How do I solve and prove this polynomial question?

Question

Factorise the polynomial f(x) = x^3 - (2p + 1)x^2 + (2p -q)x + q where p and q are constants. If the equation f(x) = 0 has three real roots, show that p^2 + q >= 0.

Full workings for it would be much appreciated. I need this for my mathematics exam revision. Thanks alot!

Answer 1

try x = 1 and put it into your polynomial.
result = 0
therefore x = 1 is a root and (x-1) is a factor of your polynomial.

f(x) = (x - 1) (x^2 -2px -q)

if it has 3 real roots, besides x = 1, there are 2 more real roots.

consider x^2 -2px -q, for it to have 2 real roots, the discriminant must be >= 0.

discriminant = (2p)^2 - 4 (-q) = 4p^2 + 4q >= 0
=> p^2 + q >= 0
note that when p^2 + q = 0, you will get repeated real roots. thus, you will get 1 distinct and 2 repeated real roots.

Answer 2

I tried finding an integer root first. Let's try x = 1.
Lo, it works!! So, by the factor theorem, x-1 is a
factor of f(x). By long division or synthetic division,
the quotient is x^2 -2px -q. Since we want
the quotient to have 2 real roots, the discriminant
of this quadratic must be nonnegative.
This gives 4p^2 + 4q >=0,
so p^2 + q >=0.

Answer 3

Run it through the formula, and you get:

x = p+x^2-q