Find all the roots of the equation x^4 - 2x^3 + 14x^2 - 18x + 45 = 0 given that 1 + 2i is one of its roots.

Question

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Answer 1

f(x) = [x - (1 + 2i)][x - (1 -  2i)](x - x1)(x - x2) = [(x - 1) - 2i][(x - 1) + 2i](x - x1)(x - x2) =

 

[(x - 1)^2 - (2i)^2](x - x1)(x - x2) = [x^2 - 2x + 1 - 4i^2](x - x1)(x - x2) = [x^2 - 2x + 1 -4(-1)](x - x1)(x - x2)

 

= (x^2 - 2x + 1 - 4)(x - x1)(x - x2) = (x^2 - 2x - 3)(x - x1)(x -  x2) = x^4 -2x^3 + 14x^2 - 18x + 45 --->

 

 

Answer 2

Since 1+2i is a root, 1-2i is also a root.  

 

So, x-(1+2i) and x-(1-2i) are factors of the given polynomial.

 

[(x-(1+2i)][x-(1-2i)] = [(x-1) - 2i][(x-1) + 2i] = (x-1)² - (2i)²

 

                                                                                 ⁼ x²-2x+1-4i²

 

                                                                     = x²-2x+5

 

Dividing the given polynomial by x²-2x+5, we get a quotient of x²+9

 

So, x⁴ 2x³+14x²-18x+45 = (x²-2x+5)(x²+9)

 

If x²+9 = 0, then x²=-9.  So, x = ±3i.

 

Roots:  1+2i, 1-2i, 3i, -3i