Answer:
[tex]\textsf{D)}\quad (x^2 - 2x + 5)(x^2-2x-1)[/tex]
Step-by-step explanation:
According to the Conjugate Root Theorem, if a polynomial with real coefficients has a complex root [tex]a + bi[/tex], then its conjugate [tex]a - bi[/tex] is also a root of the polynomial. Therefore, if [tex]1 - 2i[/tex] is a root of the polynomial, then [tex]1 +2i[/tex] is also a root.
The Factor Theorem states that if x = c is a root of a polynomial, then (x - c) is a factor of that polynomial. Therefore, two factors of the 4th degree polynomial are:
[tex](x-(1-2i))=(x-1+2i)[/tex]
[tex](x-(1+2i))=(x-1-2i)[/tex]
Multiply the factors:
[tex]\begin{aligned}(x-1+2i)(x-1-2i)&=x^2-x-2ix-x+1+2i+2ix-2i-4i^2\\&=x^2-x-x+2ix-2ix+2i-2i+1-4i^2\\&=x^2-2x+1-4(-1)\\&=x^2-2x+1+4\\&=x^2-2x+5\end{aligned}[/tex]
Therefore, a quadratic factor of the 4th degree polynomial is:
[tex](x^2 - 2x + 5)[/tex]
This means we can discount answer options A and C.
As the third root is 1 - √2, then 1 + √2 is likely to be the fourth root.
Therefore, the factors would be:
[tex](x-(1-\sqrt{2}))=(x-1+\sqrt{2})[/tex]
[tex](x-(1+\sqrt{2}))=(x-1-\sqrt{2})[/tex]
Multiply the factors:
[tex]\begin{aligned}(x-1+\sqrt{2})(x-1-\sqrt{2})&=x^2-x-\sqrt{2}x-x+1+\sqrt{2}+\sqrt{2}x-\sqrt{2}-2\\&=x^2-x-x+\sqrt{2}x-\sqrt{2}x+\sqrt{2}-\sqrt{2}+1-2\\&=x^2-2x-1\end{aligned}[/tex]
So, the other quadratic factor of the 4th degree polynomial is:
[tex](x^2-2x-1)[/tex]
Therefore, the 4th degree polynomial with roots of [tex]1 - 2i[/tex] and [tex]1 - \sqrt{2}[/tex] is:
[tex]\Large\boxed{\boxed{(x^2 - 2x + 5)(x^2-2x-1)}}[/tex]
