Question

How do you factor #4y=x^3-4x^2-11x+30 #?

Answer 1

`4y = (x-2)(x-5)(x+3)`

Explanation:

By the rational roots theorem, any rational zeros of `x^3-4x^2-11x+30` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `30` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-5, +-6, +-10, +-15, +-30`

If `x=2` then we find:

`x^3-4x^2-11x+30 = (8)-4(4)-11(2)+30 = 8-16-22+30 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^3-4x^2-11x+30 = (x-2)(x^2-2x-15)`

To factor the remaining quadratic, find a pair of numbers with difference `2` and product `15`. The pair `5, 3` works, hence:

`x^2-2x-15 = (x-5)(x+3)`

Putting it all together, we have:

`4y = (x-2)(x-5)(x+3)`