maths - 3 Questions on polynomial

First, there is some typos in Q 19. It should be 12x^3 - 4x^2 - 3x +1.

18) Let f(x) be 8x^3-20x^2+6x+9
Differentiate it with respect to x to get f'(x) = 24x^2-40x+6
Solve f'(x) = 0 to get 3/2, 1/6
Use factor theorem and find that f(3/2) = 0 and f(1/6)= 4/9 which is not 0.
Therefore, 3/2 is one of the root.
By using long division, f(x)/(3x-2)^2 = (2x + 1) ==> another root is -1/2.

19) We know that there are 3 roots, namely x1, x2 and x3 and W.L.O.G., let x1 + x2 = 0
Sum of roots = x1 + x2 + x3 = 4/12 = 1/3
x3 = 1/3
By using long division, (12x^3 - 4x^2 - 3x + 1)/(3x - 1) = 4x^2 - 1.
Solve it to get the other 2 roots which is 1/2 and -1/2.

20) Let the roots be x1, x2, x3 and x4 and x1 + x2 = 0
Sum of roots = x1 + x2 + x3 + x4 = -5/6 ==> x3 + x4 = -5/6
x1x2x3 + x1x2x4 + x1x3x4 + x2x3x4 = 15/6
x1x2(x3 + x4) + x3x4(x1 + x2) = 15/6
-5(x1x2)/6 = 15/6
x1x2 = -3
x1(-x1) = -3
x1 = sqrt[3] (Assume x1 > 0)
x2 = -sqrt[3]
x1x2x3x4 = 3
x3x4 = -1
Solving x3 + x4 = -5/6 and x3x4 = 1, we get x3 = -3/2 and x4 = 2/3.