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## Precalculus

### Course: Precalculus > Unit 3

Lesson 4: Identities with complex numbers# Factoring polynomials using complex numbers

Sal shows how to factor a fourth degree polynomial into linear factors using the sum-product rule and the sum of squares identity. Created by Sal Khan.

## Want to join the conversation?

- The
**FIRST**mistake is in writing out the problem. The polynomial given in the problem is x^4 + 5x^2 + 4. But the polynomial that Amat factored is x^4 + 10x^2 + 9.(57 votes) - how did amat in the question above came up from x^4+5x^2+4 to x^4+10x^2+9(20 votes)
- They just made a mistake in the actaully problem itself. It was really meant to be x^4+10x^2+9, but somehow they accidentally wrote x^4+5x^2+4.(7 votes)

- Why is it (x +3i) and (x-3i) instead of (x+3) (x-3) for step 4?(2 votes)
- When you are using the zero product property, set each part equal to 0. So x^2 + 9 = 0 gives x^2 = -9, and there is no real number that can be squared to give a negative answer, so we have to go into the imaginary numbers. To get your answer, you need a difference of perfect squares (x^2 - 9).(4 votes)

- Hello, can you explain this part? how is this correct?

x^2 + a^2 = (x +ai)(x-ai)

Thank youu(1 vote)- (x + ai) * (x - ai)

= x * (x - ai) + ai * (x - ai)

= x^2 - xai + xai - (ai)^2

= x^2 - (ai)^2

= x^2 - (a^2 * i^2)

= x^2 - (a^2 * -1)

= x^2 + a^2(2 votes)

## Video transcript

- [Instructor] We're told that Amat tried to write x to the fourth
plus 5x squared plus 4 as a product of linear factors. This is his work. And then they'd tell us all
of the steps that he did. And then they say, in what step did Amat make his first mistake? So pause this video and see
if you can figure that out. All right, now let's work
through this together. So we're starting with x to
the fourth plus 10x squared plus 9. And it looks like Amat
tried to factor that into x squared plus 9
times x squared plus 1. And this indeed does make sense, because if we said that let's say, u is equal to x squared, we could rewrite this right over here as u squared plus 10u plus 9. The whole reason why you would do this is so that you could write
this higher order expression in terms of a second degree expression. And then we've learned
how to factor things like this many times. We look, we say, okay, what two numbers when I add them I get 10, and when I multiply them I get nine, and it would be nine and one? And so you could write this as u plus 9 times u plus 1. And of course, if u is equal to x squared, this would be x squared plus
9 times x squared plus 1. Which is exactly what
Amat has right over here. So step one is looking great. All right, now let's think
about what Amat did in step two. They didn't do anything
to x squared plus 9 but it looks like they tried to further factor x squared plus 1. And this does seem right. We just have to remind ourselves just as you have a difference of squares if you're dealing with
non-complex numbers, so we could rewrite this right over here as x plus a times x minus a. We could have a sum of squares if we're thinking about complex numbers. This is going to be x
plus ai times x minus ai. And in this situation while the x is x and then our a would be one. So we're going to have x plus i, x plus i times x minus i . So step two is looking great. And now let's go do step three. So in step three, no change to this part of the expression. And it looks like Amat is trying to factor x squared plus 9
based on the same principle. Now x squared plus 9 is the same thing as x squared plus 3 squared. So if you use this exact same idea here, if you factored it should be
x plus 3i times x minus 3i. But what we see over here is Amat took the square root of three, instead of just having a three here. Amat treated it instead
of having a nine here as if we actually had a three so they made a little
bit of an error there. So this is the step where
Amat makes his first mistake and we're done.