Are there any exceptions to the upper and lower bounds theorem?

Question

Hi there :) I'm new to this site so please let me know if there's anything wrong with my question.

I was trying to find the roots of the polynomial expression $P(x) = 8x^5 - 14x^4 - 22x^3 +57x^2 - 35x + 6$. I first found $(x+2)$ using the Rational Zeros Theorem and synthetic division. Since the result of $\frac{P(x)}{x+2}$ is $Q(x) = 8x^4 - 30x^3 + 38x^2 - 19x + 3$, I assumed $-2$ is a lower bound, because of the alternating signs. And from the graph I can see that's true, since there are no x-intercepts smaller than $-2$.

graph of P(x)

Then, I found $(x - 1)$ as a root. The result $\frac{Q(x)}{x-1} = 8x^3-22x^2+16x-3 = R(x)$ implies that, according to the theorem, 1 is a lower bound of $Q(x)$, but this time, according to the graph, that is not true. I supposed that roots themselves cannot be considered upper and lower bounds.

graph of Q(x)

However, when dividing $R(x)$ by $(x-1)$ I get $8x^2-14x+2 - \frac{1}{x-1}$, which is not a root, but should be a lower bound according to the theorem, and that does not make sense looking at the graph.

graph of R(x)

Am I doing something wrong? Or maybe this is indeed an exception to the theorem? Also, can you tell me whether roots can or cannot be considered as bounds themselves?

Thank you in advance! :)

Edit: I first uploaded the wrong graph of Q(x). Thanks to Noah Schweber for letting me know!

Answer 1

The relevant theorem is:

Suppose $f$ is a polynomial, $c<0$ is a zero of $f$, and the coefficients of the quotient polynomial $f(x)\over x-c$ have alternating signs. Then every zero of $f$ is $\ge c$.

This has two hypotheses on $c$ which must be met: $c$ must be a zero of $f$ and $c$ must be negative. This explains each failure in your question.