Solve x^4-4x^3+4x^2-9=0 given that 1+isqrt2 is a root.

Question

I need all of the steps.
Thanks

Answer 1

x^4-4x^3+4x^2-9=0
Given 1+i√2 is a root, then
1-i√2 is also a root (conjugate).

The product of the two roots gives a real factor of
(x-1-i√2)(x-1+i√2)
=x²-2x+3

Do a long division of
(x^4-4*x^3+4*x^2-9)÷(x²-2x+3)
=x²-2x-3
and solve by factoring to get
(x-3)(x+1)=0

Alternatively,
note that
x^4-4x^3+4x^2 is a perfect square
=(x²-2x)²
so the left-hand-side can be factorized:
(x²-2x)²-3²=0
or
(x²-2x+3)(x²-2x-3)=0
which can be further factorized as:
(x²-2x+3)(x-3)(x+1)=0
Solve the first factor by the quadratic formula to get the conjugate complex roots, and the remainder for x=3, x=-1.