STUDENT’S COMPANIONS IN MATH: NINETH Polynomials and Algebraic Equations A polynomial is a function of the form n n−1 p(x) = a0x + a1x + ··· + an−1x + an. (1) where a0, a1, a2, . , an are some constants, called the coefficients of p. Using the summation notation, we may write Xn n−k p(x) = akx . k=0 In (1), when a0 6= 0, we say that the degree of p is n and we call a0 the leading coefficient. When p is a nonzero constant, we say that the degree of p is zero. When p is equal to the constant zero, it is customary assign ∞ to be the degree of p. We are mainly interested in nonconstant polynomials and there is no need to worry about this convention. By an algebraic equation we mean an equation of the form p(x) = 0. By a root of the polynomial p, or a solution to the algebraic equation p(x) = 0, we mean a (complex) number a such that p(a) = 0. QUESTION 1. If 1 and 2 are roots of x3 − x2 − cx + d, what are c and d? EXERCISE 2. Prove the following statement: if α and β are roots of x2 + ax + b and α 6= β, then −a = α + β, b = αβ. (2) (Actually, the assumption α 6= β is unnecessary. We will mention a complete set of identities giving the relation between roots and coefficients of a polynomial, traditionally called Vieta’s formula; see (12) below.) If the degree of p is two, we call p(x) = 0 a quadratic equation, if three, we call a cubic equation, if four, we call a quatric equation, and if five, we call a quintic equation. Next we discuss the general quadratic equation p(x) = 0, where p(x) = ax2 + bx + c with a 6= 0. An important technique for dealing with a polynomial of degree two is called completing square, which is based on the identity a2 + 2ab + b2 = (a + b)2. 1 EXAMPLE 3. We are asked to do the “completing square” for the following polynomials: x2 + 6x + 7, 2x2 + 2x and 3x2 − 2x + 1. Here they are: x2 + 6x + 7 = x2 + 2 . 3x + 32 − 32 + 7 = (x + 3)2 − 2, 2x2 + 2x = 2(x2 + x) = 2(x2 + 2(1/2)x + (1/2)2 − (1/2)2) = 2(x + 1/2)2 − 1/2, 3x2 − 2x + 1 = 3(x2 − 2(1/3)x + (1/3)2 − (1/3)2) + 1 = 3(x − 1/3)2 + 2/3. As you can see, “completing square” is like a tailor’s job of fitting a given expression into the form (x + a)2, keeping but without worrying what is left over. Now we return to the general polynomial of degree two: p(x) = ax2 + bx + c, with a 6= 0. In the following discussion we assume that a, b, c are real numbers. We perform “completing square” for this polynomial: µ ¶ b p(x) ≡ ax2 + bx + c = a x2 + x + c a à ½ ¾ ½ ¾ ! b b 2 b 2 = a x2 + 2 x + − + c 2a 2a 2a à ½ ¾ ! b b 2 b2 = a x2 + 2 x + − a + c 2a 2a 4a2 Or µ ¶ b 2 4ac − b2 p(x) = a x + + . (3) 2a 4a For convenience, let us write 2 p(x) = a(x − x0) + m, (4) 2 2 where x0 = −b/2a and m = (4ac − b )/4a. Notice that p(x0) = m and (x − x0) is 2 always nonnegative. So, when a > 0, we have p(x) = a(x−x0) +m ≥ 0+m = m = p(x0). Thus, when a > 0, p has a unique (global) minimum m attained at x0. Similarly, 2 when a < 0, we have p(x) = a(x − x0) + m ≤ 0 + m = p(x0) and hence p has a unique maximum m attained at x0. QUESTION 4. How do you use calculus to draw the same conclusion? QUESTION 5. How do you use (3) or (4) to derive the following formula √ −b ± b2 − 4ac x = (5) 2a 2 for finding the roots of ax2 + bx + c = 0? We return to the general theory of polynomials. You are assumed to know long division for polynomials. See if you can do the following EXERCISE 6. Use long division to divide f(x) = 2x4 − 7x3 + 14x + 4 by g(x) = x − 2. Now we have to be more careful about numbers that we are allowed to use. To avoid technicality, we restrict ourselves to the following three number systems: rationals, reals and complex numbers. The standard notation for them is: Q = the set of all rational numbers R = the set of all real numbers C = the set of all complex numbers. In what follows, we use letter F (this is not a standard notation) to stand one of the above: Q, R or C. We use this letter F here because it is a field according to some technical definition. When a polynomial all coefficients a0, a1, . , an of a polynimial n n−1 p(x) = a0x + a1x + ··· + an−1x + an are in F, then we say that p is a polynomial over F. Given polynomials f(x) and g(x) over F, we can divide f(x) by g(x), using long division, to get a quotient q(x) and a remainder r(x), which are also polynomials over F. The relation between f, g and q, r is given by the identity f(x) = g(x)q(x) + r(x). Here, the degree of r(x) is strictly less than the degree of g(x), or r(x) is identically zero. (Note: the degree of a nonzero constant polynomial is 0 but the degree of the zero polynomial is defined to be infinity.) In case r(x) ≡ 0 so that we have f(x) = g(x)q(x), we say that g is a factor of f or g divides f. When f has no factors over F other than the trivial ones, that is, constants or constant multiples of f, we say that f is an irreducible polynomial over F. Notice that reducibility of a polynomial often depend on which field F we choose. For example, x2 + 1 is a polynomial irreducible over Q or R. But it is reducible over C because it has the following proper factorization: x2 + 1 = (x + i)(x − i). QUESTION 7. Why is x2 − 2 is irreducible over Q but reducible over R? Now we state the basic theorem concerning factorization of polynomials: A noncon- stant polynomial can be written as a product of finitely many irreducible polynomials 3 which are unique up to multiples of nonzero elements in F. The proof of this is similar to the unique factorization of integers. A good book on abstract algebra should have a theory (about something called PID, that is, principal ideal domain) covering the unique factorization theorem for both polynomials and integers. PROBLEM 8. Why is x4 + 1 irreducible over Q? Recall that a number a is called a root of a polynomial if p(a) = 0. Given a polynomial p(x) and a number a, we can divide p(x) by x − a to get p(x) = (x − a)q(x) + r, where q is a polynomial with a degree smaller than that of p and the remainder r is a number. Now suppose that a is a root of p(x): p(a) = 0. Letting x = a in p(x) = (x − a)q(x) + r, we obtain 0 = r. Hence p(x) = (x − a)q(x). We have shown that if a is a root of p(x), then x − a is a factor of p(x). QUESTION 9. Why is the converse “if x − a is a factor of p(x), then a is a root of p(x)” also true? Next, suppose that b is another root: p(b) = 0 and b 6= a. Putting x = b in p(x) = (x − a)q(x), we get 0 = (b − a)q(b). As b − a 6= 0, we have q(b) = 0. Thus b is also a root of q(x) and hence x − b is a factor of q(x), say q(x) = (x − b)Q(x). Thus p(x) = (x−a)(x−b)Q(x), telling us that (x−a)(x−b) is a factor of p(x). More generally, if a1, a2, . , am are distinct roots of a polynomial p(x), then (x−a1)(x−a2) ··· (x−am) is a factor of p(x). EXERCISE 10. Give a careful proof of the last assertion by inductiion on m. One consequence of our discussion here is: If p(x) is a polynomial of degree n, then p(x) cannot have more than n roots. To see this, suppose that p(x) has more than n roots, say a1, a2, . , am with m > n. Then, according to what we have just learned, f(x) ≡ (x − a1)(x − a2) ··· (x − am) is a factor of p(x). This cannot happen because the degree of f(x) is m, which is greater than the degree of p(x), which is n. Indeed, the degree of a factor of p cannot be greater than the degree of p. Next we study the multiplicity of a root. Suppose that a is a root of a polynomial p, that is, p(a) = 0. The above discussion tells that p(x) = (x − a)q(x) for some polynomial q. Now we ask the question: is a also a root of q? The answer can be found by computing 4 q(a), the value of q at a, to see if it is zero.