How do you use the rational root theorem to list all possible roots for `x^3+5x^2-2x-15=0`?

Given:

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

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Rational root theorem

The rational root theorem will only tell you about possible rational roots - not irrational or Complex ones.

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 `-15` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1, +-3, +-5, +-15`

However, note that all of these are odd, resulting in all of the terms of `f(x)` being odd, except `-2x`, which will be even. So for any of these values we have:

`f(x) = "odd" + "odd" + "even" + "odd" = "odd" != 0`

So `f(x)` has no rational zeros.

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Bonus

Let us see what else we can find out about `f(x)`'s zeros...

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Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=5`, `c=-2` and `d=-15`, so we find:

`Delta = 100+32+7500-6075+2700 = 4257`

Since `Delta > 0` this cubic has `3` Real zeros.

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Descartes's Rule of Signs

The signs of the coefficients of `f(x)` are in the pattern `+ + - -`. With one change of sign, by Descartes' rule of signs, we can deduce that `f(x)` has exactly one positive Real zero. So the other two Real zeros are negative.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

`0=27f(x)=27x^3+135x^2-54x-405`

`=(3x+5)^3-93(3x+5)-65`

`=t^3-93t-65`

where `t=(3x+5)`

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Trigonometric substitution

Look for a substitution of the form:

`t = k cos theta`

where `k` is chosen such that the resulting cubic in `cos theta` contains:

`4 cos^3 theta - 3 cos theta = cos 3 theta`

Let `k = 2sqrt(31)`

Then:

`0 = t^3-93t-65`

`color(white)(0) = k^3cos^3 theta - 93k cos theta - 65`

`color(white)(0) = 2sqrt(31)(124 cos^3 theta - 93 cos theta) - 65`

`color(white)(0) = 62sqrt(31)(4 cos^3 theta - 3 cos theta) - 65`

`color(white)(0) = 62sqrt(31)cos 3theta - 65`

Hence:

`cos 3 theta = 65/(62 sqrt(31)) = (65sqrt(31)) / 1922`

So:

`3 theta = +-cos^(-1) ((65sqrt(31)) / 1922) + 2kpi`

`t = 2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)`

Then `x = 1/3(t-5)` and hence the zeros of our original cubic are:

`x_k = 1/3(2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)-5)`

`k = 0, 1, 2`