Given that 4x^3-16x^2+21x-27 has the solution 3 find the conjugate complex pair of roots.

We have to find the roots of the polynomial
4x^3-16x^2+21x-27.

Let's equate this to 0 and find the
roots.

4x^3-16x^2+21x-27
=0

=> (x- 3) (4x^2 - 4x + 9)
=0

Therefore x1 = 3

x2 = [-b +
sqrt (b^2 - 4ac)]/2a

=>x2 = [ 4 + sqrt(16 -
144)]/8

=> x2 = 1/2 + sqrt (-128)
/8

=> x2 = 1/2 + i (sqrt
128)/8

=> x2 = 1/2 + i (sqrt
2)

x3 = [-b - sqrt (b^2 -
4ac)]/2a

=>x2 = [ 4 - sqrt(16 -
144)]/8

=> x2 = 1/2 - sqrt (-128)
/8

=> x2 = 1/2 - i (sqrt
128)/8

=> x3 = 1/2 - i (sqrt
2)

Therefore the roots are 3, 1/2 + i (sqrt
2) and 1/2 - i (sqrt 2)