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\$ & factored it like this:

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

I want an easier way to lớn 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) 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 hold.

Bạn đang xem: Solve inequalities with step Since the polynomial is symmetric around \$x+2.5\$ let us mix \$y=x+2.5\$.

Then\$\$(x+1)(x+2)(x+3)(x+4)-24=(y-frac32)(y-frac12)(y+frac12)(y+frac32)-24\=(y^2-frac14)(y^2-frac94)-24=y^4-frac52y^2-frac37516=(y^2+frac154)(y^2-frac254)\$\$ \$(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)\$. Thanks for contributing an answer to usogorsk.comematics Stack Exchange!

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