Partial fraction decomposition exercise 2

[ducmod]

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!

 

[ehild]

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!

 

[ducmod]

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]

-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 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 \\<br /> 4B+C = 20 \\<br /> 4A+4C=-68<br />
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.