Question

What are all the possible rational zeros for `f(x)=4x^5-2x^4+30x^3-15x^2+50x-25`?

Answer 1

See explanation...

Explanation:

`f(x) = 4x^5-2x^4+30x^3-15x^2+50x-25`

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 `25` and `q` a divisor of the coefficient `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5, +-25/4, +-25/2, +-25`

In fact, note that the signs of the coefficients alternate, with no powers of `x` missing. Hence (by Descartes rule of signs) there are `1`, `3` or `5` positive Real zeros and no negative Real zeros.

So the only possible rational zeros are:

`1/4, 1/2, 1, 5/4, 5/2, 5, 25/4, 25/2, 25`

Trying each in turn, we find:

`f(1/2) = 4(1/32)-2(1/16)+30(1/8)-15(1/4)+50(1/2)-25`

`color(white)(f(1/2))=(1-1+30-30+200-200)/8 = 0`

So `x=1/2` is a zero and `(2x-1)` a factor:

`4x^5-2x^4+30x^3-15x^2+50x-25 = (2x-1)(2x^4+15x^2+25)`

We can factor the remaining quartic as a quadratic in `x^2`

`2x^4+15x^2+25 = 2x^4+10x^2+5x^2+25`

`color(white)(2x^4+15x^2+25) = 2x^2(x^2+5)+5(x^2+5)`

`color(white)(2x^4+15x^2+25) = (2x^2+5)(x^2+5)`

Hence zeros:

`x = +-sqrt(10)/2i" "` and `" "x = +-sqrt(5)i`