A.) x² + 2xy + y² + 3

B.) x² + 2xy + y²

C.) x² + 2xy + 3

D.) x² + 3 + y

E.) x² + y²

then f(x+ y)

= (x + y)² + 3 {substituted (x + y), in for x, into f(x) = x² + 3}

= (x + y)(x + y) + 3 {when squaring a binomial, multiply it by itself}

= x² + 2xy + y² + 3 {used FOIL method / distributive property to multiply}

**A.) x² + 2xy + y² + 3**

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