Quadratic Type

When approaching this problem it is important to pair terms that produce the same $\ x^2 \ and \ x$ coefficients so we can make a substitution to break down the problem. Here's how it goes... $\ (x+1)(x+3)(x-5)(x-7) = (x+1)(x-5) \times\ (x+3)(x-7)$ $= (x^2 -4x-5)(x^2 -4x -21)$ Now we can form a difference of two squares by using $\ u = x^2 -4x -13$ ... $(u-8)(u+8) = 192 \Rightarrow\ u^2 = 256 \ so \ u = \pm 16$ $\ u= 16 \Rightarrow\ x^2 -4x -29 = 0$ $\therefore\ x = 2 \pm \sqrt{33}$ . Similarly, for u = -16: $\ x^2 -4x +3 = (x-1)(x-3) = 0 \therefore\ x= 1,3$