2011/051) if a+ib is root of equation x^3+px + q = 0 the form an equation whose root is 2a

and p and q are real

a + ib is a root and coefficients are real so another root = a - ib

sum of roots = 0 (coefficient of x^2 =0) so 3rd root = -2a

now -2a is root of f(x) = x^3+ px + q = 0

so 2a is root of f(-x) = (-x)^3 - px + q = 0 or x^3+px -q = 0