Question

i have a weird question maybe someone can help.... q^2+1...how do u factor?

Answer 1

you cannot with real numbers

Answer 2

That's true..

Answer 3

so does that mean its prime?

Answer 4

\[(q-1)(q-1)\]

Answer 5

that is not correct

Answer 6

No, if you are to solve for q, you have \[\sqrt{-1}=\sqrt{i}\]
This requires that you are familiar with complex numbers.

Answer 7

but then the middle would be -2q

Answer 8

\[(q+1)(q+1)\]

Answer 9

then be +2

Answer 10

i mean +2q

Answer 11

does (q-i)(q+i) work

Answer 12

no

Answer 13

no cause then it would be -1 myininaya

Answer 14

q^2+iq-iq-i^2
q^2+0-(-1)
q^2+1
yes it works ! :)

Answer 15

but its q^2+1

Answer 16

listen

Answer 17

i=sqrt(-1)

Answer 18

q^2+1=(q-i)(q+i)

Answer 19

that is called a complex number

Answer 20

has a lot of interesting bits

Answer 21

-1*+1 is still -1

Answer 22

nooooooooooooooooooooooooooooooooooooooooo

Answer 23

(q-i)(q+i)=q^2-qi+qi-i^2
agreed?

Answer 24

-i*i=-i^2=-(-1)=1

Answer 25

its not i its 1(one)

Answer 26

q^2+0-(-1)
q^2+1

Answer 27

i^2=-1

Answer 28

\[(q+1)(q+1) = q ^{2}+1^{2}\]

Answer 29

lol this is correct to say q^2+1=(q-i)(q+i)

Answer 30

NOOOOO it is not

Answer 31

either im super stupid or im just not comprehending how that works

Answer 32

shark thats wrong
(q+1)(q+1)=q^2+2q+1

Answer 33

sharkattack you still get q^2 2q+1 as the answer

Answer 34

As I said: you cannot with real numbers! (you need to introduce complex numbers with i to solve it)

Answer 35

so andras does that mean its prime?

Answer 36

\[(X+Y)(X+Y) = X ^{2} + Y ^{2}\]

Answer 37

no it does not

Answer 38

ok Bmartin i^2=-1
(q-i)(q+i)=q^2+qi-qi-i^2
=q^2-(-1)
=q^2+1

Answer 39

for example 3^2+1=10

Answer 40

bmartin do you see the negative in front of the i^2?

Answer 41

not prime

Answer 42

so -1*+1 =+1?

Answer 43

i dont believe that

Answer 44

sorry but I'm right

Answer 45

-(-1)=1

Answer 46

i^2=-1
not 1

Answer 47

myiniaya you are just confusing them with complex numbers

Answer 48

you can factor over complex numbers

Answer 49

Sharkattack is right!!! Lets vote him for president!!!

Answer 50

lol

Answer 51

:-)

Answer 52

so the answer is (q+1)(q-1)?
i just really dont see how that works....

Answer 53

no!
it is (q-i)(q+i)

Answer 54

:)

Answer 55

no there is no answer

Answer 56

why does it make a difference where the signs go?

Answer 57

no

Answer 58

with normal (real) numbers there is no answer

Answer 59

oooooooooooooh im soooooo confused

Answer 60

I can imagine, what is not clear?

Answer 61

\[(X+Y)(X+Y) =X ^{2} +Y ^{2}\]

Answer 62

lol
(q-i)(q+i)=q^2-qi+qi-i^2=q^2-(-1)=q^2+1
how is not one seeing this except me
i dont want to be on a debate team
accept my answer or not
you can either say not factorable over the real numbers or you can say the answer is
(q-i)(q+i)

Answer 63

shark it is not: it is x^2+2xy+y^2

Answer 64

\[(q+1)(q+1) =q ^{2} +1^{2} =q ^{2} +1\]

Answer 65

i just keep seeing that answer as q^2-1 myininaya

Answer 66

myininaya you are right but if someone never learnt about complex numbers than you cannot explain them in 2 min

Answer 67

i^2=-1
-i^2=1

Answer 68

they have the same base

Answer 69

do you know the bmartin that -i^2=1?

Answer 70

sharck when you expand a bracket you have to multiply every term with every term so (x+1)(x+1)=x^2+x+x+1

Answer 71

mmmmm....

Answer 72

bmartin, do you know that -i^2=1?

Answer 73

no

Answer 74

do you about imaginary numberts>

Answer 75

no

Answer 76

i'm right in my law

Answer 77

ok the sum of two real squares is not factorable over the reals

Answer 78

wait

Answer 79

ok

Answer 80

if you knew of imaginary numbers then we could factor it

Answer 81

factorize x^2 +1
X^2+2xy+Y^2

Answer 82

hopefully i get to that chapter soon

Answer 83

lol sorry andras i thought they were talking about imagary numbers so i just figured they knew lol

Answer 84

no prob

Answer 85

well thanks everyone!

Answer 86

this took longer than I thaught

Answer 87

but

Answer 88

by the way sean the square root of -1 is i
not the square root of i

Answer 89

yeah, true :-)

Answer 90

I'm right

Answer 91

lol

Answer 92

shark is so cute

Answer 93

you are as always, go for politician you would do great

Answer 94

so persistant...you shoud be a lawyer

Answer 95

i felt as i was debating against someone such tough debater

Answer 96

i bet he was on the debate team!

Answer 97

can we convince you shark that you are wrong?

Answer 98

what about if we
9^2+1
is this equal to
9^2+9(1)+1^2