Question

What are all the possible rational zeros for `f(x)=6x^3-x^2-13x+8` and how do you find all zeros?

Answer 1

This cubic has "possible" rational zeros:

`+-1/6, +-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8`

and actual zeros:

`1" "` and `" "-5/12+-sqrt(217)/12`

Explanation:

`f(x) = 6x^3-x^2-13x+8`

By the rational roots theorem, any rational zero of `f(x)` is expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `8` 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, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8`

There is a shortcut in that the sum of the coefficients is `0`.

That is:

`6-1-13+8 = 0`

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

`6x^3-x^2-13x+8 = (x-1)(6x^2+5x-8)`

The remaining quadratic factor is in the form `ax^2+bx+c` with `a=6`, `b=5` and `c=-8`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 5^2-4(6)(-8) = 25+192 = 217`

Since `Delta > 0` this quadratic has two Real zeros, but `Delta` is not a perfect square, so those zeros are irrational.

We can find the zeros using the quadratic formula:

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

`color(white)(x) = (-b+-sqrt(Delta))/(2a)`

`color(white)(x) = (-5+-sqrt(217))/12`

`color(white)(x) = -5/12+-sqrt(217)/12`