Question

Find roots

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120$

Answer 1

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120$

$x^4+10x^3+35x^2+50x+24=120$

$x^4+10x^3+35x^2+50x-96=0$

$(x-1)(x^3+11x^2+46x+96)=0$

$(x-1)(x+6)(x^2+5x+16)=0$

$(x-1)(x+6)(x-\frac{-5+\sqrt{39}i}{2})(x+\frac{-5-\sqrt{39}i}{2})=0$

So

$x=1,-6,\frac{-5\pm \sqrt{39}i}{2}$