Question

How do you solve `2x^3 – 5x^2 – 3x + 2 = 2`?

Answer 1

`x=-1/2,0,3`

Explanation:

Begin by subtracting `2` from both sides of the equation.

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

Factor an `x` from the terms on the left hand side.

`x(2x^2-5x-3)=0`

The quadratic is factorable into

`x(2x+1)(x-3)=0`

Now that you have these three terms being equal to `0`, you know that only one of them has to equal `0` for the entire expression to equal `0`. Thus, the three solutions can be found by setting each term equal to `0`:

`x=0`

`2x+1=0" "=>" "x=-1/2`

`x-3=0" "=>" "x=3`