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Precalculus
Course: Precalculus > Unit 3
Lesson 4: Identities with complex numbersFactoring sum of squares
CCSS.Math:
Sal factors 36a^8+2b^6 as (6a^4-i*√2b^3)(6a^4+i*√2b^3). Created by Sal Khan.
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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 )?(13 votes)
Both answers are correct - your answer (2/(w + 3i)) would typically be the correct answer when it is asked to simplify the original thing.
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(7 votes)
I'm confused to as why 2 became i√2 ?(5 votes)
We wanted to write the expression as a diference of squares.
We had:
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.(6 votes)
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)(7 votes)
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.(0 votes)
- 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?(4 votes)
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?(2 votes)
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(5 votes)
How would I factorize a^2+b^2?(2 votes)
it is factorable using complex numbers. See this video: https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-identities-with-complex-numbers/v/factoring-sum-of-squares(3 votes)
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?(2 votes)- 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.(3 votes)
Is the expression that Sal got equivalent to(6a^4+b^3*sqrt(-2))(6a^4-b^3*sqrt(-2))(2 votes)- Yes, your answer is equivalent. You just haven't yet fully simplified the radicals.(3 votes)
is a^n + b^n = (a)^n - (ib)^n when n is odd, too? or only when n is even?(2 votes)- The exponent must be even to have a sum of squares.
There is a sum of cubes that is factored completely differently. See this video: https://www.khanacademy.org/math/algebra2/polynomial-functions/advanced-polynomial-factorization-methods/v/factoring-sum-of-cubes(3 votes)
So two isn't prime since it can be factored as (1+i)(1-i)(1 vote)- No, two is prime. The definition of a prime number is: "a positive integer with exactly two positive integer factors". In other words, a prime number is a positive integer which is the product of exactly one set of exactly two different positive integers.
As the factors you mentioned are not integers, they have no relevance to whether 2 is prime. All numbers can be expressed as the product of infinitely many other numbers. Primes and composite numbers are about integers only, not other numbers.(4 votes)
Video transcript
Voiceover:Let's see if we can factor 36a to the eighth, plus
2b to the sixth power and I encourage you to pause the video and try it out on your own. So let's see if we can express this, or re-express this as
the difference of squares using imaginary numbers. So we can rewrite 36 as, six squared, and a to the 8th is the same thing as, a to the fourth squared,
and so let me actually just rewrite it this way. We can rewrite it as 6a
to the fourth squared. That's this first term right over here. And then the second
term we can write it as, actually just let me
write it this way first. Let me just write it as a square. Plus, the square root of 2, b
to the third power, squared. Now we wanted to write it
as a difference of squares. So instead of writing it this way, let's get rid of this plus, and let's - so let me clear that out,
and I could write it as subtracting a negative one times that, and negative one, we
know, is the same thing as i squared, so we can rewrite this whole thing as 6a to the fourth, squared. And then we have this minus
right over here, minus. And so this is i squared. Negative one is i squared. So we can rewrite this
in this pink color as i times the square root of
two, times b to the third, all of that squared. Notice i squared is negative one. Square root of two squared, is two. B to the third, squared,
is b to the sixth power. If I raise something to an exponent, and then raise it to another exponent, I would multiply the two exponents. And so now I've expressed it as a difference of squares,
so we're ready to factor. This is going to be equal
to 6a to the fourth, minus i times the square root of two, times b to the third, times - and let me get myself space here, times 6a to the fourth,
plus all of this business, i times the square root of 2, times b to the third
power, and we are done.