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.

Use Descarte's rule of signs on the cubic. It will allow you to bypass exact calculations. Let $$f(x)=x^3+14 x^2+37 x-84$$
Then for $f(-x)$, we see that there are $2$ sign changes. This means that there are either $2$ or $0$ negative roots. Now, note that $f(-1)<0$, while $f(-8)=4>0$, and also $f(-10)<0$. Hence by intermediate value theorem there is atleast $1$ root in $(-10,-8)$ and $(-8,-1)$ respectively. This means that there are exactly $2$ negative roots, since we earlier concluded that there can't be more.
Thus, we get $2$ possible roots in the given interval for the biquadratic too.
