Question

How do you use the rational root theorem to find the roots of `8x^3-3x^2+5x+15`?

Answer 1

`(1,2,4,8)/(1,3,5,9,15,45)=1,1/3,1/5,1/9,1/15,1/45,2,2/3,2/5,2/9,2/15,2/45,4,4/3,4/5,4/9,4/15,4/45,8,8/3,8/5,8/9,8/15,8/45`

Explanation:

The general form for a polynomial is as follows:
`px^n+a_n-1+...a_1x+q`

The rational roots theorem states that, to find any potential zeroes of a given polynomial function, the formula is: `("factors of p")/("factors of q")`, where `p` and `q` are the first and last coefficients of the given function. In this case, the factors of `p` are 1,2,4,and 8, and the factors of q are 1,3,5,9,15,and 45.

Once you have all the potential zeroes of the function, you just have to test them using synthetic division.

Answer 2

You can't.

You can determine that `8x^3-3x^2+5x+15=0` has one Real irrational root in `(-1, -3/4)` and no rational roots.

Explanation:

By the rational root theorem, any rational roots of `8x^3-3x^2+5x+15 = 0` must be expressible in lowest terms in the form `p/q` where `p, q in ZZ`, `q != 0`, `p` a divisor of the constant term `15` and `q` a divisor of the coefficient `8` of the leading term.

The prime factorisation of `8` is `2*2*2`.
The prime factorisation of `15` is `3*5`.

That means that the only possible rational roots are:

`+-1/8`, `+-1/4`, `+-3/8`, `+-1/2`, `+-5/8`, `+-3/4`, `+-1`, `+-5/4`, `+-3/2`, `+-15/8`, `+-5/2`, `+-3`, `+-15/4`, `+-5`, `+-15/2`, `+-15`

That's quite a lot of possibilities to try, so let's narrow it down.

Let `f(x) = 8x^3-3x^2+5x+15`

Then `f'(x) = 24x^2-6x+5`, which has a negative discriminant.

So we can deduce that `f(x)` is strictly monotonic increasing.

`f(-1) = -8-3-5+15 = -1`

`f(-3/4) = -8*3^3/4^3-3*3^2/4^2+5*3/4+15`

`= -27/8-27/16+15/4+15`

`= (-54-26+60+240)/16 = 220/16 = 55/4`

So `f(x) = 0` has one Real root in `(-1, -3/4)`, but it is not rational.