How can one prove that the following function has 4 real roots? $$x^4 -6x^3 +x^2 +10x+1=0$$
The problem is that roots don't seem to be possible to compute by hand.
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2Have you heard of the Intermediate Value Theorem? or Descartes' Rule of Signs? – JimmyK4542 Dec 01 '14 at 01:35
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1Then try evaluating it at a few small integers, and see if you can find 4 sign changes. – JimmyK4542 Dec 01 '14 at 01:39
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thanks I was confused with Rolle theorem – GorillaApe Dec 01 '14 at 01:47
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Here is a kind of cheaty proof:
$$f(-5) = 1351 \\ f(-1) = -1 \\ f(1) = 7 \\ f(4) = -71 \\ f(7) = 463$$
So there are zeroes in the intervals $(-5,-1), (-1,1), (1,4)$ and $(4,7)$, because polynomials are continuous functions.
brick
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Another cheaty proof. $p(x)=0$ is equivalent to: $$f(x)=(x+1)^2(x-4)^2 = 14x+15 =g(x).$$ Since for $x\in\{-\infty,-1,0,4,+\infty\}$ the relations between the LHS and the RHS are $>,<,>,<,>$ we must have four real roots in the intervals $(-\infty,-1),(-1,0),(0,4),(4,+\infty)$.
By the Rouché theorem, all the roots lie in the ball having center $x=\frac{3}{2}$ and radius $\rho=\frac{9}{2}$, hence in $(-3,6)$, so the previous intervals become $(-3,-1),(-1,0),(0,4),(4,6)$.
Jack D'Aurizio
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Where did you get $f(x)$? You mentioned $p(x), f(x), g(x)$ without explanation. Equivalence? As in modular arithmetic? I dont know Rouche's theorem, but the wiki article is rather convoluted and doesnt really look like it relates to polynomials. Vaguely. – CogitoErgoCogitoSum Dec 01 '14 at 11:59
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