c^2 - 8c + 16 - y^2

Is the problem

c^2 - 8c + 16 - y^2 = 0?

no

it is

c^2 -8c + 16 - y^2

and it says to factor completely

Sorry, I am not sure what they are asking. Someone else here might understand it.

c^2 - 8c + 16 - y^2 = (c-4)^2 - y^2

= (c-4+y)(c-4-y)

Thanks Guys.

Yes, your factorization is correct.

To obtain this factorization, you applied the formula for factoring a quadratic expression in the form of c^2 - 8c + 16.

The formula is:

a^2 - 2ab + b^2 = (a - b)^2

In this case, a = c and b = 4. Plugging these values into the formula, we get:

c^2 - 8c + 16 = (c - 4)^2

Therefore, (c - 4)^2 is the correct factorization of c^2 - 8c + 16.

To further simplify the expression (c - 4)^2 - y^2, you can apply the difference of squares formula:

a^2 - b^2 = (a + b)(a - b)

In this case, a = (c - 4) and b = y. Plugging in these values, we get:

(c - 4)^2 - y^2 = ((c - 4) + y)((c - 4) - y)

Simplifying further, we have:

(c + 4 - y)(c + 4 + y)

So your factorization of (c^2 - 8c + 16) - y^2 is indeed (c + 4 - y)(c + 4 + y).