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

##### 1 Answer

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

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

We can observe that:

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

So the positive Real zero is greater than

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

#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

#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#