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

Question

how do I divide this?

Answer 1

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

 

 

 

Answer 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-

Answer 3

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.

Answer 4

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