What are the roots of 38r3−27r2−27r−27=0 ?

1 Answer
George C.
Feb 5, 2017

Real root:

x=32

Complex roots:

x=−1538±3√5138i

Explanation:

f(x)=38r3−27r2−27r−27

By the rational roots theorem, any rational zeros of f(x) are expressible in the form pq 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:

±138,±119,±338,±319,±938,±12,±2738,±1,±2719,±32,±3,±92,±9,±272,±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(32)=38(278)−27(94)−27(32)−27=27(388−94−32−1)

=2719−9−6−44=0

So x=32 is a zero and (2x−3) a factor:

38r3−27r2−27r−27=(2x−3)(19x2+15x+9)

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

x=−b±√b2−4ac2a

x=−15±√152−4(19)(9)2⋅19

x=−15±√225−68438

x=−15±3√51i38

x=−1538±3√5138i

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