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)=x^4+2x^3-12x^2-40x-32`?

Answer 1

Possible rational zeros:

`+-1, +-2, +-4, +-8, +-16, +-32`

Descartes gives us that `f(x)` has one positive and `3` or `1` negative zeros.

Actual zeros: `-2, -2, -2, 4`

Explanation:

Given:

`f(x) = x^4+2x^3-12x^2-40x-32`

Rational roots theorem

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 `-32` and `q` a divisor of the coefficient `1` of the leading term.

That means the the only possible rational zeros are:

`+-1, +-2, +-4, +-8, +-16, +-32`

Descartes' Rule of Signs

The pattern of signs of the coefficients of `f(x)` is `+ + - - -`. With one change of signs, Descartes' Rule of Signs tells us that `f(x)` has exactly one positive real zero.

The pattern of signs of coefficients of `f(-x)` is `+ - - + -`. With `3` changes of sign, Descartes' Rule of Signs allows us to deduce that `f(x)` has `3` or `1` negative real zero.

Bonus - Find the actual zeros

Note that the coefficient of `x^4` is odd, but all the other coefficients are even. Therefore any integer zero of `f(x)` must be even.

We find:

`f(-2) = (-2)^4+2(-2)^3-12(-2)^2-40(-2)-32 = 16-16-48+80-32 = 0`

So `-2` is a zero and `(x+2)` a factor:

`x^4+2x^3-12x^2-40x-32 = (x+2)(x^3-12x-16)`

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

`(-2)^3-12(-2)-16 = -8+24-16 = 0`

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

`x^3-12x-16 = (x+2)(x^2-2x-8)`

Finally, to factor the remaining quadratic, note that `4*2 = 8` and `4-2=2`, so we find:

`x^2-2x-8 = (x-4)(x+2)`

So:

`f(x) = (x+2)^3(x-4)`

has zeros `-2` with multiplicity `3` and `4` with multiplicity `1`.