If p_n is positive, then f(x)=0 has a root lying between 0 and -\infty, and if p, is negative f(x) =0 has a root lying between 0 and + \infty.
601. Every equation which is of an even degree and has its last term negative has at least two real roots, one positive and one negative.
For in this case
S(+\infty)=+\infty, f(0)=p_n, f(-\infty)=+\infty;
but p, is negative; hence f(x) = 0 has a root lying between 0 and +\infty, and a root lying between 0 and -\infty.
602. If the expressions f(a) and f(b) have contrary signs, an odd number of roots of f(x) = 0 will lie between a and b; and if f(a) and f(b) have the same sign, either no root or an even number of roots will lie between a and b.
Suppose that a is greater than b, and that c, d,e, k represent all the roots of f(x) =0, which lie between a and b. Let \phi(x) be the quotient when f(x) is divided by the product (x — c) (x — d) (x — e) --- (x—k); then
f(x)=(x — c) (x — d) (x — e) --- (x—k)\phi(x). Hence f(a) =(a — c) (a — d) (a— e) --- (a—k) \phi(a). f(b)= (b — c) (b— d) (b — e) --- (b—k) \phi(b).
Now \phi(a) and \phi(b) must be of the same sign, for otherwise a root of the equation \phi(x)=0, and therefore of f(x) =0, would lie between a and b [Art. 599], which is contrary to the hypothesis. Hence if f(a) and f(b) have contrary signs, the expressions
(a—c)(a—d) (a—e) --- (a—k),
(b —c)(b—d) (b—e) --- (b—k).
must have contrary signs. Also the factors in the first expressions are all positive, and the factors in the second are all negative; hence the number of factors must be odd, that is, the number of roots c, d, e, ---k must be odd.
Similarly if f(a) and f(b) have the same sign the num-
