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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you cannot with real numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so does that mean its prime?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, if you are to solve for q, you have \[\sqrt{1}=\sqrt{i}\] This requires that you are familiar with complex numbers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but then the middle would be 2q

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no cause then it would be 1 myininaya

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2q^2+iqiqi^2 q^2+0(1) q^2+1 yes it works ! :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is called a complex number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0has a lot of interesting bits

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nooooooooooooooooooooooooooooooooooooooooo

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2(qi)(q+i)=q^2qi+qii^2 agreed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(q+1)(q+1) = q ^{2}+1^{2}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2lol this is correct to say q^2+1=(qi)(q+i)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0either im super stupid or im just not comprehending how that works

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2shark thats wrong (q+1)(q+1)=q^2+2q+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sharkattack you still get q^2 2q+1 as the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As I said: you cannot with real numbers! (you need to introduce complex numbers with i to solve it)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so andras does that mean its prime?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(X+Y)(X+Y) = X ^{2} + Y ^{2}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2ok Bmartin i^2=1 (qi)(q+i)=q^2+qiqii^2 =q^2(1) =q^2+1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2bmartin do you see the negative in front of the i^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0myiniaya you are just confusing them with complex numbers

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2you can factor over complex numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sharkattack is right!!! Lets vote him for president!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the answer is (q+1)(q1)? i just really dont see how that works....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no there is no answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why does it make a difference where the signs go?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0with normal (real) numbers there is no answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oooooooooooooh im soooooo confused

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can imagine, what is not clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(X+Y)(X+Y) =X ^{2} +Y ^{2}\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2lol (qi)(q+i)=q^2qi+qii^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 (qi)(q+i)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shark it is not: it is x^2+2xy+y^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(q+1)(q+1) =q ^{2} +1^{2} =q ^{2} +1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just keep seeing that answer as q^21 myininaya

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0myininaya you are right but if someone never learnt about complex numbers than you cannot explain them in 2 min

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they have the same base

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2do you know the bmartin that i^2=1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sharck when you expand a bracket you have to multiply every term with every term so (x+1)(x+1)=x^2+x+x+1

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2bmartin, do you know that i^2=1?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2do you about imaginary numberts>

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2ok the sum of two real squares is not factorable over the reals

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2if you knew of imaginary numbers then we could factor it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factorize x^2 +1 X^2+2xy+Y^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hopefully i get to that chapter soon

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2lol sorry andras i thought they were talking about imagary numbers so i just figured they knew lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well thanks everyone!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this took longer than I thaught

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2by the way sean the square root of 1 is i not the square root of i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are as always, go for politician you would do great

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so persistant...you shoud be a lawyer

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2i felt as i was debating against someone such tough debater

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i bet he was on the debate team!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can we convince you shark that you are wrong?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2what about if we 9^2+1 is this equal to 9^2+9(1)+1^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2oops i mean 9^2+2(9)(1)+1^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ha ha well yall have a good night!! thanks for your help! ill beback for more debate later!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factorize (x+y)^2 and X^2+Y^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2what do you mean? (x+y)^2 is already in factorable form x^2+y^2 cannot be factored over the reals

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2let 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
 5 years ago
Best ResponseYou've already chosen the best response.213 isn't the same as 25*

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they are have different factorization

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2they are not them same i just gave you a counterexample

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0shark you are the best!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factorize (X+Y)^2 and X^2 + Y^2

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2if 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

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.2(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
 5 years ago
Best ResponseYou've already chosen the best response.0i know bec. (X+Y)^2 = X^2 +2xy +Y^2 but X^2 +Y^2 = (X+Y) (X+Y) or (XY) (XY) factorize X^2 Y^2
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