Determine if the binomial is a perfect square binomial. If so, show the original monomial squared.
X^2 + 16
X^2 + 4
neither one is.
(a+b)^2 = a^2 + 2ab + b^2
neither one of these has a middle term.
No idea what the "original monomial" could be.
x? 4? 2?
To determine if a binomial is a perfect square binomial, we need to compare it to the general form of a perfect square binomial, which is (a + b)^2 = a^2 + 2ab + b^2.
In this case, we have X^2 + 16. To see if it fits the form of a perfect square binomial, we can look for a perfect square term and twice the product of two identical terms.
The first term, X^2, is a perfect square itself because it can be written as (X^1)^2.
Now let's look at the constant term, 16. We need to find two numbers that multiply to give 16 and double that product to get the middle term. However, in this case, there are no two numbers that fit the criteria. Therefore, X^2 + 16 is not a perfect square binomial.
If we want to find a perfect square binomial that is similar to X^2 + 16, we can look at the closest perfect square binomial: X^2 + 4.
To verify if it is a perfect square binomial, we can use the same process by checking for a perfect square term and twice the product of two identical terms.
In this case, X^2 is a perfect square, and 4 = 2^2. So we can rewrite X^2 + 4 as (X + 2)^2, which is the original monomial squared.
Therefore, the original monomial squared is (X + 2)^2.