x^3-(p+q-r)x^2+(pq-pr-qr)x+pqr

Question

How to factor this??

Answer 1

x^3-(p+q-r)x^2+(pq-pr-qr)x+pqr
If it factors, the p, or q, or r must be solutions to
x^3-(p+q-r)x^2+(pq-pr-qr)x+pqr = 0
let f(x) = x^3-(p+q-r)x^2+(pq-pr-qr)x+pqr
try x = p
f(p) = p^3 -p^2(p+q-r) + (pq - pr - qr)(p) + pqr
= p^3 - p^3 - qp^2 + rp^2 + qp^2 - rp^2 - prq + pqr
= 0
so (x-p) is a factor
in the same way, I can show that f(q) = 0
so x-q is a factor

leaving me with
(x-p)(x-q)(x .... r) = 0
realizing where the first and last terms come from
x^3 comes from (x...)(x...)(x....)
and pqr comes from (.. -p)(.. -q)(.. ?? r)
so it must be +r

x^3-(p+q-r)x^2+(pq-pr-qr)x+pqr
=(x-p)(x-q)(x+r)

Answer 2

also check out the way the coefficients are sums of subsets of the roots:

http://mathforum.org/library/drmath/view/61024.html