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How would you calculate something like this:? $(a+b)(c+d)$

I have been out of maths for a while and I can remember we got tought to do: $$ac + ad + bc + bd$$ but how does that work? Are there other methods?

Mazzy
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    I'm not sure what you mean by "doesn't seem to work"; that's the correct answer, provided that $a,b,c,d$ are numbers. – vadim123 Dec 12 '13 at 18:11
  • Nvm, you're right! – Mazzy Dec 12 '13 at 18:13
  • There's no special rule that I know of for evaluating $(a+b)(c+d)$, unless $a=c$ and $b=d$, in which case it would become $(a+b)^2$. The special rule for evaluating that would be $a^2+2ab+b^2.$ – Sujaan Kunalan Dec 12 '13 at 18:22

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The more general approach is to use the distributive law twice: $$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bc$$ This works no matter how many terms you have inside the parentheses. FOIL is useful because we often see two terms in each group.

Ross Millikan
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What you need is the FOIL method. This Wikipedia article explains the method well and even shows how it applies to your exact situation.

Now some might say that it is a bit silly to have a FOIL method when we are just using the distributive property. This basic property says that when ever you ave three real numbers, say $x, y$ and $z$. Then $$ \color{red}x\cdot (y+z) = \color{red}xy + \color{red}xz. $$ So for example you have $$ 3\cdot(7+2) = 3\cdot 7 + 3\cdot 2 = 21 + 6 = 27. $$ (Note that this is the same as $3\cdot 9 = 27$.)

So now let us consider four real numbers $a,b,c$ and $d$. Then $(a+b)$ is just another real number. So $$\begin{align} \color{red}{(a+b)}(c+d) &= \color{red}{(a+b)}c + \color{red}{(a+b)}d\\ &= c\color{red}{(a+b)} + d\color{red}{(a+b)}. \\ &= \color{red}c(a+b) + \color{red}d(a+b) \\ &= ca + cb + da + db \\ &= ac + ad + bc+ bd. \end{align} $$ Here we have also used that for two real numbers $x$ and $y$: $xy = yx$ and $x+y = y+x$.

BlueRaja - Danny Pflughoeft
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Thomas
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    +1 for the "bit silly to have a FOIL method." I never had heard the term until I reached college calculus, and my algebra hasn't suffered in the least. :) – apnorton Dec 12 '13 at 19:57
  • @anorton I remember it helping some of my peers in pre-algebra (algebra 1?) in Junior High; but IIRC most of the big beneficiaries were struggling with algebra generally and in future years ended up dropping back to less challenging math classes. – Dan Is Fiddling By Firelight Dec 12 '13 at 20:59
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The idea here is that you have to multiply each term in the first set of parentheses by each term in the second set of parentheses, and then you have to sum them. (If you want to read up on the theory behind this, this is distributivity of multiplication over addition.) To do this:

You use the FOIL method, which stands for First, Outer, Inner, Last.

Take $(a+b)(c+d)$.

First: multiply $a$ and $c$ = $ac$.
Outer: multiply $a$ and $d$ = $ad$.
Inner: multiply $b$ and $c$ = $bc$.
Last: multiply $b$ and $d$ = $bd$.

Add them all up, and there you are: $ac+ad+bc+bd$. Note that we're adding because the operation inside the brackets is addition, and the signs of all our variables are positive.

Newb
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Remember that, say, $\color{red}{(a+b)}$ is just another number, so you can apply the distributive law to $\color{red}{(a+b)}(c+d)$ to give $\color{red}{(a+b)}c+\color{red}{(a+b)}d$ but then the distributive law (in the other direction) applied twice gives $a\color{blue}{c}+b\color{blue}{c}+a\color{purple}{d}+b\color{purple}{d}$.

Watch this MinutePhysics video for another perspective.

Shaun
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We use the distributive law: If x, y, z are real numbers, then $x(y+z)=xy+xz$.

Notice that (a+b) is a real number. Therefore: $(a+b)(c+d)=(a+b)c + (a+b)d$.

Now using the commutative law twice: $(a+b)c + (a+b)d=c(a+b) + d(a+b)$.

Now using the distributive law twice: $c(a+b) + d(a+b)=ca+cb+da+db$

Finnaly, using the commutative law four times: $ac+bc+ad+bd$

If you rearrange the sums you will get the expression you typed.

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This might not work for everyone, but for this special case I was taught to draw a happy face with eyes, a nose and a mouth. Each line connects a pair of numbers that are in the final expression.

enter image description here

Chris Taylor
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