Question

(x+1)(x+2)(x+3)(x+4)-8=0

Answer 1

$(x + 1)(x + 4)(x + 2)(x + 3) - 8 = 0 \\ ({x}^{2} + 5x + 4)( {x}^{2} + 5x + 6) - 8 = 0$
Let,
${x}^{2} + 5x = y$
Then,
$(y + 4)(y + 6) - 8 = 0 \\ {y}^{2} + 10y + 16 = 0 \\ (y + 2)( y+ 8) = 0 \\ y = - 2 \: and \: -8$