Explanation for 2 inequalities with same solutions. While studying, I read that $(x-a)^{2k \pm 1}f(x)>0$ has the same solution as $(x-a)f(x)>0$. I do not get why this is true. Could someone please explain why these two inequalities have the same solution?
 A: Consider the pair of inequalities: 
$(x-a) f(x) > 0$
$(x-a)^{2k+1} f(x) >0$
In inequality 1, Assume $f(x) > 0$ 
It follows that $(x-a)$ must then also be greater than 0. 
This is true when $ x > a$ 
in intersection with x such that 
$f(x) > 0$
For the second inequality, assume $f(x) >0$ It follows that $(x-a)^{2k+1} >0$ 
Taking the 2k+1 root, it follows that $ x >a$, and the solution is the same as above.  Thus, in this case, the two have the same solution. 
The interesting part comes in case 2: 
Assume $f(x) < 0$ It follows in inequality 1, that $x-a <0 \Rightarrow x < a$, in intersection with whenever $f(x) < 0$
In inequality 2, if we assume $f(x) < 0$, then $(x-a)^{2k+1} <0$ Again, taking the 2k+1 root gives the same solution as above. 
But what's special about 2k+1? That essentially means that (x-a) must have an odd power. If (x-a) had an even power, then $(x-a)^{2k} <0$ would never be satisfied, whereas it would be satisfied in inequality 1. 
What this means is that for all integer k, $(x-a)^{2k+1} f(x) >0$ will have the same solution set. 
A: Suppose that $k$ is a positive integer.
If we have
$$(x-a)^{2k+1}f(x)=(x-a)(x-a)^{2k}f(x)\gt 0,$$
then $(x-a)^{2k}=((x-a)^k)^2$ is strictly positive (since $x=a$ is not a solution), we can divide the both sides by $(x-a)^{2k}\gt 0$ to get $$(x-a)(x-a)^{2k}f(x)\gt 0\Rightarrow (x-a)f(x)\gt 0.$$
On the other hand, if we have
$$(x-a)f(x)\gt 0,$$
then $(x-a)^{2k}=((x-a)^k)^2$ is strictly positive (since $x=a$ is not a solution), we can multiply the both sides by $(x-a)^{2k}\gt 0$ to get
$$(x-a)f(x)\gt 0\Rightarrow (x-a)^{2k+1}f(x)\gt 0.$$
Thus, we have
$$(x-a)^{2k+1}f(x)\gt 0\iff (x-a)f(x)\gt 0.$$
