1. f(x) = x^3 - 7x^2 + 17x - 15 ; 2+i
2. f(x) = x^3 + 6x + 20 ; 1-3i
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 ; 1-3i
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 2-i 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 -9-3i 19+3i
------------------
1 -5+i 8-3i 4+3i
Since 2+i is given as a zero, the remainder should be zero, and not (4+3i) Using 2-i gives similar results. I'm not understanding what I'm doing wrong.
Have you tried expanding
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.
Now divide that into the given polynomial.
[x-(2+i)][x-(2-i)]
x^2-(2+i)x-(2-i)x-(2+i)(2-i)
x^2-2x+ix-2x-ix-(4-i^2)
x^2-4x-(4+1)
x^2-4x-3
How did I do the algebra incorrectly?
Personally, I think it's much easier to expand something of the formQuote:
Originally Posted by Rocker1414
by recognising it as the factorisation of the sum of two squares:
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.