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

Expert Answers
justaguide eNotes educator| Certified Educator

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)

giorgiana1976 | Student

The polynomial has 3 roots. If one root is known, the other 2 roots are x2 ,x3.

We'll use Viete's relations for finding the other 2 conjugate roots.

3 + x2 + x3 = 16/4

3 + x2 + x3 = 4

We'll subtract 3:

x2 + x3 = 4-3

x2 + x3 = 1

3*x2*x3 = 27/4

We'll divide by 3:

x2*x3 = 9/4

We'll write the equation of the quadratic when knowing the sum and the product of the roots:

x^2 - x + 9/4 = 0

We'll multiply by 4:

4x^2 - 4x + 9 = 0

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

x2 = (4 + 8sqrt2)/8

x2 = (1+2isqrt2)/2

x3 = (1-2isqrt2)/2

The complex roots of the equation are :{(1-2isqrt2)/2 ; (1+2isqrt2)/2}.