How do you solve (x^3-8)/(x-2)?

2007-08-20 07:35:37 · 2 answers · asked by Samich 3 in Education & Reference ➔ Homework Help

Factor x^3 -8 into (x-2)(x^2 +2x +4) by using the difference of cubes formula.
Then cancel the (x-2)'s leaving you with x^2 +2x + 4.

2007-08-20 07:43:00 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋

x^3-8/x-2

(x-2)(x^2+2x+4)/x-2

x^2+2x+4

2007-08-20 07:46:24 · answer #2 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋

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