# Partial fraction decomposition exercise 2

## Homework Statement

Hello!
Here is my second post on the subject partial fraction decomposition. The subject looks pretty easy to learn, but when I try exercises, I do not get to the correct answer. Please, take a look at the exercise below and help me to see my mistakes.

## Homework Equations

(-2x^2 + 20x - 68) / (x^3 + 4x^2 + 4x + 16)

## The Attempt at a Solution

Step 1: create a form for partial fraction decomposition by factoring the denominator:
x^3 + 4x^2 + 4x + 16 = (x^2 + 4) ( x + 4)
x^2 + 4 is in irreducible quadratic form, thus I will work with the above factors
Step 2: clear denominators
(-2x^2 + 20x - 68) (x^2 + 4) ( x + 4) / (x^3 + 4x^2 + 4x + 16) = A (x^2 + 4) ( x + 4) / (x + 4) + (Bx + C) (x^2 + 4) ( x + 4) / (x^2 + 4)
-2x^2 + 20x - 68 = x^2 (A + B) + x ( C + 4B) + 4C + 4A
Step 3: find values of A, B, C
A + C = -2
C + 4B = 20
4C + 4A = -68 (A + C) = -17
Matrix A
1 0 1
0 1 4
1 0 1
Matrix B
-2
20
-17
And the determinant of matrix A is zero, hence I can't solve the task using Cramer's rule. Obviously, there are mistakes in my approach. What are they? Please, help me to see them.
Thank you!

Last edited:

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ehild
Homework Helper

## Homework Statement

Hello!
Here is my second post on the subject partial fraction decomposition. The subject looks pretty easy to learn, but when I try exercises, I do not get to the correct answer. Please, take a look at the exercise below and help me to see my mistakes.

## Homework Equations

(-2x^2 + 20x - 68) / (x^3 + 4x^2 + 4x - 68)

## The Attempt at a Solution

Step 1: create a form for partial fraction decomposition by factoring the denominator:
x^3 + 4x^2 + 4x - 68 = (x^2 + 4) ( x + 4)
The two sides are not equal. Check if you copied the problem correctly.
x^2 + 4 is in irreducible quadratic form, thus I will work with the above factors
Step 2: clear denominators
(-2x^2 + 20x - 68) (x^2 + 4) ( x + 4) / (x^3 + 4x^2 + 4x - 68) = A (x^2 + 4) ( x + 4) / (x + 4) + (Bx + C) (x^2 + 4) ( x + 4) / (x^2 + 4)
-2x^2 + 20x - 68 = x^2 (A + B) + x ( C + 4B) + 4C + 4A
Step 3: find values of A, B, C
A + C = -2
C + 4B = 20
4C + 4A = -68 (A + C) = -17
Matrix A
1 0 1
0 1 4
1 0 1
Matrix B
-2
20
-17
And the determinant of matrix A is zero, hence I can't solve the task using Cramer's rule. Obviously, there are mistakes in my approach. What are they? Please, help me to see them.
Thank you!

The two sides are not equal. Check if you copied the problem correctly.
sorry, it's a typo; +16 instead of -68 in the denominator, but all calculations are based on +16 value

ehild
Homework Helper
-2x^2 + 20x - 68 = x^2 (A + B) + x ( C + 4B) + 4C + 4A
Step 3: find values of A, B, C
A + C B= -2
C + 4B = 20
4C + 4A = -68 (A + C) = -17
There is a mistake in your equations. The first one is A+B=-2.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Hello!
Here is my second post on the subject partial fraction decomposition. The subject looks pretty easy to learn, but when I try exercises, I do not get to the correct answer. Please, take a look at the exercise below and help me to see my mistakes.

## Homework Equations

(-2x^2 + 20x - 68) / (x^3 + 4x^2 + 4x + 16)

## The Attempt at a Solution

Step 1: create a form for partial fraction decomposition by factoring the denominator:
x^3 + 4x^2 + 4x + 16 = (x^2 + 4) ( x + 4)
x^2 + 4 is in irreducible quadratic form, thus I will work with the above factors
Step 2: clear denominators
(-2x^2 + 20x - 68) (x^2 + 4) ( x + 4) / (x^3 + 4x^2 + 4x + 16) = A (x^2 + 4) ( x + 4) / (x + 4) + (Bx + C) (x^2 + 4) ( x + 4) / (x^2 + 4)
-2x^2 + 20x - 68 = x^2 (A + B) + x ( C + 4B) + 4C + 4A
Step 3: find values of A, B, C
A + C = -2
C + 4B = 20
4C + 4A = -68 (A + C) = -17
Matrix A
1 0 1
0 1 4
1 0 1
Matrix B
-2
20
-17
And the determinant of matrix A is zero, hence I can't solve the task using Cramer's rule. Obviously, there are mistakes in my approach. What are they? Please, help me to see them.
Thank you!
$$\frac{A}{4+x} + \frac{Bx+C}{x^2+4} = \frac{(A+B)x^2 + (4B+C)x +(4A+4C)}{(x+4)(x^2+4)}$$
Thus, the equations are
$$A+B = -2 \\ 4B+C = 20 \\ 4A+4C=-68$$
Why bother with matrices? It is much easier just to use elementary elimination: the second equation gives $C = 20-4B$, and putting that into the other two equations gives two linear equations in $A$ and $B$ alone. Solving 2x2 linear systems is pretty easy.