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UPDATE 9/29/18: The solution to this problem is that the statement is an "if-then" situation. Unfortunately I was interpreting the theorem in the converse: I thought that any upper bound will satisfy the criteria. However, it only says that IF the selected point satisfies the test, then we can surely say that it will be an upper bound for the zeros. It doesn't mean that any upper bound will satisfy the test.


While trying to understand the upper and lower bounds of real zeros of a polynomial, I have come across something that seems to go against the logic in the textbook.

Let $f(x)$ be a polynomial with real coefficients and a positive leading coefficient. Suppose that $f(x)$ is divided by $x-c$ using synthetic division.

  • If $c > 0$ and each number in the bottom row is either positive or zero, $c$ is an upper bound for the real zeros of $f$.
  • If $c < 0$ and the numbers in the bottom row are alternately positive and negative (zero entries count as positive or negative), then $c$ is a lower bound for the real zeros of $f$.

I'm trying to find the real zeros of $f(x) = 4x^4 - 20x^3 + 37x^2 - 24x + 5$. According to WolframAlpha, there is only one real zero at $x = {1\over2}$ (with multiplicity $2$). This would mean that anything after that would not be a zero according to the Rational Zero Theorem. For example, if I use synthetic division on one of the possible rational zeros, ${5\over4}$, then clearly ${1\over2} < {5\over 4}$ and $$\begin{array}{cccccc}\boxed{5\over4} & 4 & -20 & 37 & -24 & 5\\ & & 5 & -{75\over4} & {365\over16}& -{95\over64}\\\hline & 4 & -15 & {73\over4} & -{19\over16} & {225\over64}\end{array}$$

The signs alternate instead of being all positive or zero. Did I make a mistake somewhere in the synthetic division? Even if, immediately at the start, I know we end up with $-15$ which kills the concept of the upper bound right then and there.

Or is there a slight subtlety I'm missing in the Upper and Lower Bound rules? Even if I apply synthetic division to the only zero:

$$\begin{array}{cccccc}\boxed{1\over2} & 4 & -20 & 37 & -24 & 5\\ & & 2 & -9 & 14& -5\\\hline & 4 & -18 & 28 & -10 & \boxed{0}\end{array}$$

The numbers in the bottom row are not all positive or zero, which isn't telling me that ${1\over2}$ is an upper bound like I'd expect it to.

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note that your polynomial has the form $$f(x)=(x^2-4x+5)(2x-1)^2$$

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  • $\begingroup$ I'm aware of this, but what about it exactly? To me, $f(x)$ matches the criterion: it has real coefficients, and a positive leading coefficient, but it isn't evident why the upper bound rule isn't working. $\endgroup$ – Decaf-Math Oct 3 '17 at 19:53
  • $\begingroup$ the upper and the lower bound is $$\frac{1}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '17 at 19:54
  • $\begingroup$ It's clear that ${1\over2}$ is both the lower and upper bound for the real zeros. However, $$\begin{array}{cccccc}\boxed{1\over2} & 4 & -20 & 37 & -24 & 5\\ & & 2 & -9 & 14& -5\\\hline & 4 & -18 & 28 & -10 & \boxed{0}\end{array}$$ Since ${1\over2}$ is positive, why aren't the numbers in the bottom row staying positive or $0$ to indicate that it is an upper bound? (according to the rules) $\endgroup$ – Decaf-Math Oct 3 '17 at 19:58
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You're trying to make the theorem say more than it really does. The converse of the theorem isn't true. Just because a number doesn't show up as an upper bound, doesn't mean that there is a real number greater than it that's a zero. The thing that I tell my students to keep in mind (especially with Descartes rule of signs) is that complex zeros will pretend to be either positive or negative, and the complex (imaginary) zeros for this function are beyond 5/4. Your example doesn't go against the theorem, you're just expecting too much out of it.

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