Please help ;-; -x^3+5x^2+6x=0 solve for x and show work

x = 0, 6, -1

Step-by-step explanation:

To solve for x, first get rid of the negative x^3 by making it positive.

x^3 - 5x^2 - 6x = 0

Now you can start by finding the possible zeros. To do so take the factors of p (in this case the 6 in 6x) and the factors of q (in this case 1 in -x^3) and divide p by q.

P = ±1, ±2, ±3, ±6

Q = 1

possible zeros = ±1, ±2, ±3, ±6

Go through your equation x^3 - 5x^2 - 6x = 0 and check each possible zero until you find one that gets the equation to equal zero.

You could either continue to try each number and the ones that work would be your answer, or you can find one that works, factor it out, then factor your new equation. The second method is faster so that's what I'll use

Check each possible zero. For example:

(1)^3 - 5(1)^2 - 6(1) = 0

-10=0 This will not work

(-1)^3 - 5(-1)^2 - 6(-1) = 0

0 = 0

-1 Works, so now you will factor out -1 using synthetic division.

-1 |            | 1    -5   -6   0

                |      -1     6   0

                 1    -6    0   0

Your new equation is x^2 - 6x = 0

Now factor your new equation:

x(x - 6) = 0

x = 0, 6, -1

I hope this all made sense, have a good day :)