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

how do I divide this?

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Rana A. | Experienced tutor for 30+ yrs, and still making it easy!Experienced tutor for 30+ yrs, and still...
4.8 4.8 (157 lesson ratings) (157)
2
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
 
 
 
Michael V. | Great Tutor in Multiple SubjectsGreat Tutor in Multiple Subjects
1
(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-
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
1
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.
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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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