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

2 Answers
Apr 18, 2018

Answer:

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

using complex numbers

#x^4-1=(x+ib)(x-ib)(x+1)(x-1)#

Explanation:

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-(ib)^2=(a+ib)(a-ib)#

#(1)rarrx^4-1=(x+ib)(x-ib)(x+1)(x-1)#

Apr 18, 2018

Answer:

#(x-1)(x+1)(x+i)(x-i)#

Explanation:

#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)#