Question

Find Value of x when (x+1)(x+2)(x+3)(x+4)=120

Answer 1

(x + 1)(x + 2)(x + 3)(x + 4) = 120
(x + 1)(x + 4)(x + 2)(x + 3) = 120
( x² + 5x + 4)( x² + 5x + 6) = 120
In this We will take x² +5x =y
so, (y+4)(y+6)=120
y² + 10y + 24 = 120
y² + 10y - 96 = 0
y² - 6y + 16y - 96 = 0
y(y-6)+16(y - 6) =0
(y + 16)(y - 6) = 0
now , y = -16 and y = 6
so, putting the value of y
x² + 5x = 6
x² + 5x - 6 =0
x² + 6x - 1x - 6 = 0
x(x + 6)- 1(x + 6) = 0
(x -1)(x+6) = 0
now, x = -6 and x = 1
so x is equal to 1, -6
now, if we put y = -16 ,then it will be complex
and on putting 1 ans. is 72 and on putting -6 ans. is 120
so, -6 is the answer.

Answer 2

x+1)(x+2)(x+3)(x+4) = 120
x+1+x+2+x+3+x+4=120
4x + 10 =120
x+10 = 120/4
x = 30 - 10 = 20
so x = 20