Question

Solve the following equations:
$(x-7)(x-3)(x+5)(x+1)=1680$.

Answer 1

$(x-7)(x-3)(x+5)(x+1)=1680\left\{\right(x-7\left)\right(x+5\left)\right\}\left\{\right(x+1\left)\right(x-3\left)\right\}=1680(x^2+5x-7x-35)(x^2-3x+x-3)=1680(x^2-2x-35)(x^2-2x-3)=1680$

Put $x^2-2x=t$

$(t-35)(t-3)=1680t^2-3t-35t+105=1680t^2-38t+105-1680t^2-38t-1575=0t^2-63t+25t-1575=0t(t-63)+25(t-63)=0(t+25)(t-63)=0t=-25,63x^2-2x=t{x}^2-2x=-25x^2-2x+25=\mathrm{0........}\left(i\right)x^2-2x=63x^2-2x-63=\mathrm{0.......}\left(ii\right)$

Now solving $\left(i\right)$

$x^2-2x+25=0$

using quadratic formula

$x=\frac{2\pm \sqrt{4-4\left(1\right)\left(25\right)}}{2}=\frac{2\pm \sqrt{4-100}}{2}x=\frac{2\pm 2\sqrt{-24}}{2}=1\pm \sqrt{-24}$

Solving $\left(ii\right)$

$x^2-2x-63=0x^2-9x+7x-63=0x(x-9)+7(x-9)=0(x+7)(x-9)=0x=-7,9$

So the values of $x$ are $-7,9,1+\sqrt{-24}$ and $1-\sqrt{-24}$