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### Multiplying and Simplifying Complex Binomials

```
Date: 8/19/96 at 1:40:24
From: Anonymous
Subject: Multiplying and Simplifying Complex Binomials

Multiply and simplify:
(2+3i)(5-i)

I could not see how they simplified this to what they say is the

I used FOIL to get:         10-2i+15i-3i^2

I then simplified to get:   10+13i-3i^2

The program said it could be simplified to:  13+13i

The value of (i) was not given but the problem said that (i^2) = -1.
I plugged numbers in for (i) to work out this equation with their
answer and the orginial binomial and it didn't work.

What did I miss?  Thanks.
```

```
Date: 8/19/96 at 2:42:18
From: Doctor Paul
Subject: Re: Multiplying and Simplifying Complex Binomials

You've almost got it right.

Let's first define 'i'. i = sqrt(-1)  so i^2 = -1

You have:

10 + 13i - 3i^2

Let's substitute for i^2...

10 + 13i - 3(-1)

10 + 13i + 3

and you can see how they got their answer.

13 + 13i

I hope this clears things up for you.  If you have any more questions,
feel free to submit them to Dr. Math.

Regards,

-Doctor Paul,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Imaginary/Complex Numbers

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