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Division of polynomials: x^3-7/x-2

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1 Answer

There are a couple of ways to attack this, but I will choose my way.
 
Let's rewrite the expression as:
 
(x^3 - 8)/(x - 2) + 1/(x - 2)
 
This is equal to our original expression.
 
When factoring a difference of two cubes you get this:
 
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
 
x^3 - 8 = (x - 2)(x^2 +2x + 4)
 
Now we have:
 
(x - 2)(x^2 + 2x + 4)/(x - 2) + 1/(x - 2)
 
You can cancel out the x-2 term in the numerator & denominator. You get:
 
x^2 + 2x + 4 + 1/(x-2)