how do I divide this?

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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

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

You can actually do this as long division (I will use )‾ as the division symbol)

x^{2} +2^{
}

x-5 )‾x^{3} - 5x^{2} + 2x - 10 x goes into x^{3 } x^{2} times

-(x^{3} - 5x^{2}) 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

x^{2} + 2 from the top for our new equation.

Time to check our work

(Remember FOIL - First Outer Inner Last)

(x^{2} + 2)(x-5) (x^{2})(x) = x^{3} (x^{2})(-5) = -5x^{2}

x^{3} - 5x^{2} + 2x - 10 (2)(x) = 2x (2)(-5) = -10

There we go!

(x^{3} - 5x^{2} + 2x - 10)/(x-5) = x^{2} + 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;

(x^{3}-5x^{2}+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 (x^{2})(x).

OUTER must be (x^{2})(-5).

INNER must be (2)(x).

LAST must be (2)(-5).

(x^{2}+2)(x-5)

Let's FOIL...

FIRST...(x^{2})(x)=x^{3}

OUTER...(x^{2})(-5)=-5x^{2}

INNER...(2)(x)=2x

LAST...(2)(-5)=-10

x^{3}-5x^{2}+2x-10

[(x^{2}+2)(x-5)]/(x-5)

(x-5) is in the numerator and denominator. It cancels...

[(x^{2}+2)(x-5)]/(x-5)

x^{2}+2

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