How do you factor x^4-1x41?

2 Answers
Apr 18, 2018

x^4-1=(x^2+1)(x+1)(x-1)x41=(x2+1)(x+1)(x1)

using complex numbers

x^4-1=(x+ib)(x-ib)(x+1)(x-1)x41=(x+ib)(xib)(x+1)(x1)

Explanation:

we make use of the difference of squares

a^2-b^2=(a+b)(a-b)a2b2=(a+b)(ab)

x^4-1=(x^2+1)(x^2-1)x41=(x2+1)(x21)

we can use dos for the second bracket once more

x^4-1=(x^2+1)(x+1)(x-1)--(1)x41=(x2+1)(x+1)(x1)(1)

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

note" "i^2=-1 i2=1

we see

a^2+b^2=a^2-(ib)^2=(a+ib)(a-ib)a2+b2=a2(ib)2=(a+ib)(aib)

(1)rarrx^4-1=(x+ib)(x-ib)(x+1)(x-1)(1)x41=(x+ib)(xib)(x+1)(x1)

Apr 18, 2018

(x-1)(x+1)(x+i)(x-i)(x1)(x+1)(x+i)(xi)

Explanation:

x^4-1" is a "color(blue)"difference of squares"x41 is a difference of squares

"which factors in general as"which factors in general as

•color(white)(x)a^2-b^2=(a-b)(a+b)xa2b2=(ab)(a+b)

"here "a=x^2" and "b=1here a=x2 and b=1

rArrx^4-1=(x^2-1)(x^2+1)x41=(x21)(x2+1)

x^2-1" is a "color(blue)"difference of squares"x21 is a difference of squares

rArrx^4-1=(x-1)(x+1)(x^2+1)x41=(x1)(x+1)(x2+1)

"we can factor "x^2+1" by equating to zero and solving"we can factor x2+1 by equating to zero and solving

x^2+1=0rArrx^2=-1rArrx=+-sqrt(-1)=+-ix2+1=0x2=1x=±1=±i

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

rArrx^4-1=(x-1)(x+1)(x-i)(x+i)x41=(x1)(x+1)(xi)(x+i)