How to solve this inequality where two zeroes of different factors of the numerator are the same? How do we solve this using sign-scheme method? $\dfrac{(e^x-1)x^{49}(1-x)(x-3)^{50}}{(x-2)(x-4)}\geqslant0$
Here, when I find zeroes of $(e^x-1)$ and $x^{49}$, both of them are zero. The solution set for this inequality in my book is given as $x\in(-\infty, 1] ∪ (2,3]$ but I can't figure out if I'm making some error or the book has a misprint. Please help me out here. Thanks in advance!
 A: EDIT:
I naïvely thought the book was correct and didn’t bother to scrutinise it - why do we get so many questions with mistaken books? Anyway, as Antonio says, the book erred and the second interval should be $(2,4)$.
OP:
There are just other zeros of the function. The zero of $x=0$ is contained in that interval given in your textbook, as are the zeroes of $x=1$ and $x=3$. The second interval is open at $2$ since $x=2$ is a pole of the inequality (division by zero) but all points just above it give positive values.
Moreover, when $x$ is negative, we have $e^x\lt1$, so the first bracket $(e^x-1)$ is negative, but then it is multiplied by $x^{49}$, an odd power, so the two negatives cancel to a positive. If you step through the remaining terms you find the whole inequality is positive for negative $x$: $(1-x)$ is positive, $(x-3)^{50}$, an even power, is still positive, the denominators $(x-2)$ and $(x-4)$ are both negative, multiplying to make a positive.
EDIT:
The graph of $e^x$ looks like this:

Notice how for all real $x$, it is greater than $0$, and also notice how it is always greater than $1$ when $x\gt0$. It grows very quickly - exponentially!
A: Since $\;x(e^x-1)>0\;,\;x^{48}>0\;$ for any $\;x\in\mathbb R\setminus\{0\}\;$ and $\;(x-3)^{50}>0\;$ for any $\;x\in\mathbb R\setminus\{3\}\;,\;$ your inequality is equivalent to
$\dfrac{1-x}{(x-2)(x-4)}\geqslant0\quad\lor\quad x=0\quad\lor\quad x=3\;.$
Moreover ,
$1-x\geqslant0\iff x\leqslant1\;,$
$(x-2)(x-4)>0\iff x<2\;\lor\;x>4\;.$
Consequently ,
$\dfrac{1-x}{(x-2)(x-4)}\geqslant0\iff x\leqslant1\;\lor\;2<x<4\;.$
Hence the solution set for your inequality is :
$(-\infty,1]\cup(2,4)\;.$
