- anonymous

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

- chestercat

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- anonymous

you cannot with real numbers

- anonymous

That's true..

- anonymous

so does that mean its prime?

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## More answers

- anonymous

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

- anonymous

that is not correct

- anonymous

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

- anonymous

but then the middle would be -2q

- anonymous

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

- anonymous

then be +2

- anonymous

i mean +2q

- myininaya

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

- anonymous

no

- anonymous

no cause then it would be -1 myininaya

- myininaya

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

- anonymous

but its q^2+1

- anonymous

listen

- anonymous

i=sqrt(-1)

- myininaya

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

- anonymous

that is called a complex number

- anonymous

has a lot of interesting bits

- anonymous

-1*+1 is still -1

- anonymous

nooooooooooooooooooooooooooooooooooooooooo

- myininaya

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

- anonymous

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

- anonymous

its not i its 1(one)

- myininaya

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

- myininaya

i^2=-1

- anonymous

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

- myininaya

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

- anonymous

NOOOOO it is not

- anonymous

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

- myininaya

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

- anonymous

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

- anonymous

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

- anonymous

so andras does that mean its prime?

- anonymous

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

- anonymous

no it does not

- myininaya

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

- anonymous

for example 3^2+1=10

- myininaya

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

- anonymous

not prime

- anonymous

so -1*+1 =+1?

- anonymous

i dont believe that

- anonymous

sorry but I'm right

- myininaya

-(-1)=1

- myininaya

i^2=-1
not 1

- anonymous

myiniaya you are just confusing them with complex numbers

- myininaya

you can factor over complex numbers

- anonymous

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

- anonymous

lol

- anonymous

:-)

- anonymous

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

- myininaya

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

- anonymous

:)

- anonymous

no there is no answer

- anonymous

why does it make a difference where the signs go?

- anonymous

no

- anonymous

with normal (real) numbers there is no answer

- anonymous

oooooooooooooh im soooooo confused

- anonymous

I can imagine, what is not clear?

- anonymous

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

- myininaya

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)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

they have the same base

- myininaya

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

- anonymous

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

- anonymous

mmmmm....

- myininaya

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

- anonymous

no

- myininaya

do you about imaginary numberts>

- anonymous

no

- anonymous

i'm right in my law

- myininaya

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

- anonymous

wait

- anonymous

ok

- myininaya

if you knew of imaginary numbers then we could factor it

- anonymous

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

- anonymous

hopefully i get to that chapter soon

- myininaya

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

- anonymous

no prob

- anonymous

well thanks everyone!

- anonymous

this took longer than I thaught

- anonymous

but

- myininaya

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

- anonymous

yeah, true :-)

- anonymous

I'm right

- anonymous

lol

- myininaya

shark is so cute

- anonymous

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

- anonymous

so persistant...you shoud be a lawyer

- myininaya

i felt as i was debating against someone such tough debater

- anonymous

i bet he was on the debate team!

- anonymous

can we convince you shark that you are wrong?

- myininaya

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

- myininaya

oops i mean 9^2+2(9)(1)+1^2

- myininaya

say no lol

- anonymous

no

- anonymous

look

- anonymous

ha ha well yall have a good night!! thanks for your help! ill beback for more debate later!

- myininaya

ok im looking

- anonymous

factorize (x+y)^2 and X^2+Y^2

- myininaya

what do you mean?
(x+y)^2 is already in factorable form
x^2+y^2 cannot be factored over the reals

- anonymous

no

- anonymous

what's your work ?

- myininaya

let x=3 and y=2
so (x+y)^2=(3+2)^2=5^2=25
and
x^2+y^2=3^2+2^2=9+4=13
13 isn't the same as 26
so
(x+y)^2 is not the same as x^2+y^2

- myininaya

13 isn't the same as 25*

- anonymous

they are have different factorization

- myininaya

they are not them same
i just gave you a counterexample

- anonymous

shark you are the best!

- myininaya

lol

- anonymous

answer me

- anonymous

factorize (X+Y)^2 and X^2 + Y^2

- myininaya

if i multiply (x+y)^2=(x+y)(x+y)=x^2+2xy+y^2
x^2+y^2 is not equal to that since it is missing the 2xy term

- anonymous

what's your work

- myininaya

(x+y)(x+y)=x^2+xy+xy+y^2=x^2+2xy+y^2
see the 2xy
2xy is not in
x^2+y^2
so (x+y)^2 does not equal x^2+y^2

- anonymous

i know
bec.
(X+Y)^2 = X^2 +2xy +Y^2
but
X^2 +Y^2 = (X+Y) (X+Y) or (X-Y) (X-Y)
factorize X^2 -Y^2

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