So basically I have been given two roots of 2+i and 3. It asks what the cubic polynomial is, and sadly I have forgotten how to do this all together. Thanks in advance.
If one root is (2 + i) then other root must be (2 - i) because the complex roots always occur in conjugate-pairs.
So, the three roots are (2 + i), (2 - i) and 3.
so, the cubic equation is
$\displaystyle [x-(2 + i)][x-(2 - i)][x-3]=0$
$\displaystyle (x-2 - i)(x-2 + i)(x-3)=0$
$\displaystyle (x^2-2x+ix-2x+4-2i-ix+2i-i^2)(x-3)=0$
$\displaystyle (x^2-4x+5)(x-3)=0$
$\displaystyle x^3-3x^2-4x^2+12x+5x-15=0$
$\displaystyle x^3-7x^2+17x-15=0$