Currently, we are learning how to solve inequalities of degrees greater than 2 such as $x^3+9x^2+26x+24 < 0$.
Our teacher told us to graph and find the intervals to solve for $x$ but is there an efficient method to find the $x$'s rather than graphing them? Our teacher also told us to find each case, but that is also inefficient.
What is the best method to find the $x$'s of any inequality greater than degree 2?
Thanks.
NOTE: We aren't going to be finding inequalities for degrees higher than 4. Therefore, if there does exist such a method that works more efficiently with relatively low degrees, then that is fine also.
Best Answer
Suppose your inequality is $P(x) < 0$. Let $r_1 < \ldots < r_k$ be the distinct real roots of $P(x)$, with corresponding multiplicities $d_1, \ldots, d_k$. Thus $P(x) = (x - r_1)^{d_1} (x - r_2)^{d_2} \ldots (x - r_k)^{d_k} Q(x)$ where $Q(x)$ has the same sign for all $x$ (the sign of the leading coefficient of $P$). As $x$ increases or decreases, the sign of $P(x)$ changes at $x = r_j$ if $d_j$ is odd but stays the same if $d_j$ is even. To find the $r_j$, you generally will need to use numerical methods.
In your example, the roots happen to be integers $-2,-3,-4$ and the multiplicities are all $1$. Obviously $P(x) > 0$ for $x > 0$, and thus also for $x > -2$. At each root the sign changes, so $P(x) < 0$ for $-3 < x < -2$, $P(x) > 0$ for $-4 < x < -3$, and $P(x) < 0$ for $x < -4$.
On the other hand, if it had been $R(x) = {x}^{4}+12\,{x}^{3}+53\,{x}^{2}+102\,x+72 = (x + 3) P(x)$, the roots would be the same except that the multiplicity of the root $-3$ is $2$ instead of $1$. Then at $-3$ the sign doesn't change, so $R(x) < 0$ for both $-3 < x < -2$ and $-4 < x < -3$, and $R(x) > 0$ for $x < -4$.