How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$? How many solutions the equation $(x-2)(x+1)(x+6)(x+9)+108=0$ has in the interval $(-10,-1)$ ?
Here is my work:
By expanding the expression we get, $$(x^2-x-2)(x^2+15x+54)+108=x^4+14x^3+37x^2-84x$$
So I got $x(x^3+14x^2+37x-84)=0$. One root is zero which doesn't lie in the interval $(-10,-1)$. But I don't know how many roots the cubic equation has in that interval.

The midpoints of $-2,9$ and $1,6$ are the same: $3.5$. Thus let $x = t -  3.5$:
$$(t-5.5)(t+5.5)(t - 2.5)(t + 2.5) + 108 = 0$$
$$(t^2 - 30.25)(t^2 - 6.25) + 108 = 0$$
$$t^4 - 36.5t^2 + \text{constant term} \ (297.0625) = 0$$
Since $t \to -t$ results in the same polynomial, it is symmetric across the line $x = -3.5$. Also, since $x = 0$ is one root, $x = -7$ is another.
The equation of symmetry in between the pairs of roots is:
$$t^2 = -\frac{-36.5}{2} \implies x_{\text{positive line of symmetry}} < -3.5 + \sqrt{50/2} = 1.5$$
hence the other negative root is strictly greater than $-7-2(1.5) = -10$. This implies that there are only $2$ roots that lie in $x \in (-10,-1)$: all without a calculator (if you work out the multiplications by hand).
Here is a graph of the polynomial and the axes of symmetry:

