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Solving Polynomials (page 1 of 2) The general technique for solving biggerthanquadratic polynomials is pretty straightforward, but the process can be timeconsuming. The first step is to apply the Rational Roots Test to the polynomial to get a list of values that might possibly be solutions to the polynomial equation. You can follow this up with an application of Descartes' Rule of Signs, if you like, to narrow down which possible zeroes might be best to check. Of course, if you've got a graphing calculator, it's a good idea to do a quick graph, since xintercepts of the graph are the same as zeroes of the equation. Seeing where the graph looks like it crosses the axis can quickly narrow down your list of possible zeroes. Once you've found a value you want to test, you use synthetic division to see if you can get a zero remainder. If you get a zero remainder, you've not only found a zero, but you've also reduced your polynomial by one degree. Remember that synthetic division is, among other things, division, so checking if x = a is a solution is the same as dividing out the linear factor x a. This means that you should not return to the original polynomial for your next computation (for finding the other zeroes); you should instead work with the output of the synthetic division. It's smaller, so it's easier to work with. You should not be surprised to see some complicated solutions to your polynomials (that is, solutions containing square roots or complex numbers, or both); these zeroes will come from applying the Quadratic Formula to the last (quadratic) factor of your polynomial. Here's how the process plays out:
First, I'll apply the Rational Roots Test Wait. Actually, the first thing I'll do is check to see if x = 1 or x = 1 is a root, because these are the simplest roots to test for. This isn't an "official" first step, but it can often be a timesaver, because you can just look at the powers and the numbers. When x = 1, the polynomial evaluates as 2 + 3 30 57 2 + 24 = 60, so x = 1 isn't a root. But when x = 1, I get 2 + 3 + 30 57 + 2 + 24 = 0, so x = 1 is a root, and I can take care of it right away: This leaves me with the smaller polynomial 2x^{4} + x^{3} 31x^{2} 26x + 24. (Since I've divided out the factor x + 1, I've reduced the degree of the polynomial by 1. That's how I know this is a degreefour polynomial.) Now I'll apply the Rational Roots Test to get a list of potential zeroes to try: From experience, I've learned that most of these exercises have their zeroes near the middle of the list, rather than at the extremes. This isn't always true, of course, but it's usually better to stay away from the larger numbers. In this case, I won't start off by trying stuff like x = 24 or x = 12. Instead, I'll start out with smaller values like x = 2. And I can narrow down my options further by "cheating" and looking at the graph: This is a fourthdegree polynomial, so it has, at most, four xintercepts, and I can see all four of them on the graph. It looks like one of the zeroes is around 3.5, but 7/2 isn't on the list that the Rational Roots Test gave me, so this must be an irrational root. I'll leave it until last. It also looks like there may be zeroes near 1.5 and 0.5. But the clearest solution looks to be at x = 4 and since whole numbers are easier to work with than fractions, x = 4 would probably be a good value to try:
The zero remainder says that x = 4 is a root. The bottom row of the synthetic division tells me that I'm now left with factoring 2x^{3} + 9x^{2} + 5x 6. Looking at the constant term "6", I can see that x = ±24, ±12, ±8, and 4 won't work as rational roots (even if I didn't already know from the graph), so I can cross them off of my list. (Always check the numbers as you go. The Rational Roots Test can give a very long list of possibilities, and it can be helpful to notice that some of those values can be ignored, especially if you don't have a graphing calculator to "cheat" with.) Comparing the remaining values on the list with the intercepts on the graph, I'll try x = 1/2: The remainder isn't zero, so that test root didn't work. This means that the zero close to x = 1/2 on the graph must be irrational; I'll find it when I apply the Quadratic Formula later. For now, I'll try x = 3/2: Copyright © Elizabeth Stapel 20052011 All Rights Reserved The division came out evenly, leaving me with the polynomial 2x^{2} + 6x 4. Since I'm looking for the zeroes of the polynomial, what I really have here is 2x^{2} + 6x 4 = 0. Dividing through by 2 to get smaller numbers gives me x^{2} + 3x 2 = 0, to which I can apply the Quadratic Formula: Then the complete solution is: Asking you to find the zeroes of a polynomial means the same thing as asking you to find the solutions to a polynomial equation. The zeroes of a polynomial are the values of x that make the polynomial equal to zero. So the above problem could have been stated along the lines of "Find the solutions to 2x^{5} + 3x^{4} 30x^{3} 57x^{2} 2x + 24 = 0" or "Find the solutions to 2x^{5} + 3x^{4} 30x^{3} 57x^{2} 2x = 24", and the answers would have been the exact same list of xvalues. You can use these same techniques to factor biggerthanquadratic polynomials.... Top  1  2  Return to Index Next >>



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