After that I solved the equation $x^3+10x^2+35x+50=0$and found the integer solution $-5$and I divided the polynomial to $x+5$ and got the answer $x^2+5x+10$ and factored it like this:

$x(x+5)(x^2+5x+10)$

I want an easier way to solve it; which way would you recommend?

Since $x=0$ and $x=-5$ are roots of the given equation,$$ (x+1)(x+2)(x+3)(x+4)-24 = x(x+5)\cdot q(x) \tag{1} $$where $q(x)$ is a monic second-degree polynomial. We may notice that, by De l”Hopital”s rule,$$ q(0) = \lim_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}=\frac{24\,H_4}{5}=10 \tag{2}$$and if $q(x)=x^2+Kx+10$, in order that the coefficient of $x^3$ is the same in both sides of $(1)$$$ 1+2+3+4 = K+5 \tag{3} $$i.e. $K=5$, has to hold.

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Since the polynomial is symmetric around $x+2.5$ let us set $y=x+2.5$.

Then$$(x+1)(x+2)(x+3)(x+4)-24=(y-\frac{3}{2})(y-\frac{1}{2})(y+\frac{1}{2})(y+\frac{3}{2})-24\\=(y^2-\frac{1}{4})(y^2-\frac{9}{4})-24=y^4-\frac{5}{2}y^2-\frac{375}{16}=(y^2+\frac{15}{4})(y^2-\frac{25}{4})$$

$(x+1)(x+2)(x+3)(x+4) – 24 =(x^2 +5x +4)(x^2+5x+6) -24=Y$

Let $x^2+5x=t$.

So, $Y=(t+4)(t+6) – 24 = t^2+10t=t(t+10)$Implying, $Y=(x^2+5x)(x^2+5x+10)=x(x+5)(x^2+5x+10)$.

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