Solutions to Problems; Chapter (^3 285)
third zero is -(r + l/r). S ince -c/a is the product of the zeros and b/a is
the sum of all products of pairs of the zeros,
a2 -= - c2
ab
1 - wd2 = 1
b/a ’
(b) The zeros of cx3+bx2+u are r, l/r, -(r+l/r)-l. Hence both cubits
are divisible by x2 +px + 1 = (x - r)(x - l/r).
7.6. If the polynomial is equal to (UX + vy + w)(qx + ry + s), then uq = 3,
ur + vq = 2p, vr = 2, us + wq = 2a, us + wr = -4, ws = 1. By multiplying
each factor by a suitable constant (the first by s, the second by w) if
necessary, we may suppose that w = s = 1. Hence uq = 3, u + q = 2u, so
that u and q are the zeros of t2 - 2at + 3 = (t - a)2 - (a2 - 3). Also vr = 2,
v + r = -4, so that v and r are zeros of t2 + 4t + 2. We may assume that
r=-2+&andv=-2-d.
Hence,
2p = ur + vq = u(r - v) + (U + q)v = u(2Jz) - 2(2 + Jz)a
3 p+2a=(u-u)JZ
- p2 + 4ap + 4a2 = 2(~ - a)” = 2(a2 - 3)
- p2+4ap+2u2+6=0.
7.7. Let the polynomial be t3 + at2 + bt + c and its zeros be r, s and rs.
Then (1 + r)(l + s) = 1 - a and rs(1 + r)(l + s) = b - c. If a # 1, then
rs = (b-c)/(l- a ) is rational and t - rs is a factor of the cubic. (Observe
that, in fact, 1 - a is a divisor of b - c, since rs must be an integer.) If
a = 1, then, say, r = -1 and b = c. In this case, t + 1 is a factor of the
cubic.
7.8. Setting x = 0 and x = 1, we find that a must divide both 90 and 92.
Hence a = -2, -1, 1 or 2. Since x2 - x - 2 = (x - 2)(x + l), and 2 and -1
are not zeros of xl3 + x + 90, a # -2. Similarly, a # 1. Thus, a = -1 or
a = 2.
If u is a zero of x2 - x - 1, then u2 = u + 1, whence u4 = 3u + 2,
ud = 21~ + 13, u12 = 144~ + 89, u l3 = 233~ + 144 # -u - 90. Hence u is
not a zero of xl3 + x + 1. Thus a # -1.
1fvisazeroofx2-x+2,thenv2=v-2,v4=-3v+2,vs=-3v-14,
V12 = 45v - 46, v13 = -v - 90. Hence both zeros of x2 - x + 2 are zeros of
xl3 + x + 90. Thus, a = 2.
Checking, we find that
x13+x+90=(x2-x+2)(x11+x10-xg-3xd-x7+5x6+7x5
- 3x4 - 17x3 - 11x2 + 23x + 45).
7.9. It is the cube of b(b - a - c).