Question

Solve $(x-5)(x-7)(x+6)(x+4)=504$.

Answer 1

$(x-5)(x-7)(x+6)(x+4)=504\left\{\right(x-5\left)\right(x+4\left)\right\}\left\{\right(x-7\left)\right(x+6\left)\right\}=504(x^2-x-20)(x^2-x-42)=504$

Put $x^2-x=t$

$(t-20)(t-42)=504t^2-62t+840=504t^2-62t+336=0t^2-6t-56t+336=0t(t-6)-56(t-6)=0(t-56)(t-6)=0t=56,6x^2-x=t$

For $t=56$

$x^2-x=56x^2-x-56=0x^2-8x+7x-56=0x(x-8)+7(x-8)=0(x+7)(x-8)=0x=-7,8$

For $t=6$

$x^2-x=6x^2-x-6=0x^2-3x+2x-6=0x(x-3)+2(x-3)=0(x+2)(x-3)=0x=-2,3$

so the values of $x$ are $-7,8,-3$ and $2$