1. f(x) = x^3  7x^2 + 17x  15 ; 2+i
2. f(x) = x^3 + 6x + 20 ; 13i
3. g(x) = x^4  6x^3 + 6x^2 + 24x  40 ; 3+i
4. g(x) = x^3  3x^2 + 9x  7 ; 1
1. f(x) = x^3  7x^2 + 17x  15 ; 2+i
2. f(x) = x^3 + 6x + 20 ; 13i
3. g(x) = x^4  6x^3 + 6x^2 + 24x  40 ; 3+i
4. g(x) = x^3  3x^2 + 9x  7 ; 1
I see that you have over thirty postings.
By now you should understand that this is not a homework service
So you need to either post some of your work on a problem or explain what you do not understand about the question.
Apologies. While these questions are indeed a few from my homework, I did not ask this simply to find the answers and get done with it, as I am sure is the source of mathhelpforum wanting to make clear that they are not a homework service. I figure that if I understand these 4, I will be able to do all of them. As per request, I'll work through as much as I can of the first one.
x^3  7x^2 + 17x  15; 2+i
By the Conjugate Pair Theorem, both 2+i and 2i are zeroes of the function. The goal is to synthetically divide by the zeroes until the function is down to a quadratic.
2+i 1 7 17 15

 1 2+i 93i 19+3i

1 5+i 83i 4+3i
Since 2+i is given as a zero, the remainder should be zero, and not (4+3i) Using 2i gives similar results. I'm not understanding what I'm doing wrong.
Have you tried expanding $\displaystyle [x(2+i)][x(2i)]?$
That will tell us what the other factor is.
That works out to x^2  4x  3 which, when applying the quadratic formula, works out to be 3 and 1.
but isn't that four answers to a function that only goes to a power of 3? That's not possible.
No you did the algebra incorrectly.
It should be $\displaystyle x^24x+5$.
Now divide that into the given polynomial.
[x(2+i)][x(2i)]
x^2(2+i)x(2i)x(2+i)(2i)
x^22x+ix2xix(4i^2)
x^24x(4+1)
x^24x3
How did I do the algebra incorrectly?
$\displaystyle (xz)(z\overline{z})=x^2(z+\overline{z})x+z\cdot \overline{z}=x^2(z+\overline{z})x+z^2$
Thanks guys. I believe I have it figured out now even by synthetically dividing by both given zeroes. Expanding the form works just as well, plus saves paper ;) Thanks again.