anonymous
  • anonymous
i have a weird question maybe someone can help.... q^2+1...how do u factor?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you cannot with real numbers
anonymous
  • anonymous
That's true..
anonymous
  • anonymous
so does that mean its prime?

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More answers

anonymous
  • anonymous
\[(q-1)(q-1)\]
anonymous
  • anonymous
that is not correct
anonymous
  • 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
  • anonymous
but then the middle would be -2q
anonymous
  • anonymous
\[(q+1)(q+1)\]
anonymous
  • anonymous
then be +2
anonymous
  • anonymous
i mean +2q
myininaya
  • myininaya
does (q-i)(q+i) work
anonymous
  • anonymous
no
anonymous
  • anonymous
no cause then it would be -1 myininaya
myininaya
  • myininaya
q^2+iq-iq-i^2 q^2+0-(-1) q^2+1 yes it works ! :)
anonymous
  • anonymous
but its q^2+1
anonymous
  • anonymous
listen
anonymous
  • anonymous
i=sqrt(-1)
myininaya
  • myininaya
q^2+1=(q-i)(q+i)
anonymous
  • anonymous
that is called a complex number
anonymous
  • anonymous
has a lot of interesting bits
anonymous
  • anonymous
-1*+1 is still -1
anonymous
  • anonymous
nooooooooooooooooooooooooooooooooooooooooo
myininaya
  • myininaya
(q-i)(q+i)=q^2-qi+qi-i^2 agreed?
anonymous
  • anonymous
-i*i=-i^2=-(-1)=1
anonymous
  • anonymous
its not i its 1(one)
myininaya
  • myininaya
q^2+0-(-1) q^2+1
myininaya
  • myininaya
i^2=-1
anonymous
  • anonymous
\[(q+1)(q+1) = q ^{2}+1^{2}\]
myininaya
  • myininaya
lol this is correct to say q^2+1=(q-i)(q+i)
anonymous
  • anonymous
NOOOOO it is not
anonymous
  • anonymous
either im super stupid or im just not comprehending how that works
myininaya
  • myininaya
shark thats wrong (q+1)(q+1)=q^2+2q+1
anonymous
  • anonymous
sharkattack you still get q^2 2q+1 as the answer
anonymous
  • anonymous
As I said: you cannot with real numbers! (you need to introduce complex numbers with i to solve it)
anonymous
  • anonymous
so andras does that mean its prime?
anonymous
  • anonymous
\[(X+Y)(X+Y) = X ^{2} + Y ^{2}\]
anonymous
  • anonymous
no it does not
myininaya
  • myininaya
ok Bmartin i^2=-1 (q-i)(q+i)=q^2+qi-qi-i^2 =q^2-(-1) =q^2+1
anonymous
  • anonymous
for example 3^2+1=10
myininaya
  • myininaya
bmartin do you see the negative in front of the i^2?
anonymous
  • anonymous
not prime
anonymous
  • anonymous
so -1*+1 =+1?
anonymous
  • anonymous
i dont believe that
anonymous
  • anonymous
sorry but I'm right
myininaya
  • myininaya
-(-1)=1
myininaya
  • myininaya
i^2=-1 not 1
anonymous
  • anonymous
myiniaya you are just confusing them with complex numbers
myininaya
  • myininaya
you can factor over complex numbers
anonymous
  • anonymous
Sharkattack is right!!! Lets vote him for president!!!
anonymous
  • anonymous
lol
anonymous
  • anonymous
:-)
anonymous
  • anonymous
so the answer is (q+1)(q-1)? i just really dont see how that works....
myininaya
  • myininaya
no! it is (q-i)(q+i)
anonymous
  • anonymous
:)
anonymous
  • anonymous
no there is no answer
anonymous
  • anonymous
why does it make a difference where the signs go?
anonymous
  • anonymous
no
anonymous
  • anonymous
with normal (real) numbers there is no answer
anonymous
  • anonymous
oooooooooooooh im soooooo confused
anonymous
  • anonymous
I can imagine, what is not clear?
anonymous
  • anonymous
\[(X+Y)(X+Y) =X ^{2} +Y ^{2}\]
myininaya
  • 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
  • anonymous
shark it is not: it is x^2+2xy+y^2
anonymous
  • anonymous
\[(q+1)(q+1) =q ^{2} +1^{2} =q ^{2} +1\]
anonymous
  • anonymous
i just keep seeing that answer as q^2-1 myininaya
anonymous
  • anonymous
myininaya you are right but if someone never learnt about complex numbers than you cannot explain them in 2 min
myininaya
  • myininaya
i^2=-1 -i^2=1
anonymous
  • anonymous
they have the same base
myininaya
  • myininaya
do you know the bmartin that -i^2=1?
anonymous
  • 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
  • anonymous
mmmmm....
myininaya
  • myininaya
bmartin, do you know that -i^2=1?
anonymous
  • anonymous
no
myininaya
  • myininaya
do you about imaginary numberts>
anonymous
  • anonymous
no
anonymous
  • anonymous
i'm right in my law
myininaya
  • myininaya
ok the sum of two real squares is not factorable over the reals
anonymous
  • anonymous
wait
anonymous
  • anonymous
ok
myininaya
  • myininaya
if you knew of imaginary numbers then we could factor it
anonymous
  • anonymous
factorize x^2 +1 X^2+2xy+Y^2
anonymous
  • anonymous
hopefully i get to that chapter soon
myininaya
  • myininaya
lol sorry andras i thought they were talking about imagary numbers so i just figured they knew lol
anonymous
  • anonymous
no prob
anonymous
  • anonymous
well thanks everyone!
anonymous
  • anonymous
this took longer than I thaught
anonymous
  • anonymous
but
myininaya
  • myininaya
by the way sean the square root of -1 is i not the square root of i
anonymous
  • anonymous
yeah, true :-)
anonymous
  • anonymous
I'm right
anonymous
  • anonymous
lol
myininaya
  • myininaya
shark is so cute
anonymous
  • anonymous
you are as always, go for politician you would do great
anonymous
  • anonymous
so persistant...you shoud be a lawyer
myininaya
  • myininaya
i felt as i was debating against someone such tough debater
anonymous
  • anonymous
i bet he was on the debate team!
anonymous
  • anonymous
can we convince you shark that you are wrong?
myininaya
  • myininaya
what about if we 9^2+1 is this equal to 9^2+9(1)+1^2
myininaya
  • myininaya
oops i mean 9^2+2(9)(1)+1^2
myininaya
  • myininaya
say no lol
anonymous
  • anonymous
no
anonymous
  • anonymous
look
anonymous
  • anonymous
ha ha well yall have a good night!! thanks for your help! ill beback for more debate later!
myininaya
  • myininaya
ok im looking
anonymous
  • anonymous
factorize (x+y)^2 and X^2+Y^2
myininaya
  • myininaya
what do you mean? (x+y)^2 is already in factorable form x^2+y^2 cannot be factored over the reals
anonymous
  • anonymous
no
anonymous
  • anonymous
what's your work ?
myininaya
  • 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
  • myininaya
13 isn't the same as 25*
anonymous
  • anonymous
they are have different factorization
myininaya
  • myininaya
they are not them same i just gave you a counterexample
anonymous
  • anonymous
shark you are the best!
myininaya
  • myininaya
lol
anonymous
  • anonymous
answer me
anonymous
  • anonymous
factorize (X+Y)^2 and X^2 + Y^2
myininaya
  • 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
  • anonymous
what's your work
myininaya
  • 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
  • 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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