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Polynomial equations

Polynomial equations
Good news, there’s a class of equations in which we are already experts. When f is a polynomial, the equation f of x equals 0 defines the roots of the polynomial. We have studied in detail the issue of finding these roots. Recall, for example, the following fact for the quadratic polynomial case. We know that there is something there, the discriminant, which will tell us an awful lot about the roots of this polynomial. Capital Delta– by the way, there’s another beautiful Greek letter– is the discriminant, b squared minus 4ac. We know already from this proposition that we’ve seen and proven that if Delta is strictly negative, there are no roots.
And if delta is positive, then there are roots of the polynomial, and we know how to calculate them from this quadratic formula that we will know till the end of our lives. Now it’s a fact that in specific examples you sometimes have to massage the given equation to turn it into polynomial form. A notable example of this is when rational functions are involved. Here’s a case in which you want to solve an equation where the sum of two rational functions equals 3. Remember, rational function means polynomial over polynomial. Now the natural domain of this equation is the whole real line, but take away the points 1 and minus 2 for obvious reasons.
We multiply both sides of the equation by the product, x minus 1, x plus 2. This will have the effect of getting rid of the denominators on the left-hand side, so, algebraically, it will simplify our equation. In doing this multiplication, by the way, we’re not multiplying by anything that can be 0, so we are preserving equivalence because this term will not be 0 in the domain as we’ve defined it. So after this multiplication, we come upon this equation. Now we have to simplify it. Our mastery of arithmetic allows us to do this with ease. And when the dust settles, you have a quadratic equation for x.
It turns out that it factors, and, therefore, its roots are evident, minus 5 and plus 2. And the solution set of our original equation is minus 5 and 2. Two points that are in the domain. Now that should be right because we haven’t done anything to change equivalence, but we’re not androids, so we check that it works in the original equation. How about polynomial equations of degree greater than 2? Let me remind you that we had a strategy for finding roots of a general polynomial. It was based upon writing the root factorization of the polynomial, that means writing it as a product of linear factors, x minus r, times a polynomial that has no further roots.
And the way to obtain such a root factorization involved order reduction and the fun of polynomial division. Well, we won’t go back all over that, but let’s look at the special case of a pure n-th order equation, an equation like this, a x to the n plus b equals 0. So a here is different from 0. And this equation is equivalent to x to the power n equals minus b over a. Since we want to find x, we are very tempted to simply take the n-th root. But remember that the n-th root of y is only defined when y is positive.
Therefore, we have to be a little more careful in solving the equation and what happens next depends entirely upon the parity of n. Is n even, or is n odd? If n is even and if minus b over a is strictly negative, then there are obviously no solutions to our equation because an even power of x can never be strictly negative. On the other hand, if minus b over a is positive, then the n-th root of that number is a solution, but since n is even, also minus the n-th root is the solution. So we have two solutions in that case.
In the remaining case, when n is odd, there will always be one solution, but it will be written somewhat differently depending upon whether minus b over a is positive or negative. When it’s positive, it’s just directly the n-th root. When it’s negative, we have to write the solution as minus the n-th root of the positive number, b over a.
Now let me give you an example in which we will use a frequently useful technique called change of variables to simplify a given equation that we want to solve. Suppose we have a polynomial equation like the example you see in which there is a square power of x and a fourth power of x, but there’s no cubic power and there’s no x. This suggests a certain useful change of variables that might simplify the equation. We’re going to introduce a new variable. I’ll call it capital X, and it’s going to be a stand-in for x squared. So that’s our change of variables. Why do we make this change of variables?
Well in the hope that when we express our original equation with the new variable it will be simplified. And you see this is the case because the new equation, in terms of capital X, is a straightforward quadratic equation, capital X squared minus capital X minus 2 equals 0. It’s an equation we can easily solve by factoring. We find that capital X must be 2 or minus 1. Now we go back to the original variable, capital X is little x squared, so that means x squared has to be 2 or minus 1. Now x squared can’t be minus 1, of course, therefore, we retain only the possibility that x squared equals 2.
And this gives us two roots, x equals plus and minus root 2. So the solution set is the set of two points plus and minus root 2. And we check our answer, of course. Let me remark that the change of variables X equals x cubed is one that will do very good work when we apply it to an equation such as x to the sixth minus x cubed minus 2 equals 0. It’s pretty much the same idea.
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Precalculus: the Mathematics of Numbers, Functions and Equations

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