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

2 Answers
Narad T.
Jul 31, 2017

The answer is =(-9/64)/(x-3)+(9/16)/(x-3)^2+(19/4)/(x-3)^3+(9/64)/(x+1)=−964x−3+916(x−3)2+194(x−3)3+964x+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)7x−2(x−3)2(x+1)=Ax−3+B(x−3)2+C(x−3)3+Dx+1

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

Let x=3x=3, =>⇒, 19=4C19=4C, =>⇒, C=19/4C=194

Let x=-1x=−1, =>⇒, -9=-64D−9=−64D, =>⇒, D=9/64D=964

Coefficients of x^3x3,

0=A+D0=A+D, =>⇒, A=-D=-9/64A=−D=−964

Coefficients of x^2x2

0=-5A+B-9D0=−5A+B−9D

=>⇒, B=5A+9D=-45/64+81/64=36/64=9/16B=5A+9D=−4564+8164=3664=916

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)7x−2(x−3)2(x+1)=−964x−3+916(x−3)2+194(x−3)3+964x+1

Nathan L.
Jul 31, 2017

(7x-2)/((x-3)^2(x+1)) = color(blue)(9/(16(x-3)) + 19/(4(x-3)^2) - 9/(16(x+1))7x−2(x−3)2(x+1)=916(x−3)+194(x−3)2−916(x+1)

Explanation:

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

  • x-3x−3

  • (x-3)^2(x−3)2

  • x+1x+1

So we set it up as

(7x-2)/((x-3)^2(x+1)) = A/(x-3) + B/((x-3)^2) + C/(x+1)7x−2(x−3)2(x+1)=Ax−3+B(x−3)2+Cx+1

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

7x-2 = A(x-3)(x+1) + B(x+1) + C(x-3)^27x−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)7x−2=A(x2−2x+3)+B(x+1)+C(x2−6x+9)

Collect the terms by power of xx:

7x-2 = x^2(A+C) + x(-2A + B - 6C) + -3A + B + 9C7x−2=x2(A+C)+x(−2A+B−6C)+−3A+B+9C

Equate the coefficients on both sides (there is no x^2x2 term on left hand side, so that we set equal to 00:

0 = A+C0=A+C

7 = -2A + B - 6C7=−2A+B−6C

-2 = -3A + B + 9C−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))

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