Question

Is it possible to factor `y=x^3+5x^2-14x`? If so, what are the factors?

Answer 1

Yes; `x=-7, 0, 2`

Explanation:

Original Equation:
`y=x^3+5x^2-14x`

Factor out an `x`:
`y=x(x^2+5x-14)`

Factor inside the parentheses (factors of `AC` that add up to `B`):
––Factors of `-14` that add up to `5`

You should get `7` and `-2`

Now plug it back into the equation:
`y=x(x-2)(x+7)`

Your factors are:
`x`
`(x-2)`
`(x+7)`

To find `x`, set each factor equal to `0` and solve.
Your final answer should be:
`x=0`
`x=2`
`x=-7`