Schaum's Outline of Theory and Problems of Beginning Calculus

52 FUNCTIONS AND THEIR GRAPHS [CHAP. 7EXAMPLES(a)3 is a root of the polynomial 2x3 - 4x2 - 8x + 6, since2(3)3 - 4(3)2 - 8(3) + 6 = 2(27) - 4(9) - 24 + 6 = 54 - 36 - 24 + 6 = 0X' - 2~ - 15(6) 5 is a zero of becausex+2(5)2- 2(5) - 15-25 - 10 - 15--=005+2 7 7Theorem 7.1: If f(x) = a, x" + a,- lx"- ++ a,x + a, is a polynomial with integral coefficients(that is, the numbers a,,, a,,-,, . . . , a,, a, are integers), and if r is an integer that is a rootoff, then r must be a divisor of the constant term a,.EXAMPLE By Theorem 7.1, any integral root of x3 - 2x2 - 5x + 6 must be among the divisors of 6, which aref 1, f 2, f 3, and f 6. By actual substitution, it is found that 1, - 2, and 3 are roots.Theorem 7.2:EXAMPLESA number r is a root of the polynomialf(x) = a,x" + a,-,xn-~ + * * * + a,x + a0if and only iff(x) is divisible by the polynomial x - r.Consider the polynomialf(x) = x3 - 4x2 + 14x - 20. By Theorem 7.1, any integral root must be a divisor of20; that is, f 1, f 2, f 4, f 5, f 10, or 20. Calculation reveals that 2 is a root. Hence (Theorem 7.2), x - 2must divide f(x); we carry out the division in Fig. 7-1qa). Since f(x) = (x - 2)(x2 - 2x + lO), its remainingroots are found by solvingApplying the quadratic formula (see Problem 5.2),x2 - 2x + 10 = 0Thus, the other two roots off(x) are the complex numbers 1 + 3- and 1 - 3 n .Let us find the roots of f(x) = x3 - 5x2 + 3x + 9. The integral roots (if any) must be divisors of 9: f 1, +, 3,f9. 1 is not a root, but - 1 is a root. Hencef(x) is divisible by x + 1 [see Fig. 7-1qb)l. The other roots off(x)must be the roots of x2 - 6x + 9. But x2 - 6x + 9 = (x - 3)2. Thus, - 1 and 3 are the roots off(x); 3 is calleda repeated root because (x - 3)' is a factor of x3 - 5x2 + 3x + 9.xz - 2x + 10x - 2 I x3 - 4x2 + 14x - 20x3 - 2x2- 2x2 + 1 4~- 2x2 + 4x1ox - 201ox - 20x2 - 6 ~ +9x+ 1 )x3-5x*+3x+9x3 + x2- 6x2 + 3~- 6x2 - 6~9x + 99x + 9Theorem 7.3:(Fundamental Theorem of Algebra): If repeated roots are counted multiply, then everypolynomial of degree n has exactly n roots.