Question

How do you find the number of possible positive real zeros and negative zeros then determine the rational zeros given `f(x)=8x^3-6x^2-23x+6`?

Answer 1

We can tell that `f(x)` has `1` negative real zero and `2` or `0` positive real zeros.

The zeros, which are all rational are: `2, 1/4, -3/2`

Explanation:

Given:

`f(x) = 8x^3-6x^2-23x+6`

`color(white)()`
Descartes' Rule of Signs

The coefficients of `f(x)` have signs in the pattern `+ - - +`. Since this has two changes (from `+` to `-` and from `-` to `+`), Descartes' Rule of Signs tells us that `f(x)` has `2` or `0` positive real zeros.

The coefficients of `f(-x)` have signs in the pattern `- - + +`. With one change of signs, we can deduce that `f(x)` has exactly one negative real zero.

`color(white)()`
Rational root theorem

By the rational root 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 `6` and `q` a divisor of the coefficient `8` of the leading term.

That means that the only possible rational zeros are:

`+-1/8, +-1/4, +-3/8, +-1/2, +-3/4, +-1, +-3/2, +-2, +-3, +-6`

Note that the rational root theorem only tells us about possible rational zeros, not about all possible zeros which may be irrational or non-real complex.

`color(white)()`
Actual zeros

It is probably easiest to try small integers first rather than fractions.

We find:

`f(1) = 8-6-23+6 = -15`

`f(2) = 8(color(blue)(2))^3-6(color(blue)(2))^2-23(color(blue)(2))+6 = 64-24-46+6 = 0`

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

`8x^3-6x^2-23x+6 = (x-2)(8x^2+10x-3)`

We can factor the remaining quadratic using an AC method:

Find a pair of factors of `AC = 8*3 = 24` which differ by `B=10`

The pair `12, 2` works.

Use this pair to split the middle term and factor by grouping:

`8x^2+10x-3 = (8x^2+12x)-(2x+3)`

`color(white)(8x^2+10x-3) = 4x(2x+3)-1(2x+3)`

`color(white)(8x^2+10x-3) = (4x-1)(2x+3)`

Hence the other two zeros are `x=1/4` and `x=-3/2`

graph{(y-(8x^3-6x^2-23x+6))(60(x-2)^2+y^2-0.2)(60(x-1/4)^2+y^2-0.2)(60(x+3/2)^2+y^2-0.2) = 0 [-5, 5, -20, 20]}