Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 4, -8, and 2 +3i

Question

anonymous · Answer

if one zero is $2+3i$ then the other must be its conjugate $2-3i$ 

you know it will look like 
$$(x-4)(x+8)(x-(2+3i))(x-(2-3i))$$ right ?  the question is, how do you multiply all this mess out without going nuts

anonymous · Answer

I think you distribute them separately

anonymous · Answer

the $(x-4)(x+8)$ part is routine
as for the other part, it is easier to work backwards

anonymous · Answer

$$x=2+3i\x-2=3i\(x-2)^2=(3i)^2=-9\x^2-4x+4=-9\x^2-4x+13=0$$

anonymous · Answer

so your quadratic from the second two factors is 
$$x^2-4x+13$$

anonymous · Answer

whats even easier is memorizing that is $a+bi$ is a zero, then the quadratic is 
$$x^2-2ax+(a^2+b^2)$$