Question

Factor x^2 + 64. Check your work.

Answer 1

Any ideas?

Answer 2

not factorisable xD

Answer 3

It is factorisable, but in the complex numbers.

Answer 4

(x+8i)(x-8i)

Answer 5

Are you sure there is suppose to be a "+" not a "-" @mariacruz07 ?

Answer 6

@skullpatrol yes thats exactly how the question looks.

Answer 7

@kc_kennylau how did you get (x+8i)(x-8i)?

Answer 8

But if it's x^4+64 it can be factorized unto (x^2+x+8)(x^2-x+8)

Answer 9

x^2+64=x^2-(-64)=x^2-(8i)^2=(x+8i)(x-8i)

Answer 10

@skullpatrol not really, thats why im confused. can you help me understand?

Answer 11

I believe that your textbook/website is wrong.

Answer 12

kc_kennylau has given the answer, go to YouTube and look up the Khan Academy for complex numbers.

Answer 13

@kc_kennylau wrong in what?

Answer 14

@skullpatrol ok thank you, i will do that now.

Answer 15

But I can teach you complex numbers :D
Complex numbers revolve around a concept, a number named \(i\), which is defined as:
\[i=\sqrt{-1}\]

Answer 16

@kc_kennylau thank you, you are so helpful!

Answer 17

The problem is do you wanna learn lol, coz at some extend it's difficult.

Answer 18

@kc_kennylau yes i do wanna learn this!!

Answer 19

and we can deduce that \(\large i^2=-1\)

Answer 20

And complex numbers are numbers in the form of \(a+bi\), where \(a\) and \(b\) are real numbers.

Answer 21

So how would you find out the value of \((1+2i)+(3+4i)\)?

Answer 22

would you substitute the i first?

Answer 23

No, i is not an unknown

Answer 24

well if substituting i as sqrt(-1) would make it easier, feel free to do so :)

Answer 25

ok im not sure how to solve that.. sorry, im kinda lost.

Answer 26

oh, since you're new to this matter, let me give you an example first:
\[(6+7i)+(5+6i)=6+7i+5+6i=11+13i\]

Answer 27

Terms with \(i\) can add together, terms without \(i\) can add together, but a term with \(i\) and a term without \(i\) cannot add together

Answer 28

ok now im starting to get it

Answer 29

So can you do mine? :)

Answer 30

yes give me a sec im gonna try it :)

Answer 31

ok :)

Answer 32

4 + 6i ?

Answer 33

exactly, i can see that you're getting it now :D

Answer 34

What about \((3+4i)-(1+2i)\)? :)

Answer 35

2 - 2i ? the "-" is getting me confused a bit

Answer 36

Just take out the brackets first :)

Answer 37

oh so it would be 2 + 6i right?

Answer 38

ok, lemme give you an example:
\[(7+6i)-(5+4i)=7+6i-5-4i=7-5+6i-4i=2+2i\]

Answer 39

so im right, its 2 + 6i ?

Answer 40

nope :/
try to do it with steps :)

Answer 41

\[\begin{array}{ll}&(7+6i)-(5+4i)\\=&7+6i-5-4i\\=&7-5+6i-4i\\=&2+2i\end{array}\]

Answer 42

This is the example I gave you

Answer 43

ok i did it like you did the example.. (3+4i) - (1 + 2i).... 3-1 + 4i - 2i = 2 + 2i ?

Answer 44

exactly :D

Answer 45

Now it's multiplication time...
It may be a little bit difficult, so I'll start with this: \(2(3+4i)\)

Answer 46

Just distribute to get your answer :)

Answer 47

6 + 8i?

Answer 48

yes :)
Now do this \(2i(3+4i)\)

Answer 49

Again, just distribute :D

Answer 50

Don't forget that \(i^2=-1\) :)

Answer 51

I'll take a little intervention here.
I'll show her much slower the steps to take, in order to solve a substraction or sum of complex numbers.
Now, you skipped one fundamental detail, the complex numbers behave as vectors, so you can't apply fully aritmethic properties because they won't work in many cases.
your example was:
\[(7+6i)-(5+4i)\]
in order to turn the sustraction into a sum, I have to sum the opposite (just like vectors):
\[(7+6i)+(-5-4i)\]
so to solve the sum, you just englobe the real numbers and the complex part also:
\[(7-5)+(6-4)i\]
giving you a result of:
\[2+2i\]
Now, as a trick, always group the numbers without "i" and the numbers with "i" on a parenthesis, so it doesn't look confusing.

Answer 52

6i + 8i^2 = 6i + 8(-1) = 6i - 8 ?

Answer 53

exactly :D You're doing very good here :)

Answer 54

Now I believe that you can calculate the value of (1+2i)(3+4i) :)
Again, just distribute :D

Answer 55

thank you, you are really teaching me in a way i understand. so i couldnt do this without your help!

Answer 56

ok let me try this problem

Answer 57

-5 + 10i or 10i - 5 ?

Answer 58

yes, and i'd prefer the first way of expression i.e. -5+10i :)

Answer 59

And I'm glad to see that you're getting it :)

Answer 60

Before we step into division, we must learn a thing called a "conjugate".

Answer 61

For any complex number \(a+bi\), its conjugate is \(a-bi\), and vice versa.

Answer 62

so its basically the opposite sign?

Answer 63

yes :)

Answer 64

For example, the conjugate of 3+4i would be?

Answer 65

3-4i

Answer 66

yep you are getting it :D

Answer 67

:)

Answer 68

It's the first time I'm teaching someone something new, because my dream is to become a Math teacher :)

Answer 69

And I'm very glad to see that my teaching method actually works :D

Answer 70

well you are very good at it, so im sure your dream will come true :)

Answer 71

Thanks :D

Answer 72

What's the conjugate of 5+2i?

Answer 73

5-2i

Answer 74

Multiply 5+2i by its conjugate, what do you get? :)

Answer 75

21?

Answer 76

yes :)

Answer 77

Noticed something strange?

Answer 78

yes there is no complex number

Answer 79

exactly :D

Answer 80

Now we can step into division :)

Answer 81

Try to find the value of \(\dfrac{1+2i}{3+4i}\) :)

Answer 82

Hint: make the denominator real :)

Answer 83

ok im stuck could u show me an example first?

Answer 84

\[\frac{5+6i}{7+8i}=\frac{(5+6i)(7-8i)}{(7+8i)(7-8i)}=\frac{83+2i}{113}=\frac{83}{113}+\frac2{113}i\]

Answer 85

And a little correction, (5+2i)(5-2i) is 29 not 21 :)

Answer 86

Do you get it? :/

Answer 87

wait, how was it 29?

Answer 88

(5+2i)(5-2i)
=25-10i+10i-4i^2
=25-4i^2
=25-4(-1)
=25+4
=29

Answer 89

oh i see

Answer 90

ok ima try the division problem now

Answer 91

Estoy correcto que hablas espanol? :)

Answer 92

De tu nombre lo veo yo

Answer 93

si

Answer 94

...

Answer 95

por que no me dijiste tu?

Answer 96

no sabia que hablabas español, i speak more english tho.

Answer 97

te gustaria si hablo ingles o espanol?