Solve the inequation for $x$

Question

Solve for $x$ :

$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $

I tried cases like : $ x-1 \le 0 $ , $ x \le 0 $ , $ x+1 \le 0 $

Answer 1

Well, let's see if we can simplify your expression first:

$$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $$ $$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$

Think, first, what qualifies $x$ to be $\le 0$? If $x$ is $0$ or a negative number, correct?

Well, our expression can be $0$ if $x = -1, 1$

Thus, now lets look when our expression yields a negative number.

Therefore, look at the powers and see which ones could be negative:

$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$

We can see that $(x^2 - 1)^{2005}$ can only be negative, since the other $2$ terms have even powers, meaning they will only be positive.

Therefore, when is $(x^2 - 1)^{2005} \le 0$?

Well,

$$x^2 - 1 \le 0$$

$$x^2 \le 1$$

$$-1 \le x \le 1$$

And thus we arrive at our answer.

Answer 2

By the rule of signs, $$\DeclareMathOperator{\sgn}{sgn}\sgn{(x-1)^{2005}x^{2006}(x+1)^{2007}}=\sgn{(x-1)(x+1)}=\sgn(x^2-1), $$ hence the expression is $\;\le 0\;$ if and only if $\;-1\le x\le 1$.