Question
I am trying to write this rational expression in its simplest form:
2-x+2x^2-x^3/x^2-4
However I have no idea how to factor the four numbers on the top. I"m used to factoring just three numbers. Can you show me how to do this? Thanks!!!
2-x+2x^2-x^3/x^2-4
However I have no idea how to factor the four numbers on the top. I"m used to factoring just three numbers. Can you show me how to do this? Thanks!!!
Answers
Reiny
use grouping
2-x+2x^2-x^3
= (2-x) + x^2(2-x)
= (2-x)(1+x^2)
so
(2-x+2x^2-x^3)/(x^2-4)
= (2-x)(1+x^2)/[(x-2)(x+2)]
= -(1+x^2)/(x+2)
2-x+2x^2-x^3
= (2-x) + x^2(2-x)
= (2-x)(1+x^2)
so
(2-x+2x^2-x^3)/(x^2-4)
= (2-x)(1+x^2)/[(x-2)(x+2)]
= -(1+x^2)/(x+2)
Shavaleir
2x+4x-3x/2x-4=
6x-3x/2x-4= 5/10x
6x-3x/2x-4= 5/10x