f(x) = 9x^3 – 33x^2 –55x – 25. Given that x = 5 is a solution of the equation f(x) = 0, use an algebraic method to solve f(x) = 0 completely.

x=5 is a solution, so (x-5) is a factor of f(x). So, to find the other solutions, factorise f(x): f(x)=(x-5)(ax^2+bx+c). We know it will be of this form because f(x) is cubic. To find the values of a, b and c, start by seeing what will multiply by x to get 9x^3. 9x^2 * x = 9x^2, so a=9, so we have f(x)=(x-5)(9x^2+...)=9x^3-45x^2+... . To get the coefficient of x^2 in f(x) to be -33 we need to add 12x^2, so b=12, f(x)=(x-5)(9x^2+12x+...)=9x^3-45x^2+12x^2-60x=9x^3-33x^2-60x. To get the coefficient of x in f(x) to be -55 we need to add 5x, so c=5. If 5 is a factor, we will get f(x) from these a, b and c values. f(x)=(x-5)(9x^2+12x+5)=9x^3+12x^2+5x-45x^2-60x-25=9x^3-33x^2-55x-25. Now we need to find the roots of 9x^2+12x+5. We cannot factorise it, so try the quadratic formula (using (+-) to represent the plus or minus symbol): x = (-12(+-)sqrt(12^2-4(95))/(29)=(-12(+-)sqrt(-36))/18=-12/18 (+-)6i/18=-2/3 (+-)i/3. Hence the solutions to f(x)=0 are x = 5, -2/3 + i/3, -2/3 - i/3.