Proof: (a + b)(c + d) = ac + ad + bc + bd

Question

When manipulating equations you will often have to use some kind of factoring or removing brackets. I have used the following principle a lot of times:

Formula:
$(a + b)(c + d) = ac + ad + bc + bd$

Even though I know it works I never really understood why. I know how to get from one form to the other but nobody really explained why those forms are equal to each other. It just seems like they want you to know it's true (without providing any proof) and then they use this to solve other problems. I have searched online for any proofs of this but I can't seem to find one for this formula.

So is there a way to mathematically prove it? Or is it more like something we just assumed?

Answer 1

If you calculate the area of a rectangle with length (a + b) and height (c + d) you get that the total area is ac + ad + bc + bd. The area of a rectangle is the product of its sides. Therefore: (a + b)(c + d) = ac + ad + bc + bd.

Answer 2

$$(a+b)(c+d)=a(c+d)+b(c+d)=$$

$$ac+ad+bc+bd $$

Answer 3

If you take either $(x+y)z=xz+yz$ or $z(x+y)=zx+zy$ as an axiom, you get the other by multiplication's commutativity. We say multiplication is distributive over addition. Then $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd.$$

Answer 4

A good way is looking at it mathematically. For example - $23\times15$. You can simplify it as $$\begin{align}(20 + 3)(10 + 5) &= (20\times10)+(20\times5)+(3\times10)+(3\times5)\\ &= 200 + 100 + 30 + 15\\ &= 345.\end{align}$$

Answer 5

$(a+b)(a+c)=ac+ad+bc+bd$ (distribution) $=ac+bc+ad+bd$ (closure, commutative law)