Solve to partial fraction

(3x²-1)/(x-2)²(x²+2x+2)

We are tasked with decomposing the expression \(\frac{3x^2 - 1}{(x - 2)^2(x^2 + 2x + 2)}\) into partial fractions.

### Step 1: Set up the partial fraction decomposition

The general form for the partial fraction decomposition of \(\frac{3x^2 - 1}{(x - 2)^2(x^2 + 2x + 2)}\) is:

\[
\frac{3x^2 - 1}{(x - 2)^2(x^2 + 2x + 2)} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} + \frac{Cx + D}{x^2 + 2x + 2}
\]

### Step 2: Multiply both sides by the denominator

To eliminate the denominator, multiply both sides by \((x - 2)^2(x^2 + 2x + 2)\):

\[
3x^2 - 1 = A(x - 2)(x^2 + 2x + 2) + B(x^2 + 2x + 2) + (Cx + D)(x - 2)^2
\]

### Step 3: Expand the right-hand side

Now we expand each term on the right-hand side.

- For \(A(x - 2)(x^2 + 2x + 2)\):
  \[
  A(x - 2)(x^2 + 2x + 2) = A[(x^3 + 2x^2 + 2x) - (2x^2 + 4x + 4)] = A(x^3 + 0x^2 - 2x - 4)
  \]
- For \(B(x^2 + 2x + 2)\):
  \[
  B(x^2 + 2x + 2) = Bx^2 + 2Bx + 2B
  \]
- For \((Cx + D)(x - 2)^2\), first expand \((x - 2)^2 = x^2 - 4x + 4\), then multiply by \(Cx + D\):
  \[
  (Cx + D)(x^2 - 4x + 4) = Cx(x^2 - 4x + 4) + D(x^2 - 4x + 4)
  \]
  \[
  = C(x^3 - 4x^2 + 4x) + D(x^2 - 4x + 4)
  \]
  \[
  = Cx^3 - 4Cx^2 + 4Cx + Dx^2 - 4Dx + 4D
  \]

### Step 4: Combine all terms

Now, combine all the expanded terms:

\[
3x^2 - 1 = A(x^3 + 0x^2 - 2x - 4) + (Bx^2 + 2Bx + 2B) + (Cx^3 - 4Cx^2 + 4Cx + Dx^2 - 4Dx + 4D)
\]
Group powers of \(x\) on the right-hand side:

\[
= (A + C)x^3 + (-4C + D + B)x^2 + (4C - 4D + 2B - 2A)x + (-4A + 4D + 2B)
\]

### Step 5: Equate coefficients

Now, equate the coefficients of corresponding powers of \(x\) on both sides:

- For \(x^3\): \(A + C = 0\)
- For \(x^2\): \(-4C + D + B = 3\)
- For \(x^1\): \(4C - 4D + 2B - 2A = 0\)
- For the constant term: \(-4A + 4D + 2B = -1\)

### Step 6: Solve the system of equations

From \(A + C = 0\), we get \(A = -C\). Now substitute \(A = -C\) into the other equations:

1. \(-4C + D + B = 3\)
2. \(4C - 4D + 2B - 2(-C) = 0\)
3. \(-4(-C) + 4D + 2B = -1\)

Simplifying these:

1. \(-4C + D + B = 3\)
2. \(6C - 4D + 2B = 0\)
3. \(4C + 4D + 2B = -1\)

This system of equations can now be solved for \(C\), \(D\), \(B\), and \(A\).

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### Related Questions:
1. How are partial fractions used in calculus, especially in integration?
2. What is the role of complex roots in partial fractions?
3. Can partial fraction decomposition be applied to higher degree polynomials?
4. What happens if the degree of the numerator is higher than the denominator in partial fractions?
5. How do you handle irreducible quadratic terms in partial fraction decomposition?

### Tip:
In partial fraction decomposition, always check if the degree of the numerator is less than the denominator before setting up the decomposition.

Solve to partial fraction (3x²-1)/(x-2)²(x²+2x+2)

Decompose the expression (3x²-1)/(x-2)²(x²+2x+2) into partial fractions with a detailed, step-by-step solution. This problem covers advanced algebraic concepts like partial fraction decomposition and the method of equating coefficients, suitable for Grades 11-12 students.

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