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Hints 79

5.3. Put over a common denominator and check the numerator for a dou-

ble zero at x = 1.

(b) Set u = x - 1 and expand binomially.

5.4. 1-x3 = (l-x)[(l-x)~+~x]. Expand the nth power of the expression

in square brackets binomiahy.

5.5. Observe that

ax3 + 3bx2 + 3cx + d = x(ax’ + 2bx + c) + (bx’ + 2cx + d)

a(ax3 + 3bx2 + 3cx + d) = (ax + b)(ax’ + 2bx + c)

+ [2(ac - b2)x - (bc - ad)]

d(ax3 + 3bx2 + 3cx + d) = (cx + d)(bx’ + 2cx + d)

+ x’[(ad - bc)x + 2(bd - c”)].

Note that a root of the equation has multiplicity exceeding 1 if and

only if it is a zero of ax2 + 2bx + c.

5.6. The right side is

Interchange the order of summation, and interpret the i-sum as part

of a binomial expansion.

5.7. If p(x) is nonconstant, then p(t) - tm has infinitely many zeros for

some value of m.

5.9. Differentiate the identity.

5.10. Use induction, noting that (z + y)(“+‘) = (x + y)(x + y - l)(m).

5.11. What must the degree of y be? Differentiate the equation three times

and work backwards.

5.13. The zero of the second derivative of the cubic is a zero also of the

cubic itself and its first derivative.

5.14. 0 can be at most a simple zero of the quintic.

5.15. Look at the function for small values of n and make a conjecture.

5.16. If r is a zero, so are r2 and (r - 1)2. What are the possible zeros of

f(x)?

5.17. If u is a zero, then so is --u. Thus, the polynomial has the form
x(x - u)(x + u)( x2 - v). Setting x =^10 indicates that we should look
for two divisors of -2967 which sum to 20.

5.18. What does the expansion remind you of? (Change x to Q.)