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# (x^3-5x^2+2x-10) / (x-5)

how do I divide this?

Break the top apart into two expressions, and then, factor it out, Cancel the top and bottom common factors.

(x^3-5x^2)+(2x-10)/x-5

x^2(x-5)+2(x-5)/x-5

Cancel ALL common factors (x-5) on both top and bottom
You end up with x^2+2

(x^3-5x^2+2x-10) / (x-5)
You can actually do this as long division (I will use )‾ as the division symbol)

x2            +2
x-5 )‾x3 - 5x2 + 2x - 10      x goes into x3   x2 times
-(x3 - 5x2)                   Now multiply x2 by x-5
0      0      2x - 10      Subtract and bring down the
-(2x - 10)       next terms and do it again
0      0      Finally we we take our terms
x2 + 2                     from the top for our new equation.

Time to check our work
(Remember FOIL - First Outer Inner Last)

(x2 + 2)(x-5)                    (x2)(x) = x3   (x2)(-5) = -5x2
x3 - 5x2 + 2x - 10                   (2)(x) = 2x (2)(-5) = -10
There we go!

(x3 - 5x2 + 2x - 10)/(x-5) = x2 + 2

-Mike-
In addition to factoring, esp. by grouping, you could divide using long or synthetic division.

(x^3-5x^2+2x-10) / (x-5)

5 | 1 -5 2 -10
. . . . 5. 0 . 10
. . .1 0 .2 | .0 = Remainder

So the quotient is x^2 + 0x + 2 = x^2 + 2.
Hi Shyanna;
(x3-5x2+2x-10) / (x-5)
We need to factor the numerator.  This seems intimidating because we have an x to the exponential of 3. Actually, these are the easiest to factor because there is no guesswork.
Second parenthetical equation must be (x-5).
FIRST must be (x2)(x).
OUTER must be (x2)(-5).
INNER must be (2)(x).
LAST must be (2)(-5).
(x2+2)(x-5)
Let's FOIL...
FIRST...(x2)(x)=x3
OUTER...(x2)(-5)=-5x2
INNER...(2)(x)=2x
LAST...(2)(-5)=-10
x3-5x2+2x-10
[(x2+2)(x-5)]/(x-5)
(x-5) is in the numerator and denominator.  It cancels...
[(x2+2)(x-5)]/(x-5)
x2+2