Question

Divide: x^4-1 by x+1

Answer 1

x^4/x + 1
Steps
Divide x^4/x + 1: x^4/x + 1 = x^3 + -x^3/x +1
= x^3 + -x^3/x +1
Divide -x^3/x + 1: -x^3/x + 1 = -x^2 + x^2/x + 1
= x^3 - x^2 + x^2/x + 1
Divide x^2/x +1: x^2/x + 1: x^2/x + 1 = x + -x/x + 1
= x^3 - x^2 + x + -x/x +1
Divide -x/x + 1: -x/x + 1 = -1 + 1/x + 1
= x^3 - x^2 + x - 1 + 1/x + 1

Answer 2

assumin you mean


`(x^4-1)/(x+1)`
recognize difference of 2 perfect squares
remember that
`a^2-b^2=(a+b)(a-b)`

so
factor the numerator

`x^4-1=(x^2)^2-(1)^2=`
`(x^2+1)(x^2-1)`
but look, another perfect square

`(x^2+1)(x^2-1)=(x^2+1)(x^2-1^2)=(x^2+1)(x+1)(x-1)`
so we end up with


`(x^4-1)/(x+1)=((x^2+1)(x-1)(x+1))/(x+1)` which simplifies to
`(x^2+1)(x-1)`