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## Algebra II (2018 edition)

### Unit 4: Lesson 7

Polynomial identities with complex numbers

# Factoring sum of squares

Sal factors 36a^8+2b^6 as (6a^4-i*√2b^3)(6a^4+i*√2b^3). Created by Sal Khan.

## Want to join the conversation?

• Regarding the following problem ( copied in relevant part ):

Which expressions are equivalent to:

`( w − 3i )/( w^2 + 9 ) + 1/( w + 3i )`?

Select all that apply.

`[ ] ( w − 3i )/( w^2 + 9 )`

`[x] ( 2w − 6i )/( w^2 + 9 )` Got it correct, but...

`[ ] 1 + ( w − 3i )/( w^2 + 9 )`

`[ ] ( w + 3i )/( w^2 + 9 )`

Hints:
Two expressions are equivalent to each other if they represent the same value no matter which values we choose for the variables.
Let's combine the terms by finding a common denominator. We get:

`( w − 3i )/( w^2 + 9 ) + 1/( w + 3i ) `
`= ( w − 3i )/( w + 3i )( w − 3i ) + 1/( w + 3i )` Good
`= ( w − 3i ) + ( w − 3i )/( w + 3i )( w − 3i ) = ( 2w − 6i )/( w2 + 9 )`

Looking at the factored first term alone where my method matched hints:

`( w^2 + 9 ) = ( w^2 - 9i^2 ) = ( w - 3i )( w + 3i )`

Substitute factored denominator `( w - 3i )( w + 3i )` back into first term of the equation:

`( w − 3i )/( w + 3i )( w − 3i )`,

I proceeded to cancel the `( w - 3i )`'s in the numerator and denominator:

`( w − 3i )/( w + 3i )( w − 3i ) = 1/( w + 3i )` which resulted in a common denominator of `( w + 3i )` to complete the addition:

`1/( w + 3i ) + 1/( w + 3i ) = 2/( w + 3i )`

I spent a significant amount of time trying to see where the answer options would determine the LCD as `( w^2 + 9 )` instead of `( w + 3i )`?

If we actually take a look at the 2nd answer, (2w - 6i)/(w^2 + 9), we can see that we can take 2 as a common factor on the top part (which will become 2(w - 3i)), and factorise the bottom part to (w - 3i)(w + 3i). Simply cancel out the (w - 3i) and we're left with your answer of 2/(w + 3i).

The question was possibly looking to see if you could change any of the given answers to a simpler version and then match with your answer. 1st, 3rd, and 4th answers all simplify respectively to: 1/(w + 3i), 1/(w^2 + 9) + 1/(w + 3i), and 1/(w - 3i).

Another way to tackle this problem (after getting 2/(w + 3i)) is to realise that all of the answers have a denominator of (w^2 + 9), and so you must try and change the answer to make it similar. Multiplying both the top part and the bottom part by (w - 3i)/(w - 3i) (essentially just multiplying by '1' but in a way which we can change the form) we'll get 2(w - 3i)/((w + 3i)(w - 3i)). Expand the brackets, and it'll definitely be the second one.

Apologies if this is confusing, especially with the linear layout of the equations. Wolfram Alpha also seems to agree too: http://www.wolframalpha.com/input/?i=%28w-3i%29%2F%28w%5E2%2B9%29+%2B+1%2F%28w%2B3i%29
• I'm confused to as why 2 became i√2 ?
• We wanted to write the expression as a diference of squares.
36a^8 + 2a^6

To write a number as a square, we just use the square root of that number and put all squared (as the square root will cancel out the square). So X = (√X)^2
For example, 9 = (√9)^2 = 3^2.

Then, to write 2a^6 as a square, we may do ( √[2a^6] )^2 = (√2 * a^3)^2
Now we have
(6a^4)^2 + (√2 * a^3)^2
It is a sum of squares. But, oh no! We wanted a diference, not a sum! Well, but we know that a + b = a - (-1 * b)

Then we would have:
(6a^4)^2 - [ -1(√2 * a^3)^2 ]
To put the -1 inside the square we would need to get his square root (has with 2a^6)
(6a^4)^2 - (√(-1) * √2 * a^3)^2
But now we have a problem! What is √(-1)? Well, there is no real number x where x^2 = -1. But there is another set of numbers (Complex numbers) we there is a number called i. i^2 = -1, so √(-1) = i. (If you don't know what is a Complex Number, i suggest you to watch videos about it here on khan academy)

Now we have
(6a^4)^2 - (i√2 * a^3)^2

That's why you see i√2.
• can you not factor out the 2 from the get go? giving you a final answer: 2(4a^4-(ib)^3)(4a^4+(ib)^3)
• Factoring out the 2 would "destroy" the perfect square of 36 (6^2 or 6*6).
Therefore, Sal left the 2 by itself as a √2.
(the square root is made by alt + 251)

Even if you did factor out the two, you couldn't factor out the b^6.
• In the practice problems for these videos I'm required to choose which factored expressions are correct after being shown the problem (say, x^4 -- 16.) Is there any easy or quick way to do this? Sal shows us how to take the expression and factor it, but he doesn't show us how to pick the correct factorizations out of 4 factorizations. I'm just asking because the only way I can think of doing this is to carefully expand all the options, and this can get complicated fast. Are there any tricks in this video that I'm missing?
• I'm confused about how step one turn to step two(I know why it's a imaginary number)
(6a^4)²-i²(√2b³)²
(6a^4)²-(i√2b³)²
How can that imaginary number go into the parentheses?
• It goes into the parentheses the same way a variable would.
Forget for a moment that we're dealing with an imaginary number. Also let's call the complicated √2b³, t
Then we have
(6a^4)²-i²(√2b³)² = (6a^4)²- i²t²
And we know from basic algebra that i²t² = (it)²
Substituting back for t gives i²(√2b³)² = (i√2b³)²
• How would I factorize a^2+b^2?
• How does (x^2 +4i)(x^2 - 4i) get to x^4 + 16?
I see that when you factor x^4 + 16 into a difference of squares this is what it becomes, but when I actually work the problem and multiply the terms I come up with x^4 - 16. I think I may not be understanding how the i unit works. Can anyone explain this or point me to where they explain it on Khan?
• Here's the multiplication steps:
`(x^2 +4i)(x^2 - 4i)` = `x^4 - 4ix^2 + 4ix^2 - 16i^2`
The middle two terms cancel out: `x^4 - 16i^2`
Remember: i*i = -1
So: `x^4 - 16i^2` = `x^4 - 16 (-1)` = `x^4 + 16`
Hope this helps.
• Is the expression that Sal got equivalent to
`(6a^4+b^3*sqrt(-2))(6a^4-b^3*sqrt(-2))`