Question

the product of four consecutive natural number is 840 find numbers

Answer 1

let the consecutive natural terms be x,x+1,x+2,x+3 (x>0)

given

x(x+1)(x+2)(x+3)=840

(x^2+x)(x^2+5x+6)=840

Answer 2

Let four consecutive natural numbers are( x -2 ) , (x-1) , x and (x+1). Acordingly:-

(x-2).(x-1) .x.(x+1) = 840

(x^2–2x)×(x^2–1)= 840

x^4–2x^3-x^2+2x-840 = 0

On putting x=6

R=1296–432–36+12–840=1308–1308=0

(x-6)is a factor

x^4–2x^3-x^2+2x-840=0

x^3(x-6)+4x^2(x-6)+23x(x-6)+140(x-6)=0

(x-6)(x^3+4x^2+23x+140) =0

Euther x-6=0 => x =6

1st number =x-2 =6–2 = 4

2nd number =x-1=6–1 = 5

3rd number =x = 6

4th number =x+1 =6+1 = 7

4 numbers are 4 , 5 , 6 and 7 . Answer.