how to factorize x2+1?

Answer 1

x^2 + 1 is unfactorisable in terms of real numbers.

Possible explanations:
1) You typed the question wrong
If you instead meant "how to factorize x^2 - 1" then it is simple.
The method is called "The difference of two squares"
Since 1 is a square number, and x^2 is (obviously) a square number, you can immediately write it as (x + 1)(x - 1)
This causes the term in x to cancel. Expand the brackets if you're not convinced.

2) You need an answer in terms of imaginary numbers.
While it is true that there are no real roots of the equation x^2 + 1 = 0, there ARE imaginary ones.
Substitute the correct coefficients into the quadratic formula and you will find that the roots of the equation are at +i and -i respectively, where i is the imaginary number sqrt(-1).

From this, we can immediately see that the eqation factorises into (x + i)(x - i) just like the difference of two squares method. Multiply this out if you need convincing, remembering that i^2 = -1.


Hope this helps!

Answer 2

In general, considering imarginary as well as real coefficients, the factorisation is as follows:
x^2 + 1
= x^2 - i^2
= (x + i)(x - i)

However, with the additional restriction of all the coefficients of the exponents in x being strictly real, the factorisation would be:

x^2 + 1
= (x+1)^2 - [sqrt(2x))] ^2
= [x+sqrt(2x) +1][x-sqrt(2x)+1]

Answer 3

by x2+1 do you mean x^2 + 1 or x^(2+1) or 2x + 1 or somehting else?

Answer 4

In the universe of the real numbers it can't be done.

If x exists in the imaginary plane, then x^2 + 1 = (x + i)(x - i).

Answer 5

Use the quadratic formula. However - it seems to me that you will only be able to factor this expression if you are open to using imaginary numbers. Limiting yourself to only the real numbers will result in no solution, I do believe.

Answer 6

expensive Clouie, x^(2)-a million The binomial could be factored utilising the version of squares formula, simply by fact the two words are appropriate squares. the version of squares formula is a^(2)-b^(2)=(a-b)(a+b). (x-a million)(x+a million) ================= 3x^(2)+12x+9 element out the GCF of three from each and every term interior the polynomial. 3(x^(2))+3(4x)+3(3) element out the GCF of three from 3x^(2)+12x+9. 3(x^(2)+4x+3) For a polynomial of the kind x^(2)+bx+c, locate 2 factors of c (3) that upload as much as b (4). in this difficulty 3*a million=3 and 3+a million=4, so insert 3 simply by fact the impressive hand term of one element and a million simply by fact the impressive-hand term of the different element. 3(x+3)(x+a million) ======================== x^(4)-a million The binomial could be factored utilising the version of squares formula, simply by fact the two words are appropriate squares. (x^(2)+a million)(x^(2)-a million) The binomial could be factored utilising the version of squares formula, simply by fact the two words are appropriate squares. the version of squares formula is a^(2)-b^(2)=(a-b)(a+b). (x^(2)+a million)(x-a million)(x+a million)

Answer 7

multiply the value of x times 2 and then add one

example: if x = 5

5 x 2 + 1
10 + 1
11

Is this what you meant?

Answer 8

that can't be done without imaginary numbers.

if you possible meant x2-1 then its (x+1)(x-1)

otherwise you have incorporate "i" which i forget how to do.

Answer 9

we could do it in the imaginary plane remembering that
i=(-1)1/2 and i^2=-1
x^2+1
=x^2-i^2
=(x+i)(x-i)

Answer 10

2x+1 is already factored.