Question

How do you write the partial fraction decomposition of the rational expression `(7x-2)/((x-3)^2(x+1))`?

Answer 1

The answer is `=(-9/64)/(x-3)+(9/16)/(x-3)^2+(19/4)/(x-3)^3+(9/64)/(x+1)`

Explanation:

Let's perform the decomposition into partial fractions

`(7x-2)/((x-3)^2(x+1))=A/(x-3)+B/(x-3)^2+C/(x-3)^3+D/(x+1)`

`=(A(x-3)^2(x+1)+B(x-3)(x+1)+C(x+1)+D(x-3)^3)/((x-3)^2(x+1))`

The denominators are the same, we compare the numerators

`(7x-2)=A(x-3)^2(x+1)+B(x-3)(x+1)+C(x+1)+D(x-3)^3`

Let `x=3`, `=>`, `19=4C`, `=>`, `C=19/4`

Let `x=-1`, `=>`, `-9=-64D`, `=>`, `D=9/64`

Coefficients of `x^3`,

`0=A+D`, `=>`, `A=-D=-9/64`

Coefficients of `x^2`

`0=-5A+B-9D`

`=>`, `B=5A+9D=-45/64+81/64=36/64=9/16`

Therefore,

`(7x-2)/((x-3)^2(x+1))=(-9/64)/(x-3)+(9/16)/(x-3)^2+(19/4)/(x-3)^3+(9/64)/(x+1)`

Answer 2

`(7x-2)/((x-3)^2(x+1)) = color(blue)(9/(16(x-3)) + 19/(4(x-3)^2) - 9/(16(x+1))`

Explanation:

We recognize that there will be three different fractions with denominators of

• `x-3`

• `(x-3)^2`

• `x+1`

So we set it up as

`(7x-2)/((x-3)^2(x+1)) = A/(x-3) + B/((x-3)^2) + C/(x+1)`

Multiply both sides by the quantity `(x-3)^2(x+1)`:

`7x-2 = A(x-3)(x+1) + B(x+1) + C(x-3)^2`

Expand the terms:

`7x-2 = A(x^2-2x+3) + B(x+1) + C(x^2-6x+9)`

Collect the terms by power of `x`:

`7x-2 = x^2(A+C) + x(-2A + B - 6C) + -3A + B + 9C`

Equate the coefficients on both sides (there is no `x^2` term on left hand side, so that we set equal to `0`:

`0 = A+C`

`7 = -2A + B - 6C`

`-2 = -3A + B + 9C`

Now we set up an augmented matrix (sorry for the poor quality):

`((1color(white)(aaa),0color(white)(aaa),1color(white)(aaa)|-2),(-2color(white)(aaa),1,-6color(white)()|7), (-3color(white)(aaa),1color(white)(aaa),9color(white)(aaa)|-2))`

In order to avoid mayhem, I won't work out the solving of the matrix (maybe you already know how, if not there's plenty of other sources, maybe some of Socratic!), but the solutions are

`A = 9/16`

`B = 19/4`

`C = -9/16`

Plugging these into the original equation, we have

`(7x-2)/((x-3)^2(x+1)) = color(blue)(ul(9/(16(x-3)) + 19/(4(x-3)^2) - 9/(16(x+1))`