## Chapter 3: Polynomial Equations## 3.1 Polynomials A A We can define operations on polynomials in the usual manner. The sum of two polynomials is obtained by taking the sums of the coefficients. If and are two polynomials, then their product is obtained by taking the sum of all the terms of the form and collecting the terms with the same powers of x. This is precisely what one would do if you just expanded the product using the distributive law. One can extend in the obvious way, these operations to operations on rational functions. Although tedious, it is straightforward to show that the rational functions of one variable with coefficients in a field F form a field.
Recall from high school algebra that one can do long division of polynomials to obtain a polynomial quotient and a polynomial remainder whose degree is smaller than that of the divisor. In summary:
In particular, any element of F has at most n -roots. So, the list of roots we found by Corollary 8 in section 2.8.2 is complete. Let f(x) and g(x) be two polynomials with coefficients in a field F. Then f(x) is a divisor of g(x) if and only if there is a polynomial q(x) with coefficients in F such that g(x) = f(x)q(x). Note that the degrees are related by
One says that f(x) is a Basically, irreducible polynomials play the same role as did prime numbers. Just as with the natural numbers, the division theorem can be used to define the Euclidean Algorithm which allows one to calculate the greatest common divisor of two polynomials and to express it as a linear combination of the two original polynomials. The only difference here is that we compare polynomials using their degrees; so, for example, the greatest common divisor is any common divisor with largest possible degree. Details are given in the exercises. One has:
## 3.2.1 Graphing near a PointLet's start off by assuming that our polynomials f(x) have real number coefficients and that we are evaluating them for real values of the variable x. - For x near zero, the graph of y = f(x) is close to the graph of the constant function y = f(0).
- If you take the constant term and the term of lowest positive degree,
then, for x close to zero, the graph of y = f(x) is approximately the
same as the graph of this much simpler function.
- For example if the coefficient of x is non-zero, then our new function looks like . This is a straight line with slope equal to the coefficient of the x term. (To see this, graph any polynomial function on your calculator -- regardless of how curved the graph looks, it straightens out as soon as you zoom in on the value at zero.)
- If the coefficient of x is zero, but the coefficient of
is
non-zero, then our new function looks like
. This
is a parabola. If
, the parabola opens upward and the graph
takes on its smallest value
at x = 0. Our original function does
the same for x sufficiently close to zero; we say that the function has
a
**local minimum**at x = 0. If , the parabola opens downward, and so the function is biggest at x = 0; one says that the function has a**local maximum**at x = 0.
The function is approximated by for x close to zero. has a local maximum at x = 0.
So, the graph of y = f(x) near x = 2 is the same as the graph of y = g(x) near zero, and it is approximated by y = 9x + 6. So, the graph of y = f(x) near x = 2 looks like a straight line with slope 9 and passing through (2, 6). If we are interested in what the graph of y = f(x) looks like near x = -1, then consider g(x) = f(-1 + x): So, the graph of y = f(x) near x = -1 is the same as the graph of y = g(x) near x = 0, which is approximated by . So, the graph of y = f(x) has a local maximum at x = -1. As a final example, check you understanding by showing that the function f(x) has a local minimum at x = 1. where
The polynomial f'(y) is called the When x is negative and large in absolute value, will be large in absolute value, but its sign will be either negative or positive depending on whether n is odd or even. - For x near 0, the function y = f(x) behaves like its low degree terms. (If |x| is small, then is smaller still, etc.) So, y = f(x) is approximated by or more generally by the sum of the constant term and term of lowest positive degree.
- For x near the complex number a, the function y = f(x) behaves like g(x) = f(a + x) near x = 0.
- When x is large in absolute value, then f(x) is dominated by its highest degree term, and so f(x) is approximated by . In particular, if f is not a constant, then |f(x)| is large as soon as |x| is.
In the case of real polynomials, |f(x)| be smallest at x = a for all x near a even though . For example, has a local minimum at x = 0 and . This only happens because we restrict the values of x to be real. For example, if we allowed x = it, then takes on values f(it) smaller than 3 for all sufficiently small real values of t. One can do the same thing for a general polynomial. If we approximate f(x) by its constant term and its term of lowest positive degree, then . Then we can write the numbers in polar form: Then one has Now, if is non-zero, we can choose so that the angle of this last equation differs from by precisely . If we did this, then the sum of the two terms would be smaller in absolute value than provided that r is not too large. What one has shown is:
## 3.2 Rational RootsIf f(x) = 0 is a non-zero polynomial equation with integer coefficients, then there are at most a finite number of possible rational number roots:
where the coefficients are integers and and . Any rational root of this equation must be expressible in the form where p is a divisor of and q is a divisor of .
Multiplying through by gives The first n terms are all multiples of a and so a divides . On the other hand, b is a divisor of the last n terms; so b divides . The result now follows from these two assertions and
ii. The equation f(x) = 1 has no rational solutions where because both f(1) and f(-1) are non-zero. iii. Consider the quartic equation . The possible roots are 10, 10/2 = 5, 10/3, 10/6, 5, 5/2, 5/3, 5/6, 2, 2/2 = 1, 2/3, 2/6 = 1/3, 1, 1/2, 1/3, 1/6 and the negatives of these numbers. Substituting these into the equation, one eventually discovers that the first of these that is a root is x = 5/2. Since it is a root, we can apply the division theorem; x - 5/2 is a divisor of the quartic. Dividing out we are reduced to a cubic equation: . Dividing out by 2, we get The possible rational roots of the new equation are: 2/1 = 2, 2/3, 1/1 = 1, and 1/3 as well as the negatives of all these numbers. Checking each of the possibilities, we see that x = 2/3 is a root. Again, we can divide by x - 2/3 to get This is a quadratic; so we know we can solve it completely. In our case the two roots are . So the original equation has rational roots 5/2 and 2/3 as well as two complex roots .
## 3.3 The Fundamental Theorem of AlgebraWe will not be ready to prove this result until we get to Chapter 5, but we can at least state the result and see a few of its consequences.
The
- If r is a real number, then . Conversely, if , then z is a real number.
- The complex conjugate of the complex conjugate of any complex number is equal to the original complex number.
Expanding the right side gives which is the same as the expansion of the left side.
Given a polynomial
with
complex coefficients, one
can define the
So, f must have no real roots. Suppose f has a non-real complex root a. Then Corollary 5 tells us that is also a root of f. Applying the Division Theorem twice, we get where g has complex coefficients. Now, Now, the coefficients of this quadratic are equal to their complex conjugate; so the last proposition says that the quadratic has real coefficients. Since f also has real coefficients, also has real coefficients. But g has degree less than the degree of f; so it is a product of zero or more linear and quadratic polynomials with real coefficients. But then so is f(x), which again is a contradiction.
## 3.4 Cubic EquationsWhereas the solution of quadratic equations was known in some form from the time of the Babylonians, the solution of the general cubic equation was first discovered perhaps by Tartaglia and published by Cardan during the Renaissance. It represented a giant leap forward; the most spectacular achievement showing for the first time in close to two millenia that the great achievement of the classical civilizations could be bettered. The problem is to find solutions of the general cubic equation where a, b, and c are complex numbers. In the case of a cubic, we can attempt the same trick. And it does help somewhat. In this case we have a cubic and so, if we cube (x + d) we would get . Comparing this with our cubic, we see that we can make the quadratic term match by taking d = a/3. This gives: This looks very complicated, but the idea is that, if we think of x + a/3 as being a new variable, say y, then the polynomial is a cubic in y without any quadratic term. We have shown:
The above reduction step was well known at the time and does not represent any great achievement. To understand the next step, again consider the case of quadratics. If has roots and , then the Division Theorem tells us that . Multiplying this out, we get Comparing these with our original coefficients, we see that
This last lemma was also a well known property of quadratic equations. Now, let's try to solve the cubic . The real achievement of del Ferro or Tartaglia was to see that the relation between the problem of solving the cubic, Lemma 3, and the formula for cubing a binomial: If you simply factor 3uv out of the two middle terms, the formula becomes: Now, thinking of u + v as being the variable, this says that a cubic is equal to a linear, i.e. it is like the equation where So, if we could find u and v so that these two equations were satisfied, then x = u + v would be the root of the cubic. Now, compare these two equations to Lemma 3. It tells us that, in order to make a quadratic with roots and , we just let the first degree term be and the constant term be . So, the quadratic is Since we know how to solve every quadratic, we can find the roots of this quadratic. If and are the two roots, then u and v are simply the cube roots of these two roots, choosing them so that 3uv = -a. But then x = u + v is a solution to the original cubic and we have found a root of the cubic! By varying the cube roots, one expects to obtain the various solutions of the equation. But something is wrong here. We know that there are three cube roots and so pairing these up with each other, one expects to get 9 different pairs (u, v). But we know that a cubic has only 3 roots! The problem is that we have created extraneous solutions. What we did was to find pairs (u, v) so that . But our actual equation was uv = -a/3. So, it is not enough to choose any pair of cube roots of and . We need to choose the pairs so that uv = -a/3. So, suppose our original pair was chosen so that uv = -a/3. If is a cube root of unity other than 1. Then the other cube roots of u are and , and similarly for v. If we want another solution, we can choose for the new u. But then we must choose for the new v in order for the product to give the same value -a/3. Similarly, if we choose for the new u, we have to choose for the v. So, we see that there are only three possible solutions and not nine. This completes the solution of the general cubic.
Our binomial expansion is . Matching up with the original equation, we would want and . The quadratic that would roots and is . The quadratic factors as (x + 4)(x - 2) = 0. So the roots of the quadratic are 2 and -4. Let u = and where we have chosen the roots so that uv = -2. One root is . To get the other roots, take a cube root of unity . Then the other two roots will be and .
The quadratic formula tells us that the roots of the quadratic are . One of the roots is positive and the second root is negative. So, if you take the positive cube root of the positive one and the negative cube root of the negative one, then their product will be negative; so their sum will be one root. To get the other two roots, first multiply one cube root by and the other by . Then multiply the first by and the second by omega. ## 3.5 Quartic EquationsFerrari, a student of Cardan, showed that one could find the roots of the general quartic equation using techniques similar to those used for cubics. The story stops here. According to Abel's Theorem, there is no general formula for expressing the roots of a general polynomial equation of degree five or higher involving only arithmetic operations and roots. Some fourth degree (quartic) polynomials are easy to solve. For example, consider: . This can be factored If the product is zero, then one factor or the other is zero. So, solving for the roots of the quartic is the same as solving two quadratic equations, which is something that one knows how to do. More generally, if we can arrange the equation to be solved into the form: where p and q are quadratic polynomials, then the roots are the roots of and . As it will turn out, every fourth degree polynomial can be put in this form. Let's start with a general fourth degree polynomial with complex number coefficients. Dividing through by the leading coefficient, one needs to solve an equation of the form . First, we can assume that a = 0. In fact, if it were not, then one could apply the same trick as we did with cubics. The first two terms look like the first two terms of the expansion of and so substituting x = y - a/4 converts our polynomial in x into one in y with no cubic term. If one could solve the new polynomial in y, then using x = y - a/4, one would get the roots of the original polynomial. So, we can assume the form of the equation is: . We can complete the square using the first two terms: . We have one square on the left, we just need to make the right side a square - but that is not promising because the right side is only linear in x. The problem is that we didn't quite choose the right constant term on the left side of the equation -- but how should we choose the right constant term? Since we don't know, we can simply adjust by adding y to the constant term. Our new equation becomes where we added the last two terms on the right in order to make the two sides equal. Now, we have a square on the left and we need to choose y so that we have a square on the right. But the right side is a quadratic in x and so, we just need to make sure that this quadratic has both roots equal. (If they were both equal to , then the right side would factor as as desired.) We need the following result which is an obvious consequence of the quadratic formula:
Applying Lemma 4 to the right side of the equation, we see that we need y to satisfy: But this is just a cubic equation in y. We can use the method of the last section to find a y which satisfies this equation. Then, we can factor the right side -- the roots are by the quadratic formula. So, our formula is just Since both sides are squares, we see that we can find the roots of the quartic by solving two quadratics, viz. This completes the solution of the general fourth degree polynomial equation. All contents © copyright 2001 K. K. Kubota. All rights reserved |