How do you factor x4−1 ?

Question

temnimam2 · Accepted Answer

We make use of the difference of squares

a^{2} - b^{2} = (a + b) (a - b)

x^{4} - 1 = (x^{2} + 1) (x^{2} - 1)

we can use dos for the second bracket once more

x^{4} - 1 = (x^{2} + 1) (x + 1) (x - 1)

- (1)
for real numbers we can proceed no further, but if we use complex numbers
note

i^{2} = - 1

we see

a^{2} + b^{2} = a^{2} - {(i b)}^{2} = (a + i b) (a - i b)

\to x^{4} - 1 = (x + i b) (x - i b) (x + 1) (x - 1)

Edward Patten · Accepted Answer

x4−1 is a difference of squares  which factors in general as  a2−b2=(a−b)(a+b)  here a=x2 and b=1  ⇒x4−1=(x2−1)(x2+1)  x2−1 is a difference of squares  ⇒x4−1=(x−1)(x+1)(x2+1)  we can factor x2+1 by equating to zero and solving  x2+1=0⇒x2=−1⇒x=±−1=±i  factors are (x−(+i))(x−(−i))  ⇒x4−1=(x−1)(x+1)(x−i)(x+i)

karton · Accepted Answer

Rewrite x4 as (x2)2 (x2)2−1 Rewrite 1 as 12 (x2)2−12 Since both terms are perfect squares, factor using the difference of squares formula, a2−b2=(a+b)(a−b) where a=x2 and b=1 (x2+1)(x2−1) Simplify. (x2+1)(x+1)(x−1)

How do you factor x^4-1 ?

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