Solve $x^3(x^2+1)^2(x-1)(x+1)>0$

Question

I need to solve this inequality:

$$x^3(x^2+1)^2(x-1)(x+1)>0$$

Note: I haven't learnt imaginary numbers. Does the $(x^2+1)^2$ affect the inequality?

Thanks

The $(x^2+1)^2$ does not effect the inequality because it is always positive. — user157227, Oct 07 '14 at 13:14
This has told you where the 0's are, and since this is a polynomial, it is continuous everywhere. It's enough to determine whether or not this is positive at a single point in each interval between 0's, since the Intermediate Value Theorem assures you there are no more. — A. Thomas Yerger, Oct 07 '14 at 13:15
So I can divide it out? What if it had just been $x^2$? Would I have to test for x=0? — Jasern. L, Oct 07 '14 at 13:15
Google "solving inequalities" and you will find the instructions. — E W H Lee, Oct 07 '14 at 13:15
@Jasern: Since $(x^2+1)^2$ is positive for all $x \in \mathbb{R}$, yes you can divide both sides to obtain an equivalent inequality. — E W H Lee, Oct 07 '14 at 13:17

score 3 · Answer 1 · answered Oct 07 '14 at 13:16

3

Hint:
You have $(x^2+1)^2\gt0$ over the reals, therefore $$x^3(x^2+1)^2(x-1)(x+1)\gt0\iff x^3(x-1)(x+1)\gt0.$$

answered Oct 07 '14 at 13:16

Hakim

10,213

score 2 · Accepted Answer · answered Oct 07 '14 at 13:15

2

Firs of all, we have $x\not =0$.

Then, $x^2(x^2+1)^2\gt 0$, so the inequality is equivalent to $$x(x-1)(x+1)\gt 0\iff x(x^2-1)\gt 0$$$$\iff "x\gt 0\ \text{and}\ x^2-1\gt 0"\ \ \text{or}\ \ "x\lt 0\ \text{and}\ x^2-1\lt 0"$$ $$\iff x\gt 1\ \ \text{or}\ \ -1\lt x\lt 0.$$

answered Oct 07 '14 at 13:15

mathlove

139,939

How do we know x cannot equal 0? – Jasern. L Oct 07 '14 at 13:17
3

@Jasern.L Try to plug in $x=0$ then your inequality yields $0\gt0$ which is false. – Hakim Oct 07 '14 at 13:18
Oh right, thanks! – Jasern. L Oct 07 '14 at 13:19

Bman72 · Answer 3 · 2014-10-07T13:26:42.727

The term $(x^2+1)^2$ does not affect the inequality, since when you square a real number you always get a positive number, and $x^2+1$ is a real number, in your case. So what remains to look at is when

$$x^3(x-1)(x+1)>0$$

Now you have three critical point, if you want: $x_1=-1,$ $x_2=0$, $x_3=1$.

You need to look at what sign the function $f(x) := x^3(x-1)(x+1)$ has in each interval. For instance

$$\begin{array}{|c|c|c|c|c|} \hline x\in&(-\infty,-1)&(-1,0)&(0,1)&(1,\infty)\\\hline \implies\text{signum $f(x)$} &-\cdot -\cdot -=-&-\cdot-\cdot+=+&+\cdot -\cdot +=-&+\cdot + \cdot +=+\\\hline\end{array}$$

Where for $x\in (-\infty,-1)$ we have for instance $x^3 <0$, $(x-1)<0$, $(x+1)<0$. (similarly for the other cases)

So you have that

$$x^3 (x^2+1)^2(x-1)(x+1)>0 \iff x\in (-1,0)\cup (1,\infty)$$

($-1, 0, 1$ are not included since there you have $f(x)=0$ which is not greater than $0$).

score 0 · Answer 4 · answered Oct 07 '14 at 13:32

IF you have not learned about complex numbers, I'll assume that $x \in \mathbb{R}$. Then, you can simplify the inequality as:

$$x^3(x-1)(x+1)>0$$ because $(x^2+1)^2>0\,\,\,\forall x\in\mathbb{R}$.

Now, the easiest way to solve this is searching for the critical points -where the expression on the LHS equals $0$- so we analize the intervals in between:

a) Search for critical points: $$x^3(x-1)(x+1)>0\Rightarrow x^3=0 \vee (x-1)=0\vee (x+1)=0 $$ $$x^3=0\Rightarrow x=0$$ $$x-1=0\Rightarrow x=1$$ $$x+1=0\Rightarrow x=-1$$

b) Make a table with the real line, and these critical points marked. Then, analyze each expression that conforms the inequality, and anaalyze where they're positive or negative:

$$\begin{array}{c|ccccccc|c} & &-1& &0& &1&&\\\hline x^3&-&|&-&|&+&|&+&x^3>0\text{ when }x>0\\ x-1&-&|&-&|&-&|&+&x-1>0\text{ when }x>1\\ x+1&-&|&+&|&+&|&+&x+1>0\text{ when }x>-1\\\hline x^3(x-1)(x+1)&-&|&+&|&-&|&+&\text{result of multiplying}\\ \end{array}$$

So, the solution is the union of all the cases where the expression is positive: $$x\in]-1,0[\,\cup\,]1,\infty[$$

Solve $x^3(x^2+1)^2(x-1)(x+1)>0$

4 Answers4