Unknown

268 Answers to Exercises and Solutions to Problems

(b) Setting u = x - 1 in the expression yields

uv2{-nu - [l - (1+ u)“]} = 2 ( ; ) uk-2 = 2 ( ; ) (x - I)?

kc2 k=2

5.4.

(1 - xs)n = [(l - x)(1 + x + x2)]”

= (1 - z)“[(l - x)” +3x]”

= (1 - x)“[(l - x)2n + n(3x)(l- x)2+-2 + ...I

which yields the result.
5.5. Let u be a root of multiplicity exceeding 1. If u = 0, the condition (a)
follows immediately. Let u # 0. Then au2 + 2bu + c = 0 implies that

O=(uu3+3bu2+3cu+d)-u(uu2+2bu+c)=bu2+2cu+d

whence

0 = u(bu2 + 2cu + d) - b(au2 + 2bu + c)

= 2(uc - b2)u + (ad - bc)

and

0 = c(bu2 + 2cu + d) - d(uu2 + 2bu + c)

= u[(bc - ud)u + 2(c2 - bd)].

Condition (a) and result (b) follow immediately.
On the other hand, assume that the condition in (a) holds. Since

u(ux3 + 3bx2 + 3cx + d) = (ax + b)(ax2 + 2bx + c)

+ [2(uc - b2)x - (bc - ad)],

any common zero of ax2 + 2bx + c and 2(uc - b2)x - (bc - ad) must be at
least a double zero of the cubic. But the quadratic and linear polynomials
have a common zero if and only if

u(bc - ud)2 + 4b(ac - b2)(bc - ad) + 4c(uc - b2)2 = 0.

This is ensured by using condition (a) to substitute for (bc - ud)2.
Remark. Consider the case that ac - b2 = 0. If the cubic equation has
a double root, then we must have ad - bc = 0. On the other hand, if (a)
holds, then ad - bc = 0 and

u2(ux3+3bx2+3cx+d) = (ux+b)3+(3ux+b)(uc-b2)+u(ad-bc) = (ux+b)3.