how to factorize x^2+1?

Answer 1

(x-i)(x+i)

Answer 2

(x+i)(x-i)
where i^2=-1

Answer 3

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

Answer 4

x² + 1 This is the same as:
x² + 0*x + 1 That is, a quadratic equation. So you can use the quadratic equation formula.
[-b ±√(b² - 4ac)] / 2a
In this question a = 1 and b = 0 and c = -1.
x = [-b ±√(b² - 4ac)] / 2a
x = [-(0) ±√((0)² - 4(1)(-1)] / 2(1)
x = [ ±√( - 4(1)(-1)] / 2(1)
x = ± √4 / 2
x = ± 2 / 2
x = ± 1

Answer 5

Hey guys. He is asking HOW?

To factor any polynomial, you need a root.

Begin with x^2+1=0
then we can rearrange this to get
x^2=-1
therefore x = j ; where j is sqrt(-1)

there is some theorem about that says if you know a root of a polynomial then (x-root) is a factor of that polynomial -i think it is called the "fundamental theorem of algebra"

so now we know that one factor is (x-j)

we need to divide x-j into x^2+1
__x__+ j_______
x-j ) x^2+1
x^x -jx
----------
1+jx
jx -(j*j)
-----------
0

therfore the other factor is (x+j)

so, x^2+1 = (x-j)( x+j)

Answer 6

(x+1)(x-1)

Answer 7

x^2 + 1 = 0
x^2 = -1 ( Taking 1 to RHS )
So, no solution possible as square of any number cannot be negative ( unless you apply imaginary numbers, where answer would be i)

Answer 8

x^2 + 1

= x^2 - i^2

= (x + i)(x - i) where i = sqrt(-1) called imaginery square root of unity

Answer 9

x(x+(1/x))

x * x is x^2

1/x * x is 1 so

x(x+(1/x)) is your answer

or (x+1)(x-1) works aswell

Answer 10

it "really" can't be factored. I can only give you an imaginary answer.

Answer 11

(x +i)(x-i)