Proving $(a + b) \cdot (e+ c) = ae+ac+be+bc$.

Question

The multiplication of two terms say $(a + b) \cdot (e+ c)$ involves multiplying corresponding elements i.e. $ae+ac+be+bc$. How was this proved?

Answer 1

This is just application of Distributive Law(s): $$a\cdot(b+c) = a\cdot b+a\cdot c$$ $$(a+b)\cdot c = a\cdot c+b\cdot c$$ Specifically \begin{align*} (a + b) \cdot (e+ c) &= (a + b) \cdot e+ (a + b) \cdot c\\ &= a\cdot e+ b\cdot e+ a\cdot c + b \cdot c\\ \end{align*}