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## Precalculus

### Course: Precalculus > Unit 3

Lesson 4: Identities with complex numbers# Complex numbers & sum of squares factorization

Learn how expressions of the form x^2+y^2 can be factored into linear factors. This would not be possible without the aid of complex numbers! Created by Sal Khan.

## Want to join the conversation?

- how is this whole thing useful . it just makes things more complicated(4 votes)
- There are many reasons we might need to fully factor a polynomial, and this provides a way of giving a more complete answer than "no real solutions". Even a simple quadratic can be very important to describe or model a phenomenon, yet not have real solutions. This method provides a way of stating
**all**the roots. It is definitely important to be able to describe the behavior of polynomials in many careers.

As an example, think of a quadratic equation such as y = x²- 2x - 6

It has two real roots, and it can describe someone's business or scientific phenomenon.

Compare that to y = x²- 2x + 10

This one looks just as promising, but it has no real roots. You might think of it as a loser, therefore. But, if it described a business, it would be one that would never experience a loss! The value of the function never goes below zero. It is useful to be able to figure out why one business model works and one doesn't, or why one model fits a scientific phenomenon, while the other would cause a space vehicle to spin out of control.(21 votes)

- Couldn't you factorize (x^2 + y^2) as (x + y)^2 - 2xy or (x - y)^2 + 2xy?(2 votes)
- Yes and no. Those are correct (and sometimes useful) expressions, but they aren't "factored" because they are not written as a product of factors -- each involves the sum of two terms.

You could factor it as (x + yi)(x - yi), using complex numbers.(23 votes)

- When you write "i", does it need to be written in a different/special way ?(3 votes)
- A lower case i is sufficient, though many people make this lower case i in italic or cursive font - notice the little curl, or serif, that Sal puts on the bottom part of the i he writes.(5 votes)

- Is there a video series or a module on imaginary and complex numbers on Khan Academy as I would like to learn more about them? By the way I am watching this on the Algebra 2 playlist.(4 votes)
- You can find the playlist on imaginary and complex numbers here: https://www.khanacademy.org/math/algebra-home/alg-complex-numbers(2 votes)

- Can you tell why do we use 'i' in these cases? Is it just for a reference or can we use other letters too?(2 votes)
- i is the symbol used to represent the imaginary number sqrt(-1).

For review:

https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers(5 votes)

- So it basically equals z times its conjugate, where z = x + yi.(4 votes)
- Why we can't just factor it like this

x^2+y^2=x(x+y^2/x) ?

In this case it's a product of factors, aren't it?(3 votes)- Yes, you can do that, but one of your resulting factors not only still have a variable squared, but also has a variable to the
`-1`

power, so it can be seen as even more complex than the original expression.

Factorization is commonly used to simplify a expression, the video shows a way in which you end up with two linear factors, which are simpler than a quadratic expression.(3 votes)

- I've taken the sum of cubes rule which is (a+b)(a^2-ab+b^2) ... Is there a similar rule for the sum(or difference) of ''4th power'' terms?(2 votes)
- The difference of fourth powers can be treated as the difference of squares. Since a^4 = (a^2)^2, we can write: a^4 – b^4 = (a^2)^2 – (b^2)^2 = (a^2 + b^2)(a^2 – b^2) = (a^2 + b^2)(a + b)(a – b). Now, a^2 + b^2, technically, can be factored over the irrational numbers: a^2 + b^2 = a^2 + 2ab + b^2 – 2ab = (a + b)^2 – 2ab = (a + b – sqrt(2ab))(a + b + sqrt(2ab)), or we can factor it over the complex numbers as shown in Sal’s video.

The sum of fourth powers can be treated as the sum of squares and factored over the irrational numbers too: a^4 + b^4 = a^4 + 2*a^2*b^2 + b^4 – 2*a^2*b^2 = (a^2 + b^2)^2 – 2(ab)^2 = (a^2 + b^2 – sqrt(2)ab)(a^2 + b^2 + sqrt(2)ab). Or we can factor it over the complex numbers.(3 votes)

- Is 2(x+y)^2 considered a sum of squares?(1 vote)
- Sum or squares would be: x^2 + y^2

Your factors will create a perfect square trinomial with an additional common factor of 2

Hope this helps.(4 votes)

- So, how would you do this if you factored a Trinomial using the Quadratic Factoring Method?(2 votes)
- you would need to know the quadratic factoring method. you would need to find what a is equal to and what b is equal and what c is equal to from the equation : ax^2 + bx +c = 0 A can not be equal to 0!

hope's

this helped,

Raechel(2 votes)

## Video transcript

Voiceover:Long ago in
algebra class, we learned to factor things like X
squared minus Y squared. We saw that this was a
difference of squares that you could factor this
as X plus Y, times X minus Y. This is a little bit of a refresher, you can multiply these two together to verify you get X
squared minus Y squared. In fact, let's just do that for fun. X times X is X squared, X times negative Y is negative XY, Y times X is positive XY, and then Y times negative
Y is minus Y squared. These two middle terms cancel out and your left with X
squared minus Y squared. What I want to tackle in this video is something that we didn't
know how to factor before and that's the sum of the squares. So if we were to factor X,
let me use a brighter color, if we want to try to factor
X squared plus Y squared, before we knew about imaginary numbers, complex numbers, we didn't
know how to factor this. But now that we do,
what I want to try to do and I encourage you
actually to pause the video and try to do it before me,
is to try to express this as a difference of squares,
using the imaginary unit I. So let's try to do this. So the first we want to
write as a difference of squares, so X square
we'll just keep as X squared, but now I want to write
this part, I want to write the second part, right over
here, as subtracting a squares. I want to subtract a square. So we could write this part as subtracting negative Y squared, obviously you subtract a negative that's the
same thing as adding. Now how does this help us? Well this is the same thing as subtracting negative one times Y squared. If we wanted to write this
whole thing as a square, how would we do it? Well we have Y squared and what's negative one the square of? Well we know by definition,
negative one is equal to I squared or that I squared
is equal to negative one. Let's rewrite it that way,
so this is going to be equal to X squared minus,
instead of negative one, I'll write that as I squared,
minus I squared Y squared. All I did was replace this
negative one with an I squared. And now this is interesting. I think you see where this is going, but I'll just make it very explicit. This is now X squared minus IY squared. And just like that, using
I, I've been able to write this sum of squares as
a difference of squares. And now we can factor
just the exact same way we factored this original expression. This thing is going to be equal to X plus IY times X minus IY. And we can verify that
if you multiply these two expressions together, you're going to get X squared plus Y squared. Let's do that, X times X is X squared. X times negative IY is
negative IXY, and then IY times X is positive
IXY, and then finally let me do this in a
color I'm not using yet, finally, IY times negative IY is equal to negative I squared Y squared. Now these middle two terms cancel out, I squared is negative one, and so we have X squared minus negative one Y squared, so this is subtracting a negative, the same thing as adding a positive, and so this simplifies to
X squared plus Y squared. So hopefully this now
gives you an appreciation of using the complex imaginary unit I, how you can actually
factor this expression into essentially the product
of two complex numbers.