Nth degree polynomial function

[outime]

Hi, I was trying to solve this problem, but so far this is what I got.

Find an nth degree polynomial function with real coefficients satisfying the given conditions.

n = 3; -1 and -2 + 3i are zeros; leading coefficient is 1

(X+1)(X-(-2+3i))(X-(-2-3i))

(X+1)(X+2-3i)(X+2+3i)

Then I'm suppose to combine those terms? I don't really know from here.

Thanks

 

[outime]

The magic happens every time.

[x - (a + bi)] * [x - (a - bi)] = x^2 - x[(a + bi) + (a - bi)] - [(a + bi)(a - bi)] =

x^2 - x(a + bi + a - bi) - [a^2 - (bi)^2] = x^2 - 2ax - {a^2 - [b^2 * i^2]} = x^2 - 2ax + {a^2 - [b^2 * (-1)]} =

x^2 - 2ax + a^2 - [ - (b^2)] =

x^2 - 2ax + a^2 + b^2. No complex terms

First of all, thanks for your answer.

Giving x^2 - 2ax + a^2 + b^2. then it will be:

X²+4x+5

So (X+1)(X²+4x+5)

The final answer will be: X³+5x²+9x+5

But there is something wrong on my calculation, given that the choices are:

A) f(x) = x³ + 3x² + 15x - 13
B) f(x) = x³ + 3x² + 9x - 13
C) f(x) = x³ + 5x² + 9x - 14
D) f(x) = x³ - 4x² + 9x - 13

Thanks