Question

What are all the possible rational zeros for `f(x)=x^3+5x^2-11x-6` and how do you find all zeros?

Answer 1

`f(x)` has zeros `2` and `-7/2+-sqrt(37)/2`

Explanation:

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

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `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.

Thats means that the only possible rational zeros are:

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

In addition, note that the signs of the coefficients of `f(x)` are in the pattern: `+ + - -`. With one change of sign, Descartes' Rule of Signs tells us that `f(x)` has exactly one positive Real zero. So let's try the positive rational possibilities first:

`f(1) = 1+5-11+6 = 1`

`f(2) = 8+5(4)-11(2)-6 = 8+20-22-6 = 0`

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

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

The remaining two zeros are roots of the quadratic: `x^2+7x+3 = 0`, which we can solve using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-7+-sqrt(7^2-4(1)(3)))/(2*1)`

`color(white)(x) = (-7+-sqrt(49-12))/2`

`color(white)(x) = -7/2+-sqrt(37)/2`