=0. So by the Zero Product Property, either x+1=0 or x-1=0, so either x=-1 or x=1. These are the two real roots.
- The quadratic x2=0 has degree 2. How many real roots does it have? It actually has only
one real root, although this root is "repeated." The root is obvious: x=0.
- The quadratic x2+1=0 has degree 2. How many real roots does it have? It has
no real roots! You can see why in two ways, either algebraic or geometric. Algebraically, if x2+1=0, then x2=-1, but it's not possible to square a real number and get a negative number, so there can be no real solutions. Geometrically, the graph of f(x)=x2+1 lies entirely above the x axis, so it has no
x intercepts and therefore has no roots.
The important lesson from that last example is that a polynomial of degree n
need not have exactly n real roots. In fact, it might not have any roots at all. This is one reason why the complex numbers were "invented" -- to make polynomials have solutions. Because even though that last quadratic x2+1 doesn't have any real roots, it DOES have two
complex roots, as the Fundamental Theorem guarantees. The two complex roots are
i and -i, where i is "the square root of -1".
Okay, so you've seen a little bit about how solving polynomial equations can be different when we're looking for
real solutions versus when we're looking for complex solutions. If we allow our solutions to be complex numbers, a polynomial of degree n will always have n solutions (except if a root is repeated), but if we restrict our solutions to the real numbers, a polynomial of degree n might not have n solutions.
So one question we often have to ask ourselves is -- exactly how many real
roots does a given polynomial have? In general that's a very hard question to answer. Descartes's Rule of Signs, together with the Rational Roots Theorem, are two powerful tools that can help us analyze a given polynomial.
Descartes's Rule says that the number of positive real roots of a polynomial is either equal to the number of "sign changes" in the polynomial's coefficients or less than the number of sign changes by a multiple of 2; it also says that the number of negative real roots of a polynomial is either equal to the number of sign changes in the coefficients
if you change the sign of every odd degree term, or less than that number be a multiple of 2. The
Rational Roots Theorem tells us that if a polynomial has a rational foot a/b, a must be a factor of the constant term of the polynomial, and b must be a factor of the leading coefficient. Together, these two facts can help us predict and in some cases even find the roots of a given polynomial.
Let's illustrate these ideas with the cubic
We know this polynomial has at most 3 real roots, but how many does it have exactly? Well, let's start with Descartes's Rule. How many sign changes are the in the coefficients? The sign changes from + to - from the second to the third term, and that's the only sign change. So Descartes's Rule says the number of positive real roots is either 1 or less than 1 by a multiple of 2 -- i.e. 1, or -1, or -3, etc. But obviously the
number of positive real roots couldn't be negative. So Descartes's Rule guarantees us that this polynomial has EXACTLY 1 positive real root.
What about negative real roots? Well, to use Descartes's Rule, we first have to change the sign of all the odd degree terms; doing so, we get
Now how many sign changes does this have? It has two sign changes, from the first to the second term and the third to the fourth term. So Descartes's Rule says the number of negative real roots is either 2 or less than 2 by a multiple of 2 -- i.e. 2, or 0, or -2, etc. So the best we can conclude is that this polynomial has either 2 or 0 negative real roots.
Can we get any more precise, or are we stuck with such a weak conclusion? The Rational Roots Theorem can help us try to determine what the positive root has to be, at least if it's rational. All the rational roots would have to have the form (factors of -3)/(factors of 1), and the only such numbers are 3, -3, 1, and -1. But if we're only interested in finding that positive root first, we should focus only on the positive numbers: 3 and 1. We can use
synthetic division or the remainder theorem to quickly check whether 3 or 1 is a root. In fact, 1 is a root, because plugging it in yields 0. So we found our one real positive root!
Using synthetic division, we can divide our polynomial by the corresponding factor (x-1). Doing the division leaves us with x2+4x+3. But now the situation is much easier to analyze, because we know to deal with quadratics -- and in fact this quadratic factors! It factors as (x+1)(x+3), yielding two negative roots of -1 and -3. Therefore Descartes's Rule was right that the number of negative real roots would be 2 or 0. It turned out to be 2.
Hopefully this sheds some light on your assignment and gets you thinking. You might try to perform a similar kind of analysis for the polynomials x5+1 and x4+3x-9.