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.

I just solved the equation with the following method:
$$\color{red}{(x-2)}\color{green}{(x+1)(x+6)}\color{red}{(x+9)}+108=0$$
$$(x^2+7x-18)(x^2+7x+6)+108=0$$
By using the substitution $t=x^2+7x$  we get,
$$t^2-12t=0\Rightarrow t_1=0\quad,t_2=12$$So we have,
$x^2+7x=0\Rightarrow \quad x_1=0 ,\quad x_2=-7$
$x^2+7x-12=0\Rightarrow\quad x_{3}=\dfrac{-7+\sqrt{97}}{2},\quad x_4=\dfrac{-7-\sqrt{97}}{2}$
Hence $x_2,x_4\in(-1,-10)$
