Originally Posted by
HallsofIvy Did you replace $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$, and $\displaystyle x_4$ with 1, 1+ i, 1- i, 2- i and 2+ i, respectively?
First, the reason red-dog said, "I think the polynomial has real coefficients" is that if you allow complex coefficients, there are an infinite number of such polynomials. Assuming complex coefficients, all roots must be in complex conjugate pairs. That is why 1- i and 2+ i must also be roots. Of course, it follows (You've solved polynomial equations by factoring haven't you. Same idea here). Saying "monic" polynomial means the leading coefficient (the coefficient of $\displaystyle x^5$ here) is 1.
That is why the polynomial must be the product, (x-1)(x-1-i)(x-1+i)(x-2+i)(x-2-i). That is the polynomial. Whether you are required to multiply that out depends on the particular statement of the question.
If you do, it's simplest to multiply the complex conjugates first: $\displaystyle (x-1-i)(x-1+i)$ can be done as "sum and difference product": $\displaystyle ((x-1)-i)((x-1)+i)= (x-1)^2- i^2= x^2- 2x+ 1+ 1= x^2- 2x+ 2$
(x-2+i)(x-2-i)= ((x-2)+i)((x-2)-i= $\displaystyle (x-2)^2- i^2= x^2- 4x+ 4+ 1= x^2- 4x+ 5$
The polynomial you want is $\displaystyle (x-1)(x^2- 2x+ 2)(x^2- 4x+ 5)$
Can you multiply that out?