I've been having trouble with this problem: Find a polynomial function of minimum degree with $-1$ and $1-\sqrt{3}$ as zeros. Function must have integer coefficients.
When I tried it, I got this:
\begin{align} (x+1)(x-(1-\sqrt{3}))=& x^2 - x(1-\sqrt{3}) + x - 1 + \sqrt{3}\\ =& x^2 - x + x(\sqrt{3}) + x - 1 + \sqrt{3}\\ =& x^2 + x(\sqrt{3}) - 1 + \sqrt{3} \end{align}
This looks completely wrong to me and it probably is as my teacher said the answer should not be a quadratic. I'm not looking for the answer to be given to me; but if I could get some guidance, it would be greatly appreciated.
$\endgroup$ 52 Answers
$\begingroup$Your polynomial does not work because it does not have integer coefficients. So no quadratic will work, because the two roots are forced to be $-1$ and $1-\sqrt 3$, and you just showed this doesn't work.
However, a cubic will work. Try the one with $-1$, $1-\sqrt 3$, and $1+\sqrt 3$ as roots.
$\endgroup$ 3 $\begingroup$If you're looking for a polynomial that has those two roots and integer coefficients, you'll need to add another root.
Substitute $w = x-1$ so that $(x+1)(x-(1-\sqrt{3})) = (w+2)(w+\sqrt{3})$.
Now multiply by $(w-\sqrt{3})$ to get
$(w+2)(w+\sqrt{3})(w-\sqrt{3}) = (w+2)(w^2-3)$.
Then substitute back in $x$ and you have your polynomial, which is a cubic with integer coefficients:
$(x+1)(x^2-2x-2) = x^3 -2x^2 - 2x + x^2 - 2x - 2 = x^3 - x^2 - 4x - 2$.
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