Algebra 2 Question Help

f(n)= n^2-n+2, find k if f(n^2+k) = f(n) * f(n+1).
Steps would be much appreciate.

If you find a shorter answer then great, but this is what I ended up doing:

First, we know that f(n) = n^2 - n + 2

Let's find f(n+1) = (n + 1)^2 - (n + 1) + 2

These two expressions must be multiplied, as indicated in the problem:

(n^2 - n + 2)[(n + 1)^2 - (n + 1) + 2]

Now that is a boring process. I just multiplied like crazy, some things get cancelled out, and then eventually you end up with:

n^4 + 3n^2 + 4

We then have to say that this is equal to f(n^2+k). So doing f(n^2+k) we get:

f(n^2+k) = (n^2 + k)^2 - (n^2 + k) + 2

It's a similar story of careful multiplying. Eventually you get:

n^4 + 2kn^2 - n^2 + k^2 - k + 2

Now here you must be just a bit resourceful and group things in a way that resembles the product we developed earlier. Like this:

n^4 + (2k - 1)n^2 + (k^2 - k + 2)

Now, since f(n^2+k) = f(n) * f(n+1), let's see what that looks like (I'm switching the order though, but of course it makes no difference):

n^4 + 3n^2 + 4 = n^4 + (2k - 1)n^2 + (k^2 - k + 2)

As you can see, I did some small tweaking, as I said, so that we can now see two things:

3 = (2k - 1) -----> k = 2

Also, you can see this:

4 = (k^2 - k + 2) -----> k^2 - k - 2 = 0 -----> (k - 2)(k + 1) = 0

We had already determined that k = 2, and here we see again that k = 2 and k = -1

However, we throw out k = -1 because it won't work for the other equation that only gives k = 2

Your answer then is k = 2

I sure hope this helps...

f(n^2+k) = (n^2+k)^2 - (n^2+k) + 2 = n^4 + (2k-1)n^2 + k^2 - k + 2 ....................................(1)

f(n)*f(n+1) = (n^2 - n + 2) ((n+1)^2 - (n+1) + 2) = (n^2-n+2)(n^2+n+2) = n^4 + 3n^2 + 4 ............(2)

Comparing coefficients of n^2 in (1) and (2), we have

2k - 1 = 3 ==> k = 2.

k^2-k+2 = 2^2-2+2 = 4 which is consistent with (2).

Answer: k=2.