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Solutions to Problems; Chapter 2 267

(b) Let q(x) = 6x5 - 15 x4 - 10x3 + 30x2 = x2(6x3 - 15x2 - 10x + 30).

The number of zeros of q(x) + k: is the number of times the line y + !C = 0

intersects the graph of y = q(x). F’r om the graph we see that q(x) + k has

exactly one simple real zero when k < -19 or JC > 8;

exactly three simple real zeros when -19 < L < -11 or 0 < JZ < 8;

exactly five simple real zeros when -11 < k: < 0;

one double and one simple real zero when k = -19, 8;

one double and three simple real zeros when k = -11, 0.

YA

(-1,191

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(2, -8)

Solutions to Problems

Chapter 2

5.1. Let f(x) = x(x - 1) .. .(x - n + 1) - k. Then f’(x) takes values
of alternate signs at x = 0, 1,... , n - 1. Hence, f’(x) has n - 1 distinct
zeros, one between each pair i - 1, i of integers (1 5 i < n - 1). Since
degf’(z) = n- 1 , each of these zeros must be simple. Hence, a zero of f(x)
has multiplicity at most 2. We can make any zero r of f’(x) a zero of f(z)
as well by choosing k = r(r - 1)... (r - n + 1).
5.2. If all three have a common zero, the result is trivial. Otherwise, the
polynomials can be written u(x-u)(x-v), b(x-u)(x-to), c(x-v)(x-w)
where a, b, c are positive and u < v < w. Their sum is positive at u and
w, negative at v, and so the sum has a zero between u and v, as well as
between v and w.
5.3. (a) Putting the expression over a common denominator (1 - x)~ yields
a numerator n(1 - x) - (1 - x”). This vanishes along with its derivative
when x = 1, so that it is divisible by (1 - x)“. We find that the given
expression is equal to

2 n-2 + 2xnm3 + 3xnm4 + 1.. + (n - 2)x + (n - 1).