# Please help soon. Due tomorrow. Can anyone help me create a polynomial function with given zeros?

## The zeros are 2, 4 + i, 4 - i The i is imaginary I can get this far: (x-2)(x-4+i)(x+4-i) But I'm confused with the signs for the second and third sets of parenthesis. I'm not sure what signs are supposed to change or if I have them correct.

Dec 5, 2016

The zeros are $2$, and $4 + i$ and $4 - i$

The factors are:
$\left(x - 2\right)$, $\text{ }$ $\left(x - \left(4 + i\right)\right)$, $\text{ }$, and $\text{ }$ $\left(x - \left(4 - i\right)\right)$

I would start by multiplying

$\left(x - \left(4 + i\right)\right) \left(x - \left(4 - i\right)\right) = {x}^{2} - x \left(4 - i\right) - x \left(4 + i\right) + \left(4 + i\right) \left(4 - i\right)$

$= {x}^{2} - 4 x + x i - 4 x - x i + \left(16 - {i}^{2}\right)$

$= {x}^{2} - 8 x + \left(16 - \left(- 1\right)\right)$

$= {x}^{2} - 8 x + 17$

Note that I used the product of a sum and difference $\left(4 + i\right) \left(4 - i\right) = 16 - {i}^{2}$. You can multiply htis out in more detail is you like.

Now multiply $\left(x - 2\right) \left({x}^{2} - 8 x + 17\right)$ to get

${x}^{3} - 8 {x}^{2} + 17 x - 2 {x}^{2} + 16 x - 34$

$= {x}^{3} - 10 {x}^{2} + 33 x - 34$