Find 3 consecutive integers such that the product of the middle and largest integers is five more then the smallest integer squared

Only option: 1,2,3

Step-by-step explanation:

3 consecutive integers:

n-1 , n  , n+1

Why does it work? They should be consecutive for any integer n.

Example with n=5:

4 ,  5  ,  6

These are consecutive integers because they happen right after each other in the list of integers.

Integers { ..., -5,-4,-3,-2,-1,0,1,2,3,4,5,...}

We are given the product of the middle and largest is 5 more than the smallest integer squared.

In other words, the product of           n    and  (n+1)    is 5 more than (n-1)^2.

As an equation that is:                        n(n+1)                =       5     +       (n-1)^2.

Let's write it more together:

n(n+1)=5+(n-1)^2

Distribute:

n^2+n=5+n^2-2n+1   (Used (x-y)^2=x^2-2xy+y^2)

Combine like terms on either side:

n^2+n=n^2-2n+6

Get everything on one side so one side is 0.

Subtract n^2 on both sides:

n=-2n+6

Add 2n on both sides:

3n=6

Divide both sides by 3:

n=2

So if n=2,

then n-1=1

and n+1=3.

So we have one option (1,2,3).

3(2)=5+(1)^2

6=5+1

6=6 is true so (1,2,3) is an option.

the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. hope this ! mark brainliest! you v much! : )

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step-by-step explanation: