I recently absolutely bombed my precalculus mid-term, and I'm reeling from that at the moment. I'm trying to find out where I went wrong.
One of the ones I'm continuing to have problems with says (and I've changed the numbers here, so I'm not posting any kind test material): find a polynomial function of lowest degree with rational coefficients and some of its zeroes at $0$, $-\sqrt{3}$, and $1+i$.
So, the first thing I did was say: $$f(x) = x(x+\sqrt{3})(x-(1-i))(x-(1+i))$$
Then... I kind of lost the plot. I multiplied everything out, hoping to reduce to a single polynomial, but what I ended with (twice) wasn't an answer choice.
so, starting with the two right side parens (excuse the crappy ASCII table): $$(x-1+i)(x-1-i)$$
x -1 +i ------------ x | x2 -x +xi | -1|-x +1 -i | -i|-xi +i +1
...which got me a total of $x^2-2x+2$
Plugging that back into the original:
$$f(x) = x(x+\sqrt{3})(x^2-2x+2)$$
Then taking the easy route and multiplying the sigleton $x$:
$$f(x) = (x+\sqrt{3})(x^3-2x^2+2x)$$
...and then another crappy ASCII table:
x +√3
------------
x3 | x4 +√3x3
|
-2x2 |-2x3 -2√3x2
|
+2x |+2x2 +2√3x
...which equals: $$f(x) = x^4-2x^3+\sqrt{3}x^3-\sqrt{3}x^2+2\sqrt{3}x$$
which I suppose could be written: $$f(x) = x^4 + (-2+\sqrt{3})x^3-\sqrt{3}x^2+2\sqrt{3}x$$
...and none of the answers was even close to this.
Where did I go wrong? Am I approaching this problem the wrong way, somehow?
The answer I ended up going with (which was marked wrong) looked like the original problem I set up: $$f(x) = x(x+\sqrt{3})(x-(1-i))(x-(1+i))$$
Please hellllllp!
EDIT
D'oh. I left out the $+\sqrt{3}$... so, putting that back in there:
x +√3
----------
x | x2 +√3x
-√3|-√3x -3
so: $$f(x) = (x^2-3)(x^3-2x^2+2x)$$
$$f(x) = x^5-2x^4-x^3+6x^2-6x$$
ARGH Thank you so much @quasi !