Any easy way to solve this equation

(x-1)x(x+1)(x+2)=24

(x-1)x(x+1)(x+2)=24
<=> x(x+1).(x+2)(x-1) = 24
<=> (x^2 + x) (x^2 + x -2) = 24 (1)
t = x^2 + x
(1) => t(t-2) = 24
=> t^2 - 2t = 24
=> t = 6 or t = -4
=> x^2 + x = 6 or x^2 + x = -4
=> x = -3, x = 2

We have:
(x-1)x(x+1)(x+2)=x^4+2x^3-x^2-2x=24
If you have a calculator, type fx570es, press:
ALPHA X^4+2X^3-X^2-2X-24 ALPHA CALC then press 0 next press SHIFT and CALC then press 1 =
Wait for 30s, then you'll see result. (If your calculator appears L-R=0, that is correct)
Good luck!!

you have (x-1)(x+1) a difference of squares = (x^2 - 1) and x(x+2) = x^2 + 2x

(x^2-1)(x^2 + 2x) = x^4 + 2x^3 -x^2 - 2x = 24

x^4 + 2x^3 -x^2 - 2x - 24 = 0

(x^2 + ax + b)(x^2 + cx + d) = 0 where bd = -24 and a+c = 2 and b+d= -1 and ad+bc = -2

so, b = -1 - d (-1-d)d = -d^2 -1 = -24 --> d^2 = 23 d = sqrt23 b = -sqrt(23) - 1

a = 2-c --> (2-c)sqrt(23) - (sqrt(23) + 1)c = -2

2sqrt23 - 2csqrt23 - c = -2

c = (-2sqrt23 - 2)/(-2sqrt23 - 1) = (2sqrt23 + 2)/(2sqrt23 + 1)

Easy way is to put x = 0, 1, 2 ......
and check the outcome for which it is = 24

(x-1)x(x+1)(x+2)=24
for x = 2
( (2-1) 2 ( 2+1) ( 2+2)
=> 1 ( 2) (3) ( 4) =24
so x = 2 is the answer

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(x-1)x(x+1)(x+2)=24

x may be equal to 2

because x-1= 2-1==1

x= 2

x+1= 3
x+2= 4
& l LHS= [2--1][2][2+1][[2+2]
=1[2][3][4]
= 24

= RHS
HENCE VERIFIED TRUE

use trial error method,
start with -1, 1 and 0
if none of these work, then start with 2 , then -2 etc
ans will be 2

(x-1)(x)(x+1)(x+2) =(1)(2)(3)(4)
Hence x=2 ...............Ans