Square of Sum with Double/Algebraic Proof 1

Theorem

$\forall a, b \in \R: \paren {a + 2 b}^2 = a^2 + 4 a b + 4 b^2$

Proof

Follows from the distribution of multiplication over addition:

\(\ds \paren {a + 2 b}^2\)

\(=\)

\(\ds \paren {a + 2 b} \cdot \paren {a + 2 b}\)

\(\ds \)

\(=\)

\(\ds a \cdot \paren {a + 2 b} + 2 b \cdot \paren {a + 2 b}\)

Real Multiplication Distributes over Addition

\(\ds \)

\(=\)

\(\ds a \cdot a + a \cdot 2 b + 2 b \cdot a + 2 b \cdot 2 b\)

Real Multiplication Distributes over Addition

\(\ds \)

\(=\)

\(\ds a^2 + 4 a b + 4 b^2\)

$\blacksquare$