# How do you factor #x^4-1#?

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#x^4-1" is a "color(blue)"difference of squares"#

#"which factors in general as"#

#•color(white)(x)a^2-b^2=(a-b)(a+b)#

#"here "a=x^2" and "b=1#

#rArrx^4-1=(x^2-1)(x^2+1)#

#x^2-1" is a "color(blue)"difference of squares"#

#rArrx^4-1=(x-1)(x+1)(x^2+1)#

#"we can factor "x^2+1" by equating to zero and solving"#

#x^2+1=0rArrx^2=-1rArrx=+-sqrt(-1)=+-i#

#"factors are "(x-(+i))(x-(-i))#

#rArrx^4-1=(x-1)(x+1)(x-i)(x+i)#

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#### Answer:

using complex numbers

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we make use of the difference of squares

we can use dos for the second bracket once more

for real numbers we can proceed no further, but if we use complex numbers

note

we see

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