Unknown

Solutions to Problems; Chapter 4 307

2(n - m) of the a;. Suppose the former occurs. Then either g(t) - 1 or

g(t) + 1 must vanish for at least m+ 1 of the oi, contradicting the fact that

deddt) f 1) = m. Hence, the factorization f = gh is not possible and f(t)

is irreducible.

Solutions to Problems

Chapter 4

7.1. The equation can be rewritten (x - u)” - 6u2x2 = 0. Factoring the left
side as a difference of squares leads to two quadratic equations.

7.2. Let u = x2 + x - 1, v = x2 - x - 2, so that u - v = 2x + 1. Let

g(x) = (x2 + x - 1)4 - (x” - 2 - 2)4 - (2x + 1)”

= u4 - v4 - (u - v)” = 2V(U - V)(2U2 - uv + v”)

Then g(x) = 0 * (1) v = 0 q x = -1 or x = 2

or (2) u - 2) = 0 =j 2 = -l/2

or (3) 0 = 2u2 - uv + v2 = 2x4 + 2x3 - x2 + x + 4.

In case (3)) we have 4~ = (1 f iJ?) v, which yields the quadratic equations

(3 - iJs)x2 + (5 + iJ5)x + (-2 + 2iJ5) = 0

(3 + h/7)x2 + (5 - il/T)x + (-2 - 2ifi) = 0

for the remaining solutions. [As a check, note that the product of the two
quadratic polynomials is equal to

(322 + 52 - 2)” + 7(x2 - x - 2)2 = 8(2x4 + 2x3 - x2 + 2 + 4).]

7.3. The substitution x = 1 - t yields the equation

t4 - 2t3 - 4t2 + 2t + 1 = 0

or (t2+tv2)-2(t-t-l)-4 = 0. Set u = t-t-‘, whence u2 = (t2+tm2)-2.
IIence u2 - 2~ - 2 = 0. We solve successively for U, t and x.

7.4. The equation can be rewritten

(x2 + 5ux + 5~2~)~ - u4 = b4,

whence x2 + 5ax + 5a2 = &/w (cf. Problem 33.20).

7.5. The equation implies z+ufi+b = x+c2-2cfi, whence (u+~c)~x =
(c2 - b)2.