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.
 A: 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 $x_2,x_4\in(-1,-10)$
A: 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.
A: 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:

A: It suffices to sketch the graph of $(x-2)(x+1)(x+6)(x+9)$, which is easy because we know the roots.
The crucial part is deciding whether the local minima are smaller than $-108$.
Since they are $-144$, the number of solutions of $f(x)=-108$ is the same as the number of solutions of $f(x)=0$.
(Also, it turns out that the derivative is a cubic that can be factored.)

