Prove that (2n-3)^2 - (2n+3)^2 is a multiple of 8 for all positive integer values of n.

Question

This is the supposed way of working it out:

Expand: (2n+3)^2 - (2n-3)^2

(2n+3)(2n+3) - (2n-3)(2n-3)

(4n^2+6n+6n+9) - (2n^2-6n-6n+9)

4n^2+6n+6n+9-2n^2-6n-6n+9

Then simplify to 4n^2+12+9-4n^2+12n-9 = 24

Then factorise 24 to 8(3n)

8(3n) is always a multiple of 8.

 

My problem is that I don't understand how, when simplified, it equals 24?

-6n-6n should make -12n, so 12n + -12n should equal 0, but I'm being told it is 24n. 

 

Answer 1

(2n + 3)² - (2n - 3)²

(2n + 3)(2n + 3) - (2n - 3)(2n - 3)
(4n² + 6n + 6n + 9) - (4n² - 6n - 6n + 9)

Mistake there. you wrote 2n² instead of 4n² (2n * 2n = 4n²)

 

4n² + 6n + 6n + 9 - 4n² + 6n + 6n - 9

The mistake here was that you only applied the subtraction to the first number instead of distributing it to everything inside. Think of it as you have to distribute a -1 to everything inside the parentheses.

4n² - 4n² = 0

6n + 6n + 6n + 6n = 24n

9 - 9 = 0

 

So that leaves us with only 24n.

You can then factor out 8 to get your 8(3n)

I hope that clarifies what is happening in the problem.