Question

How do you find all rational zeroes of the function using synthetic division `f(x)=2x^5+x^4-23x-16`?

Answer 1

`f(x)` has three irrational real zeros and two non-real complex zeros.

Explanation:

Given:

`f(x) = 2x^5+x^4-23x-16`

By Descartes' Rule of Signs `f(x)` has exactly one positive real zero, since the signs of its coefficients `+ + - -` have only one change of signs.

`f(-x)` has signs in the pattern `- + + -`. With `2` changes of sign, we can deduce that `f(x)` has `0` or `2` negative real zeros.

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 `-16` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

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

We find:

`f(-2) = -64+16+46-16 = -18 < 0`

`f(-1) = -2+1+23-16 = 6 > 0`

`f(-1/2) = -1/16+1/16+23/2-16 = -9/2 < 0`

`f(1) = 2+1-23-16 = -36 < 0`

`f(2) = 64+16-46-16 = 18 > 0`

Hence `f(x)` has irrational zeros in `(-2, -1)`, `(-1, -1/2)` and `(1, 2)`

The remaining `2` roots are non-real complex.

We cannot immediately deduce it from what we have found, but actually none of the zeros are expressible in terms of `n`th roots (i.e. square, cube roots etc.).