Question

Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and `sqrt(3)` are two of its zeros

Answer 1

(i) Given that `1 + 2"i", sqrt(3)`

Another roots be `1 - 2"i", - sqrt(3)`

Sum of roots = 1 + 2i + 1 – 2i

Product roots = (1 + 2i)(1 – 2i)

1² + 2² = 1 + 4 = 5

x² – 2x + 5 = 0

(ii) Sum of roots = `sqrt(3) - sqrt(3)`

Product roots = `(sqrt(3))(- sqrt(3))`

x² – 0x – 3 = 0

x² – 3 = 0

(x² – 2x + 5)(x² – 3) = x⁴ – 2x³ + 2x² + 6x – 15

x⁶– 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135

= (x⁴ – 2x³ + 2x² + 6x – 15)(x² + px – 9)

Equate of co-efficient of x on both sides

– 39 = – 54 – 15 p

– 39 + 54 = – 15 p

15 = – 15 p

P = – 1

∴ x² – x – 9 = 0

x = `(1 +- sqrt(1 - 4(1)(- 9)))/(2(1))`

= `(1 +- sqrt(1 + 36))/2`

= `(1 +- sqrt(37))/2`

Roots are `(1 + sqrt(37))/2, (1 - sqrt(37))/2, 1 + 2"i", 1 - 2"i"` and `sqrt(3), -sqrt(3)`.