Question

How do you use the rational root theorem to list all possible roots for `5x^3-11x^2+7x-1=0`?

Answer 1

The "possible" rational roots are: `+-1/5, +-1`

The actual zeros are: `1, 1, -1/5`

Explanation:

`f(x) = 5x^3-11x^2+7x-1`

By the rational root theorem, the only possible rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-1` and `q` a divisor of the coefficient `5` of the leading term.

That means that the only possible rational zeros are:

`+-1/5`, `+-1`

There could be other zeros, but they will be irrational or non-Real Complex. The rational root theorem only tells us possible rational zeros.

Note that the sum of the coefficients of `f(x)` is `0`. That is:

`5-11+7-1 = 0`

So `f(1) = 0`, `x = 1` is a zero and `(x-1)` a factor:

`5x^3-11x^2+7x-1 = (x-1)(5x^2-6x+1)`

The same is true of the remaining quadratic:

`5 - 6 + 1 = 0`

So we have another zero `x=1` and factor `(x-1)`:

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

The remaining factor `(5x+1)` corresponds to zero `x=-1/5`.