Question

How do you use the rational roots theorem to find all possible zeros of `F(x) = 6x^3 - 20x^2 + 11x + 7`?

Answer 1

Zeros: `7/3`, `1/2 +-sqrt(3)/2`

Explanation:

`F(x) = 6x^3-20x^2+11x+7`

By the rational roots theorem, any rational zeros of `F(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `7` and `q` a divisor of the coefficient `6` of the leading term.

That means that the only possible rational zeros are:

`+-1/6, +-1/3, +-1/2, +-1, +-7/6, +-7/3, +-7/2, +-7`

Trying each in turn, we (eventually) find:

`F(7/3) = 6(343/27)-20(49/9)+11(7/3)+7`

`=(686-980+231+63)/9 = 0`

So `x=7/3` is a zero and `(3x-7)` a factor:

`6x^3-20x^2+11x+7`

`=(3x-7)(2x^2-2x-1)`

Use the quadratic formula to find the zeros of the remaining quadratic:

`x = (2+-sqrt((-2)^2-4(2)(-1)))/(2*2)`

`=(2+-sqrt(4+8))/4`

`=(2+-2sqrt(3))/4`

`=1/2+-sqrt(3)/2`