How do you factor #x^4-1#?
2 Answers
using complex numbers
Explanation:
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
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)#