Two roots of the equation x^3+px+q are -1and 3.The third root is?

Answer 1

let f(x) = x^3 + px + q
-1 is a root, so
f(-1) = 0 = -1 -p+ q or
-p + q = 1
p - q = -1

also f(3) = 0
27 + 3p + q = 0
3p + q = -27

solve for p and q, I would add them to get p,
then you can find q
sub back into the original and then factor to get the last root

or

since -1 and 3 are roots, let the third root be a
(x+1)(x-3)(x-a) = x^3 + px + q
(x-a)(x^2 - 2x - 3) = x^3 + px + q
x^3 - 2x^2 - 3x - ax^2 + 2ax + 3a = x^3 + px + q
x^2(-2 - a) + x(-3 + 2a) + (3a) = px + q

since there was no x^2 term, -2-a = 0, a = -2
alos -3 + 2a = p
p = -3 - 4 = -7
q = 3a = -6

we would have x^3 - 7x - 6

check: What is (x+1)(x-3)(x+2)
= (x+2)(x^2 - 2x - 3)
= x^3 + 2x^2 - 3x - 2x^2 - 4x - 6
= x^3 - 7x - 6
as expected