Question

What are the roots of `38r^3-27r^2-27r-27 = 0` ?

Answer 1

Real root:

`x = 3/2`

Complex roots:

`x = -15/38+-(3sqrt(51))/38i`

Explanation:

`f(x) = 38r^3-27r^2-27r-27`

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 `-27` and `q` a divisor of the coefficient `38` of the leading term.

That means that the only possible rational zeros are:

`+-1/38, +-1/19, +-3/38, +-3/19, +-9/38, +-1/2, +-27/38, +-1, +-27/19, +-3/2, +-3, +-9/2, +-9, +-27/2, +-27`

That's rather a lot of possibilities to try, so let's see how we can narrow down the search...

First note that the pattern of signs of the coefficients of `f(x)` is `+ - - -`. By Descartes' Rule of Signs, since this has one change, we can deduce that `f(x)` has exactly one positive real zero.

We can observe that:

`f(1) = 38-27-27-27 = -43 < 0`

So the positive Real zero is greater than `1`.

We also find:

`f(3) = 38(27)-27(9)-27(3)-27 = 27(38-9-3-1) = 27*25`

`= 675 > 0`

So let's try:

`f(3/2) = 38(27/8)-27(9/4)-27(3/2)-27 = 27(38/8-9/4-3/2-1)`

`=27(19-9-6-4)/4 = 0`

So `x=3/2` is a zero and `(2x-3)` a factor:

`38r^3-27r^2-27r-27 = (2x-3)(19x^2+15x+9)`

We can find the zeros of the remaining quadratic using the quadratic formula with `a=19`, `b=15`, `c=9`...

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (-15+-sqrt(15^2-4(19)(9)))/(2*19)`

`color(white)(x) = (-15+-sqrt(225-684))/38`

`color(white)(x) = (-15+-3sqrt(51)i)/38`

`color(white)(x) = -15/38+-(3sqrt(51))/38i`