For C, factor out an x on the LHS. That corresponds to the solution x = 0. Then what remains is a quadratic, for which there are numerous methods of solution. For D, each factor could be zero, so solve them separately.
Make sense?
C) x^3 - 8x^2 + 15x = 0
I donīt know how to solve this the formulae I usually use I canīt use it since it is x^3 and not x^2... I donīt know what the formulae is called in English.
D) 4(3-3x) (8-2x^2)= 0
Help anyone? :/