Question

The largest root of the equation $(x-5)(x-7)(x+6)(x+4)=504$ is

Answer 1

$(x-5)(x-7)(x+6)(x+4)=504\Rightarrow (x-5)(x+4)(x-7)(x+6)=504\Rightarrow (x^2-x-20)(x^2-x-42)=504$
Assuming $y=x^2-x-20$, we get
$\Rightarrow y(y-22)=504\Rightarrow y^2-22y-504=0\Rightarrow (y-11{)}^2-121-504=0\Rightarrow (y-11{)}^2=625\Rightarrow y=11\pm 25\Rightarrow y=-14,36\Rightarrow x^2-x-20=-14,36$
So, we get two quadratic equation,
$x^2-x-6=0\text{and}{x}^2-x-56=0$
Solving them,
$x^2-x-6=0\Rightarrow (x-3)(x+2)=0\Rightarrow x=-2,3x^2-x-56=0\Rightarrow (x-8)(x+7)=0\Rightarrow x=-7,8$

Hence, the largest root is $8.$