Question

The product of 3 consecutive odd numbers is 3,315. Enter a simplified expression for finding the numbers. Let n be the first odd number. The simplified expression for finding the consecutive odd numbers is .

Answer 1

Answer:

$n^3+6n^2+8n=3315$

Step-by-step explanation:

Let n be the first odd number.

The 2nd consecutive odd number would be $n+2$ and 3rd consecutive odd number would be $n+2+2=n+4$.

We have been given that the product of 3 consecutive odd numbers is 3,315. The product of 3 consecutive odd numbers would be $n(n+2)(n+4)$.

Now we will equate the product with 3315 as:

$n(n+2)(n+4)=3315$    

Let us simplify the left side of equation.

$(n\cdot n+n\cdot 2)(n+4)=3315$

$(n^2+2n)(n+4)=3315$

Now, we will apply FOIL to find the product of left side as:

$n^2\cdot n+n^2\cdot 4+2n\cdot n+2n\cdot4=3315$

$n^3+4n^2+2n^2+8n=3315$

$n^3+6n^2+8n=3315$

Therefore, our required equation would be $n^3+6n^2+8n=3315$.