posted by Connor .
Determine K and solve the equation x^3-kx^2+3x+54=0, if one of its zeros is triple of another.
let the roots be a, 3a and b
(x-a)(x-3a)(x-b) = x^3 - kx^2 + 3x + 54
looking at the last term -3a^2b = 54
a^2b = -18 , the only square factors of -18 are 1 and 9
case 1: a=1, b = -18
(x-1)(x-3)(x+18) would be the expression
this give us x^3 + 14x^2 - 69x + 54
which would not match up the x terms
case 2: a=3, b=-2
(x-3)(x-9)(x+2) would be the expression for
x^3 - 10x^2 + 3x + 54
this matches if k = 10
and the roots are 3,9 and -2
(there should be an easier way to do this)