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 $x+5$ & got the answer $x^2+5x+10$ and factored it like this:


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



Since $x=0$ & $x=-5$ are roots of the given equation,$$ (x+1)(x+2)(x+3)(x+4)-24 = x(x+5)cdot q(x) ag1 $$where $q(x)$ is a monic second-degree polynomial. We may notice that, by De l"Hopital"s rule,$$ q(0) = lim_x o 0frac(x+1)(x+2)(x+3)(x+4)-24x(x+5)=frac24,H_45=10 ag2$$and if $q(x)=x^2+Kx+10$, in order that the coefficient of $x^3$ is the same in both sides of $(1)$$$ 1+2+3+4 = K+5 ag3 $$i.e. $K=5$, has to lớn hold.

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



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

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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