Question

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

Answer 1

(i) Given that ${\displaystyle 1+2\text{i},\sqrt{3}}$

Another roots be ${\displaystyle 1-2\text{i},-\sqrt{3}}$

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 = ${\displaystyle \sqrt{3}-\sqrt{3}}$

Product roots = ${\displaystyle \left(\sqrt{3}\right)(-\sqrt{3})}$

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 = ${\displaystyle \frac{1\pm \sqrt{1-4\left(1\right)(-9)}}{2\left(1\right)}}$

= ${\displaystyle \frac{1\pm \sqrt{1+36}}{2}}$

= ${\displaystyle \frac{1\pm \sqrt{37}}{2}}$

Roots are ${\displaystyle \frac{1+\sqrt{37}}{2},\frac{1-\sqrt{37}}{2},1+2\text{i},1-2\text{i}}$ and ${\displaystyle \sqrt{3},-\sqrt{3}}$.