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Is there any way to simplify the expression:

$$(a\pm b)(c\pm d)\quad ?$$ At first, I tried just distributing the values, and through FOIL I achieved:

$$ac \pm ad \pm bc \pm bd$$

This works when the signs multiply to positive, but if one of the signs is negative and the other positive, it gives the wrong lower bound.

The problem is that the plus-or-minus bd "depends" on the signs of b and d.

Is there a concise and succinct way to write this expression?

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The sequence of symbols $$(a\pm b)(c\pm d)$$ is an expression, not an equation (there's no equals sign). –  Zev Chonoles Feb 3 '13 at 3:08
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2 Answers 2

up vote 5 down vote accepted

Using two $\pm$ signs in an expression is inherently ambiguous, because at least in some contexts and according to some conventions, they are assumed to be linked, i.e. either both plus or both minus.

If you're really set on writing everything in one equation, I'd recommend $$(a+(-1)^rb)(c+(-1)^sd)=ac+(-1)^rbc+(-1)^sad+(-1)^{r+s}bd$$ and then say whatever it is you intend about $r$ and $s$.

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That's a pretty smart way to do it! Thanks. I would prefer one equation so that I can express my errors in measurements. –  John Feb 3 '13 at 3:08
    
Glad I could help! –  Zev Chonoles Feb 3 '13 at 3:10
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Edit: As Novice has pointed out in a comment below, my answer includes the expressions $(a+b)(c-d)$ and $(a-b)(c+d)$ which, one could argue, shouldn't be included. As the link in Zev Chonoles's answer explains, using two $\pm$ signs in the same expression is ambiguous as it is not clear whether there are two or four possibilities. If you take $(a\pm b)(c\pm d)$ to mean only $(a+b)(c+d)$ and $(a-b)(c-d)$, then we have $$(a\pm b)(c\pm d) = ac \pm ad \pm bc + bd.$$


I don't think there is a concise way to combine them. First note that the expression $(a\pm b)(c\pm d)$ covers four possibilities:

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

The coefficient of the $ac$ term is always positive ($ac$ may be negative, but the coefficient is always $1$).

The coefficient of the $ad$ term is positive if we choose $+$ in the second bracket (this is because $a$ always has a positive coefficient and now we are choosing $d$ to have a positive coefficient); likewise, the $ad$ term has negative coefficient if we choose $-$ in the second bracket (now we are choosing $d$ to have a negative coefficient).

Similarly, the coefficient of $bc$ is positive if we choose $+$ in the first bracket and negative if we choose $-$.

As you point out, the problem occurs because of the $bd$ term. If the same sign ($+$ or $-$) in both brackets, the coefficient is positive, otherwise it is negative.

As Zev Chonoles has pointed out (and I intended to), you can combine the four expressions using the fact that $a + b = a + (-1)^0b$ and $a - b = a + (-1)^1b$; note that $0$ can be replaced by any even integer, and $1$ can be replaced by any odd integer. If there was only one $\pm$ you have the distribution rules:

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

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

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$(a - b)(c + d)$ and $(a + b)(c - d)$ are two possibilities when we have $(a \pm b)(c \mp d)$, but they are not the possibilities for $(a \pm b)(c \pm d)$. :-) –  Parth Kohli Feb 3 '13 at 3:35
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