Expand x^3/x^2 + x^2/x + 7/-3 by partial fractions. Answer f(x) + a/x-1 + b/ x+2, where f(x) =_____a? And b =____?

Answer 1

I think you meant this:

Expand:

(x^3 + x^2 + 7)

------------------ by partial fractions.

(x^2 + x - 2)

You can't equate it to (x^3/x^2) + (x^2/x) + (7/-2) --- (this would just be 2x-7/2 since each of the first two fractions simply works out to an x by itself)

We have to complete the long-division first since the order of the numerator is 3 which is greater than that of the denominator (2).

X^2 +stuff goes into x^3 + other stuff more than x times, so we put x up top and multiply the divisor by x to see what to subtract from the denominator. We end up subtracting x^3 + x^2 - 2x from (x^3 + x^2 + 7) which leaves us with 2x + 7

x^2 doesn't go into 2x+7 so 2x+7 is the remainder and our long division is finished. The quotient was x and the remainder was 2x+7 ....e.g. the original fraction is equal to x plus a new fraction of 2x+7 over the original divisor.

Therefore f(x) = x.

The denominator I was forced to assume you meant can be factored into (x+2)*(x-1). So the new fraction has some equivalent A/(x-1) + B/(x+2).

To easily find A, set x = 1 (so A's denominator equals zero) and multiply both sides by (x-1) in the overall equation:

2x+7 (x-1) A * (x-1) B(x-1)

==== · = ---------------- + ----------------

(x-1)(x+2) (x-1) x+2

After cancelling and simplifying, this leaves us with:

2x+7 A 0

-------- = +

x+2

Evaluating the left-hand side (since x = 1), tells us that A = 9/3 = 3 --> A = 3

To easily find B, set x = -2 (so B's denominator equals zero) and multiply both sides by (x+2) in the overall equation:

2x+7 A * (x+2) B (x+2)

==== · (x+2) = ---------------- + --------------

(x-1)(x+2) (x-1) x+2

After cancelling and simplifying, this leaves us with:

2x+7 0 B

-------- = + = B

x-1

Evaluating the left-hand side (at x = -2), tells us that B = 3/(-3) = -1 ---> B = -1