How do you factor x^4-1x4−1?

2 Answers
sjc
Apr 18, 2018

x^4-1=(x^2+1)(x+1)(x-1)x4−1=(x2+1)(x+1)(x−1)

using complex numbers

x^4-1=(x+ib)(x-ib)(x+1)(x-1)x4−1=(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)a2−b2=(a+b)(a−b)

x^4-1=(x^2+1)(x^2-1)x4−1=(x2+1)(x2−1)

we can use dos for the second bracket once more

x^4-1=(x^2+1)(x+1)(x-1)--(1)x4−1=(x2+1)(x+1)(x−1)−−(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)(a−ib)

(1)rarrx^4-1=(x+ib)(x-ib)(x+1)(x-1)(1)→x4−1=(x+ib)(x−ib)(x+1)(x−1)

Jim G.
Apr 18, 2018

(x-1)(x+1)(x+i)(x-i)(x−1)(x+1)(x+i)(x−i)

Explanation:

x^4-1" is a "color(blue)"difference of squares"x4−1 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)∙xa2−b2=(a−b)(a+b)

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

rArrx^4-1=(x^2-1)(x^2+1)⇒x4−1=(x2−1)(x2+1)

x^2-1" is a "color(blue)"difference of squares"x2−1 is a difference of squares

rArrx^4-1=(x-1)(x+1)(x^2+1)⇒x4−1=(x−1)(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=0⇒x2=−1⇒x=±√−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)⇒x4−1=(x−1)(x+1)(x−i)(x+i)

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