Question

How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descarte's rule of signs to list the possible positive/negative zeros of `f(x)=36x^4-12x^3-11x^2+2x+1`?

Answer 1

`f(x)` has zeros `1/2` and `-1/3`, each with multiplicity `2`.

Explanation:

Given:

`f(x) = 36x^4-12x^3-11x^2+2x+1`

By the rational zeros 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 `1` and `q` a divisor of the coefficient `36` of the leading term.

That means that the only possible rational zeros are:

`+-1/36, +-1/18, +-1/12, +-1/9, +-1/6, +-1/4, +-1/3, +-1/2, +-1`

Note that the pattern of signs of the coefficients of `f(x)` is `+ - - + +`. With `2` changes of sign, Descartes' Rule of Signs tells us that `f(x)` has `2` or `0` positive real zeros.

The pattern of signs of the coefficients of `f(-x)` is `+ + - - +`. With `2` changes of sign, Descartes' Rule of Signs tells us that `f(x)` has `2` or `0` negative real zeros.

Note that if we reverse the order of the coefficients, we get a polynomial whose zeros are the reciprocals of the zeros of `f(x)`:

`g(x) = x^4+2x^3-11x^2-12x+36`

This has possible rational zeros:

`+-1, +-2, +-3, +-4, +-6, +-9, +-12, +-18, +-36`

(I prefer doing arithmetic with whole numbers)

We find:

`g(2) = (2)^4+2(2)^3-11(2^2)-12(2)+36 = 16+16-44-24+36 =0`

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

`x^4+2x^3-11x^2-12x+36 = (x-2)(x^3+4x^2-3x-18)`

We find that `2` is also a zero of the remaining cubic:

`(2)^3+4(2)^2-3(2)-18 = 8+16-6-18 = 0`

So `(x-2)` is a factor again:

`x^3+4x^2-3x-18 = (x-2)(x^2+6x+9)`

We can recognise the remaining quadratic factor as a perfect square trinomial:

`x^2+6x+9 = (x+3)^2`

So the last two zeros of `g(x)` are `x=-3` (with multiplicity `2`)

So `g(x) = (x-2)^2(x+3)^2`

and `f(x) = (2x-1)^2(3x+1)^2` has zeros `1/2, 1/2, -1/3, -1/3`