c^2 - 8c + 16 - y^2
(c + 4) (c + 4)
(c + 4)^2 - y^2
(c +4 +y) (c + 4 - y)
Is this correct?
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).