Question

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given `f(x)=x^4+2x^3-9x^2-2x+8`?

Answer 1

The zeros of `f(x)` are:

`1, -1, -4, 2`

Explanation:

Given:

`f(x) = x^4+2x^3-9x^2-2x+8`

Note that the pattern of the signs of the coefficients is `+ + - - +`. By Descartes' Rule of Signs, since this has `2` changes of sign, `f(x)` has `2` or `0` positive Real zeros.

The pattern of the signs of `f(-x)` is `+ - - + +`. Since this has `2` changes of sign, `f(x)` has `2` or `0` 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 `8` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

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

In addition, note that the sum of the coefficients of `f(x)` is `0`. That is:

`1+2-9-2-8 = 0`

Hence `f(1) = 0`, `x=1` is a zero and `(x-1)` a factor:

`x^4+2x^3-9x^2-2x+8 = (x-1)(x^3+3x^2-6x-8)`

Note that `x=-1` is a zero of the remaining cubic, since:

`-1+3+6-8 = 0`

So `(x+1)` is a factor:

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

Finally, note that `4-2 = 2` and `4*2 = 8`. So we find:

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

So the zeros of `f(x)` are:

`1, -1, -4, 2`