# How to Use Algebra in Arithmetic

## USING ALGEBRA IN ARITHMETIC

It’s interesting that algebra can be useful in arithmetic.

Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic.

However, you can apply your algebra skills to arithmetic problems.

Following is an example.

Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy.

One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together.

Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58.

You may not have realized it at the time, but your teacher was applying the distributive property of mathematics.

When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364.

Now let’s see how algebra can supply an alternative solution.

I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator.

So let’s write 58 as 60 minus 2.

Then the problem is 58 × 58 = (60 – 2) × (60 – 2).

Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.”

One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y².

Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2.

58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2).

Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2).

58 × 58 = 3600 – 240 + 4 = 3364.

Note that (–2) × (–2) = +4 because the two minus signs cancel out.