You can certainly do basic arithmetic like 12 times 28
without algebra, but it’s interesting that algebra can be utilized to make some
arithmetic calculations more efficient. This can be really handy when you don’t
have a calculator around (or when you don’t want to waste your cell phone
battery for math), or if you take a test where a calculator isn’t allowed.
As one example, consider the algebraic expression (x + 1)(x –
1). When you foil this out, you get x2 + x – x – 1. The cross-terms
vanish, leaving just x2 – 1.
Knowing that (x + 1)(x – 1) = x2 – 1 can actually
be useful when doing multi-digit multiplication.
For example, consider 19 times 21. This is the same as (20 –
1)(20 + 1), which equals 202 – 1 or 399. It’s much easier to do 20
squared in your head than it is to work out 19 times 21.
Suppose you wish to multiply 26 by 34. You can write this as
(30 – 4)(30 + 4) = 302 – 42 = 900 – 16 = 884.
Let’s try 13 times 17. You can turn this into (15 – 2)(15 +
2) = 152 – 22 = 225 – 4 = 221. Here, it helps to know the
perfect squares of 11 thru 20.
The cross-term doesn’t always vanish, though. Consider 18
times 23. This becomes (20 – 2)(20 + 3) = 202 – 2(20) + 3(20) – 6 =
400 – 40 + 60 – 6 = 414.
As another example, 11 times 13 can be written as (10 +
1)(10 + 3) = 102 + 10 + 30 + 3 = 143.
When the cross-term doesn’t vanish, you’re probably not
saving anything by using algebra – the usual arithmetic of 18 times 23 or 11
times 13 will be just as much work.
However, if you want to multiply 1001 times 1050, it may be
simpler to write (1000 + 1)(1000 + 50) = 10002 + 1000 + 50000 + 50 =
1,051,050, provided that you’re good at counting the zeroes when you work with
multiples of 10 and good at keeping track of decimal positions when you add (if
not, the conventional method helps you stay organized).
Challenge yourself: Can you think of any arithmetic problems
that would ordinarily be quite tedious, which algebra would make much simpler?
Chris McMullen, author of the Improve Your Math Fluency series of workbooks
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