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Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?
There are a couple of reasons.Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?
Let ##p(x)\in \mathbb{R}[x]## be our polynomial. If we had a root, say ##r_1##, then with the Euclidean algorithm, a long division, we get ##p(x)=q(x) \cdot (x-r_1)## since ##p(r_1)=0##. There is at least one real root, as the graph of ##p(x)## has to cross the ##x-##axis at least once, because it comes from ##+\infty## and goes to ##-\infty## or vice versa. Now we can go on with ##q(x)## which has degree ##6##. Either has ##q(x)## also a real root, which does not have to the case, or it has not. If it has, say ##r_2##, we continue the division by ##(x-r_2)##. But then we get a polynomial of degree ##5##, which thus again has to have a real root, because the degree is odd and the graph has again to cross the ##x-##axis. At the end, we will get an odd number of roots.Why is it that for a 7th degree polynomial, the number of real roots is either 1, 3, 5, or 7?