Solutions to Problems; Chapter 2 271
x = 0 yields g(0) = 1, g’(0) = 3, g”(0) = 12. Since degg = 2, use Taylor’s
Theorem to obtain g(x) = 6x2 + 3x + 1. Hence, f(x) = 6x5 - 15x4 + 10x3.
5.13. Suppose usx3 + azx2 + ~13: + uc is the cube of a linear polynomial.
Then it must have a triple zero u which is a zero of its first and second
derivatives. Hence u = -u2/3us. Since 3usu2 + 2uzu + al = 0, we find that
ui = 3uius. Finally, a3u3 + u2u2 + uiu + uo = 0 leads to 9uou3 = aluz.
On the other hand, suppose that the conditions are satisfied. Let p =
a3 1’3, q = a0 1’3. Then, it is rea dily checked that the cubic is equal to (px+q)3.
5.14. (x - 2)(x - 1)x(x + 1)(x + 2) = x5 - 5x3 + 4x has three nonzero
coefficients. If p(x) has only two nonzero coefficients, it must have one of
the forms x5 - a, x5 - bx, for otherwise 0 would be at least a double root.
But neither possibility has five integer solutions. Hence k = 3.
5.15. For small values of n, we find that fc(x) = 1, fi(x) = x, fz(x) =
x(x + 2), f3(2) = x(x + 3)2. A s an induction hypothesis, let n^2 1 and
sunldpose that fn(x) = x(x + n)“-‘. Then f,,(x) = (x + n)” - n(x + n)“-’
fA+l(x) = (n + 1)(x + n + 1)” - (n + l)n(x + n + 1),-l.Hence
fn+l(x) = (x + 12 + l)n+l - (n + 1)(x + n + l)n = x(x + n + l)n.In particular, ficc(l) = 101”.
5.16. If f(x) is constant, then f(x) = 0 or -1. Otherwise, let r be a zero
off. Then r2, r4,... must also be zeros. Since f has at most finitely many
zeros, Irl can take only the values 0 and 1. Also, (r - 1)” is a zero of f,
so Ir - 11 = 0 or 1. Consulting a sketch of the complex plane assures us
that the only possibilities are r = 0 and r = 1. Hence f(x) = ax”‘(x - 1)“.
Trying this out leads to a = -1, m = n.
5.17. The polynomial must have the form x5 + ax3 + bx. If u is a non-
trivial zero, then so is -u, and the polynomial can be factored to x(x - u)
(x + u)(x2 - v) where u2v = b, u2 + v = --a. Substituting x = 10 yields
(10 - u)( 10 + u)( 100 - v) = -2967 = -3.23.43. Now 10 - u and 10 + u are
a pair of divisors of 2967 which sum to 20; the only possibilities are given
by u = 13 and u = 33. Thus, there are two polynomials which satisfy the
conditions: x5 - 226x3 + 9633x and x5 - 1186x3 + 105633x.
5.18. The expression is equal to the Taylor expansion of f(0) about the
point 2. If f(0) = 0, i t vanishes identically and the degree is -oo. If f(0) #
0, it is constant and the degree is 0.
5.19. (a) p(t) = t2 + 1.
(b) For convenience, let f(0) = x, f’(0) = y, f”(0) = z; g(0) = u,
g’(0) = v, g”(0) = w. That cf is in A is obvious. Suppose that x and u are