Question

What is the quotient and remainder, written as partial fractions, of 2x^3-25x+40/x^2+2x-8

Answer 1

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Answer:

  (2x -4) +1/(x -2) -2/(x +4)

Step-by-step explanation:

The attached shows the quotient is (2x -4) and the remainder expressed as a fraction is ...

  r = (8 -x)/(x^2 +2x -8)

So, the problem is to write the partial fraction expansion of this remainder. It will be of the form ...

  r = A/(x -2) +B/(x +4)

where (x-2)(x+4) is the factorization of the divisor quadratic.

The value of A can be found by evaluating (x -2)r at x=2.

 (x -2)r for x=2 is (8 -2)/(2 +4) = 6/6 = 1

The value of B can be found by evaluating (x +4)r at x=-4.

  (8 -(-4))/(-4-2) = 12/-6 = -2

Then the quotient and remainder, written as partial fractions is ...

  = (2x -4) +1/(x -2) -2/(x +4)

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Additional comment

In the form ...

  $r=\dfrac{8-x}{x^2+2x-8}$

we can see that (x -2)r will be ...

  $(x -2)r = \dfrac{(x-2)(8-x)}{(x-2)(x+4)}=\dfrac{8-x}{x+4}$

This can be evaluated at x=2, as we have done above. Similar factor cancellation works to give (x+4)r = (8-x)/(x-2).

This method of arriving at the "A" and "B" values may not pass technical scrutiny regarding where r is defined or undefined--but it works. One could consider the work to be finding a limit, rather than evaluating a rational function at points where it is undefined.