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Mathematics
General Math
Polynomial degree and root relationship
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[QUOTE="fresh_42, post: 6010183, member: 572553"] 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. Another way is to directly look at possible factorizations ##p(x)=q_1(x)\cdot \ldots \cdot q_n(x)##. The only polynomials, which cannot be factored further are either of the form ##x+a## which give us roots ##-a##, or quadratic. Now gathering all factors, we get $$ p(x)=(x-r_1)\cdot \ldots \cdot (x-r_m)\, \cdot \, q_1(x) \cdot \ldots \cdot q_k(x) $$ with quadratic polynomials ##q_i(x)##. So the degree equation reads: ##\deg p(x)=2\cdot k + m##. So if ##\deg p(x)=7## or any other odd number, ##m## has to be odd, too. [/QUOTE]
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Polynomial degree and root relationship
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