Follow how the inequality $$ -x^4+10x^3+11x^2>0 $$ is solved.
(1) First we rewrite the given inequality to: $$ \begin{gather} -x^2(x^2-10x-11)>0 \cr x^2(x^2-10x-11)<0 \end{gather} $$
(2) Now we factor the trinomial within the parentheses: $$ \begin{gather} D =(-10)^2-4\cdot (-11) \cr x_1=\frac{10-\sqrt{144}}{2}=-1,x_2=\frac{10+\sqrt{144}}{2}=11 \cr x^2-10x-11=(x+1)(x-11) \end{gather} $$ (3) Finally we rewrite the inequality as a product of three factors: $$ x^2(x+1)(x-11)<0 $$
(4) Let $P(x)=x^2(x+1)(x-11)$. We immediately observe that $P(x)=0$ when $x=-1$, $0$ and $11$. These zero points divide the number line into intervals: $$ (- \infty;-1),~(-1;0),~(0;11),~(11;+\infty) $$ (5) We know, that $P(x)$ changes sign whenever $x$ passes a zero point. $P(x)>0$ on the first interval $(- \infty;-1)$. Thus, we can conclude that the solution set of the inequality is the set of all $$x\in (-1;0)\cup (11;+\infty).$$ Is there an error in the solution, or is everything all right?
The solution is correct.
The error is in step (2). The trinomial must be factored this way: $$ x^2-10x-11=(x-1)(x+11) $$
The error is in step (4). Polynomial $P(x)=x^2(x+1)(x-11)$ should be divided by the squared factor $x^2$. This results in the intervals: $(- \infty;-1)$, $(-1;11)$, $(11;+\infty)$.
The error is in step (5). Solution is not correct.
A polynomial does not always change sign whenever $x$ passes a zero point. This is exactly the case of the polynomial $P(x)=x^2(x+1)(x−11)$. See the table:
| $x$ | $(- \infty;-1)$ | $(-1;0)$ | $(0;11)$ | $(11;+\infty)$ |
|---|---|---|---|---|
| $x^2$ | $+$ | $+$ | $+$ | $+$ |
| $x+1$ | $-$ | $+$ | $+$ | $+$ |
| $x-11$ | $-$ | $-$ | $-$ | $+$ |
| $P(x)$ | $\oplus$ | $\ominus$ | $\ominus$ | $\oplus$ |
From the table, it is evident that the solution set to the inequality is the set of all $$x\in (−1;0)\cup (0;11)$$
Importantly, the polynomial $P(x)=x^2(x+1)(x−11)$ has the squared factor $x^2$, making $x=0$ a double root of $P(x)=0$. In such cases, the polynomial does not change sign as $x$ passes through this zero point.