Let $\iota$ be a root of the equation $x^2+1=0$ and let $\omega$ be a root of the equation $x^2+x+1=0$. Construct a polynomial
$$f(x)=a_0+a_1x+\dots+a_nx^n$$
where $a_0,a_1,\dots,a_n$ are all integers such that $f(\iota+\omega)=0$.
My approach was to substitute $(\iota + \omega)$ in place of $x$ and try some expansion but that resulted in nothing. This question was asked in a 10 marks question of a very popular entrance examination at the 10+2 level. A rigorous answer is appreciable. Thank you.
Hint:
suppose we have:
$p(x)=x^2-1=(x-1)(x+1)$; $x_p=1, -1$
$q(x)=x^2-5x+6=(x-3)(x-2)$; $x_q=2, 3$
And we want a polynomial its root are $x_p+x_q$, we must have:
$x_1=1+2=3$, $x_2=1+3=4$, $x_3=-1+2=1$ and $x_4=-1+3=2$
So our polynomial is:
$P(x)=(x-3)(x-4)(x-1)(x-2)$
Now consider:
$P(x)=(x-a)(x-b)(x-c)(x-d)=x^4-(a+b+c+d)x^3+(ab+cd)x^2+abcd$
Therefore to avoid confusion and mess it is better you first find the coefficients and then construct the polynomial.In your question:
$x_1=i+\omega$, $x_2=i+\overline{\omega}$, $x_3=-i+\omega$, $x_4=-i+\overline{\omega}$
Let $\bar\omega$ be the other root of $x^2+x+1$. Compute $$(x-i-\omega)(x-i-\bar\omega)(x+i-\omega)(x+i-\bar\omega).\tag{1}$$
The first factor of (1) is $x-i-\omega$, so it should be clear that when $x=i+\omega$, the whole product is zero as required. The only remaining issue is whether the coefficients of (1) will be real integers, which we can verify by direct computation.
To do this, it will help to observe that $$(x-\omega)(x-\bar\omega) = x^2+x+1\tag{2}.$$ Now group expression (1) above as follows:
$$((x-i)-\omega)((x-i)-\bar\omega)\cdot((x+i)-\omega)((x+i)-\bar\omega)$$
Then by (2), the product of the left two factors is $$(x-i)^2 + (x-i)+1$$ and the product of the right two factors is $$(x+i)^2 + (x+i)+1.$$
Can you take it from there?
