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Hey , Im struggling on a few questions and was wondering if you could help me out so that i can see how it done correctly, the question is on Partial Fractions and reads

Resolve the following into partial fractions

a) 3(2x^2-8x-1)/(x+4)(x+1)(2x-1)

b)3x^2-8x-63/x^2-3x-10

Any help would be appreciated!

Any rational function f(x) = P(x) / Q(x) can be split up into "partial fractions." These are components of the, like:

2/(x^2-1) = 1/(x-1) - 1/(x+1)

In general, you take the denominator Q(x) and factor it into linear and quadratic factors. Every linear factor will correspond to a term A/(x-r), while quadratic factors will correspond to (Ax+B)/(x^2+px+q).

So for (a), you have the factors, and you know:

3(2x^2-8x-1)/(x+4)(x+1)(2x-1) = A/(x+4) + B/(x+1) + C/(2x-1)

In order to solve this, you can just plug in some "good" values of x. In particular, you should clear out the denominators by multiplying both sides by Q(x).

3(2x^2-8x-1) = A(x+1)(2x-1) + B(x+4)(2x-1) + C(x+4)(x+1)

then plug in x = -1, -4, and 1/2. That gives you zeroes for most of your terms:

x = -1 gives you: 27 = A*0 + B*3*(-3) + C*0
x = -4 gives you: 189 = A*(-3)*(-9) + B*0 + C*0
x = 1/2 gives you: -27/2 = A*0 + B*0 + C*(9/2)*(3/2)

These all instantly give you:

B = -3
A = 7
C = -2

And so you conclude:

3(2x^2-8x-1)/(x+4)(x+1)(2x-1) = 7/(x+4) - 3/(x+1) - 2/(2x-1)

Now, it is VERY important that in your function f(x)=P(x)/Q(x) that the degree of P(x) is LESS than Q(x). In part (b) it is not true. You must do polynomial division to first obtain:

(3x^2-8x-63)/(x^2-3x-10) = 3 + (x-33)/(x^2-3x-10)

Then you split up:

(x-33)/(x^2-3x-10) = A/(x-5) + B/(x+2)

I'll leave the rest to you, since this looks like homework.

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