When we're dealing with an inequaltiy, what do we do when there is a common factor in the numerator and the denominator and it completely cancels the one in denominator?
For example $${{(x-1)(x+5)^4}\over {(x+5)}}\leq0?$$
Do we cancel it and write $x \in [-5,1]$ for an answer?
Or we do not, for it makes the denominator zero when $x=-5$ making the original expression undefined? What happens to the numerator if we exclude the the -5 part from our answer? Does it become
$$(x-1)(x+5)^3 ≤0$$
$$\frac{(x-1)(x+5)^4}{(x+5)}\leq0\\ \iff \left[(x-1)(x+5)^3 \leq0 \text { and } x\neq-5\right].$$
In other words, we are allowed to cancel out the common factor (as explained in Thomas's answer), but in doing so, the fact that the solution set excludes $-5$ needs to be explicated, for this information is no longer encoded within the resulting expression (whereas it was implicit in the original expression).
A typical alternative presentation is to first note that $x\neq-5,$ then write $$\frac{(x-1)(x+5)^4}{(x+5)}\leq0\\ (x-1)(x+5)^3 \leq0$$ and solve, then adjust the solution in the final step. In this case, the connective that is implicitly between those two inequalities is $\implies$ (implies) instead of $\iff$ (if and only if); as such, the resulting solution is only provisional, part of which may need to be excluded in forming the final solution.
As long as $b\neq 0,$ $\frac{ab}{b}=a.$
So if $\frac{ab}{b}\leq 0$, we can conclude $a\leq 0,$ since $\frac{ab}{b}$ and $a$ are the same number.
You. might be confused because, if $a\leq b,$ we can only conclude $ca\leq cb$ if $c\geq 0.$
But that is very different case - the second inequality is about different numbers.
