Question

How can you use the rational zeros theorem with this problem to find all real zeros??

`f(x)=2x^3-x^2+2x-1`

Answer 1

The only real zero is `x=1/2`

Explanation:

Given:

`f(x) = 2x^3-x^2+2x-1`

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 `-1` and `q` a divisor of the coefficient `2` of the leading term.

So the only possible rational zeros are:

`+-1/2`, `+-1`

In addition, note that the pattern of the signs of the coefficients is `+ - + -`. With `3` changes of sign, Descartes' Rule of Signs tells us that `f(x)` has `3` or `1` positive real zeros. The pattern of the signs of the coefficients of `f(-x)` is `- - - -`. With no changes of sign, we can tell that `f(x)` has no negative real zeros.

So the only possible rational zeros are:

`1/2`, `1`

We find:

`f(1/2) = 2(1/8)-(1/4)+2(1/2)-1 = 0`

`f(1) = 2-1+2-1 = 2`

So the only rational real zero is `x=1/2`

This has a corresponding factor `2x-1` and we find:

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

Note that `x^2+1 != 0` for all real values of `x`. So we have found all of the real zeros.

Answer 2

There is only one Real zero occurs at: `color(blue)(x=1/2)`

Explanation:

According to the Rational Zero Theorem:
For a polynomial of the standard form `f(x)=a_nx^n+...+a_0, a_i in ZZ AAi`
any rational zero must be of the form `p/q`
where `p` is an integer factor of `a_0`
and `q` is an integer factor of `a_n`

For the given polynomial: `f(x)=2x^3-x^2+2x-1`
this means that the only candidates as possible rational zeros are `{1, -1, 1/2, -1/2}`

Evaluating the given function for each of these values:
#color(white)("XXX"){: (ul(x)," | ",ul(2x^3-x^2+2x-1)), (+1," | ",color(white)("xxxxxxx")2), (-1," | ",color(white)("xxxxx")-6), (+1/2," | ",color(white)("xxxxxxx")0), (-1/2," | ",color(white)("xxxxx")-2.5) :}#

So the only rational zero is at `x=1/2`

Note that the Rational Zero Theorem does not indicate if there are or are not non-rational, Real zeros.

To determine this for the given case:
Note that if `x=1/2` is a zero then `(x-1/2)` is a factor of the polynomial; which in turn implies `(2x-1)` is a factor of the polynomial.

Performing the division:
`color(white)("XXX")color(blue)(""(2x^3-x^2+2x-1))divcolor(green)(""(2x-1))=color(red)(""(x^2+1))`
So (for Real values) if `color(green)(""(2x-1))!=0`
then `color(red)(""(x^2+1))=0`

But `color(red)(""(x^2+1))=0`
`rArr x^2=-1`
which has no Real solutions.

Therefore (among Real solutions) `color(magenta)(x=1/2)` is the only Real zero.