Question

The value of -1 is a lower bound for the zeros of the function show below.

Answer 1

To perform this check, you must use the following theorem: $a$ is a lower bound for the zeroes of $f(x)$ if, when divide $f(x)$ by $(x-a)$, the quotient and the remainter alternate signs.

Since the long division yields the result

$\dfrac{x^4+x^3-11x^2-9x+18}{x+1}=x^3-11x+2+\dfrac{16}{x+1}$

You can see that the signs don't alternate (the last two terms are positive). So, -1 is not a lower bound.

In fact, the actual roots of the polynomial are -3, -2, 1, 3, so as you can see there are roots smaller than -1.