Question

How do you use the rational roots theorem to find all possible zeros of `P(x) = x^3 - 4x^2 + x + 6`?

Answer 1

Zeros: `-1, 2, 3`

Explanation:

`P(x) = x^3-4x^2+x+6`

By the rational roots theorem, any rational zeros of `P(x)` are expressible as `p/q` for integers `p, q` with `p` a divisor of the constant term `6` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-6`

We find:

`P(-1) = -1-4-1+6 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

`x^3-4x^2+x+6 = (x+1)(x^2-5x+6)`

We could simply evaluate `P(x)` for the remaining possible zeros, but `x^2-5x+6` factors fairly easily:

`x^2-5x+6 = (x-2)(x-3)`

So the other two zeros are `x=2` and `x=3`.