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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!

Try to write the fraction as a sum of fractions over each of the factors of the denominator.

a)

(6x²-24x-3)/(x+4)(x+1)(2x-1) = a/(x+4)  + b/(x+1) + c(2x-1)

where a,b,c are constants we need to find .

Add the fractions on the right side, then equate numerators from both sides.

This gives

6x²-24x-3 = a(x+1)(2x-1) + b(x+4)(2x-1) + c(x+4)(x+1)

set coefficients of the same powers of x from the left and right sides equal to each other

2a+2b+c = 6

a+7b+5c = -24

-a-4b+4c = 3

Solve the system and get

a = 65/9  , b = -11/3 , c = -10/9

so

(6x²-24x-3)/(x+4)(x+1)(2x-1) = 65/9(x+4)  - 11/3(x+1) - 10/9(2x-1)

b)

You can always start by getting the degree of the numerator smaller than the degree of the denominator , if it isn't already.

(3x²-8x-63)/(x²-3x-10) = [3(x²-3x-10) + x-33]/(x²-3x-10) = 3 + (x-33)/(x²-3x-10)

So we want to write (x-33)/(x²-3x-10) as a sum of partial fractions.

Procede as in part a)

(x-33)/(x²-3x-10) = (x-33)/(x-5)(x+2) = a/(x-5) + b/(x+2)

a(x+2) + b(x-5) = x-33

a+b = 1

2a-5b = -33

a=-4 , b=5

so

(3x²-8x-63)/(x²-3x-10) = 3 - 4/(x-5) + 5/(x+2)

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