Answer
$f(x)=x^{4}-6x^{3}+10x^{2}+2x-15$
(sample answer with $a=1$)
Work Step by Step
Complex zeros come in conjugate pairs (conjugate zeros theorem).
$2-i$ is a zero $\Rightarrow 2+i$ is also a zero.
If $k$ is a zero, then $(x-k)$ is a factor of $f(x)$ ... (factor theorem)
$(x-3)$ is a factor of f,
$(x+1)$ is a factor of f,
$(x-(2-i))=[(x-2)+i]$ is a factor of f,
$(x-(2+i))=[(x-2)-i]$ is a factor of f.
We assume multiplicity 1, so we have
$f(x)=a(x-3)(x+1)[(x-2)+i][(x-2)-i]$
We are asked for $a$ polynomial, so we choose the one for which $a=1.$
$f(x)=(x-3)(x+1)[(x-2)+i][(x-2)-i]$
... Use FOIL for the first two factors
... Recognize a difference of squares in the last two factors
$f(x)=(x^{2}+x-3x-3)[(x-2)^{2}-i^{2}]$
$=(x^{2}-2x-3)[x^{2}-4x+4-(-1)]$
$=(x^{2}-2x-3)(x^{2}-4x+5)\quad$... distribute,
$=x^{2}(x^{2}-4x+5)-2x(x^{2}-4x+5)-3(x^{2}-4x+5)$
$=x^{4}-4x^{3}+5x^{2}-2x^{3}+8x^{2}-10x-3x^{2}+12x-15$
$f(x)=x^{4}-6x^{3}+10x^{2}+2x-15$
