Solve x^3+20x^2=125x

The expression can be rearranged to equal 0 by taking the 125 to the other side. To do this the opposite thing has to be done, so 125 must be subtracted to get x^3+20x^2-125=0 . All parts in this function include the letter x and therefore you can divide through by x to factorise the x out, to get x(x^2+20x-125)=0. By looking at this it can be seen that either x or the quadratic expression in the brackets must = 0, and therefore one solution is x=0. The other solutions must be found by factorising the quadratic expression.
To factorise x^2+20x-125=0, we must find two numbers that both multiply to give -125 and add or subtract to give 20. It can be seen that these two numbers in this case are -5 and 25, as 25x-5=-125 and 25-5=20. The coefficient of the x^2 term is 1 and therefore, we know the brackets must also only have the coefficient of x as 1. Therefore, it can factorise to (x-5)(x+20)=0, which can then be checked by expansion to make sure it gives the same quadratic expression. This factorisation shows that the two other solutions are x=5 and x=-20, as well as x=0.