Let f(x) = 3-x (1-2x)(1+x²)

a) Express f(x) in partial fractions.

b) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x².

### a) Express \( f(x) = \frac{3 - x}{(1 - 2x)(1 + x^2)} \) in partial fractions:

First, let's assume that the function can be decomposed into partial fractions of the following form:

\[
\frac{3 - x}{(1 - 2x)(1 + x^2)} = \frac{A}{1 - 2x} + \frac{Bx + C}{1 + x^2}
\]

Here, \( A \), \( B \), and \( C \) are constants that need to be determined.

Multiply both sides by the denominator \( (1 - 2x)(1 + x^2) \):

\[
3 - x = A(1 + x^2) + (Bx + C)(1 - 2x)
\]

Expand both terms:

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

Now, compare the coefficients of \( 1 \), \( x \), and \( x^2 \) on both sides:

1. Constant term: \( A + C = 3 \)
2. Coefficient of \( x \): \( B - 2C = -1 \)
3. Coefficient of \( x^2 \): \( A - 2B = 0 \)

Solve this system of equations:

- From \( A - 2B = 0 \), we get \( A = 2B \).
- Substitute \( A = 2B \) into \( A + C = 3 \):
  \[
  2B + C = 3 \quad \text{(1)}
  \]
- From \( B - 2C = -1 \), we have:
  \[
  B = 2C - 1 \quad \text{(2)}
  \]

Substitute equation (2) into equation (1):

\[
2(2C - 1) + C = 3
\]
\[
4C - 2 + C = 3
\]
\[
5C = 5
\]
\[
C = 1
\]

Now substitute \( C = 1 \) into equation (2):

\[
B = 2(1) - 1 = 1
\]

Substitute \( B = 1 \) into \( A = 2B \):

\[
A = 2(1) = 2
\]

Thus, the partial fraction decomposition is:

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

### b) Obtain the expansion of \( f(x) \) in ascending powers of \( x \):

Now, expand each term separately in ascending powers of \( x \).

1. **Expand** \( \frac{2}{1 - 2x} \) using the binomial series:
   \[
   \frac{2}{1 - 2x} = 2(1 - 2x)^{-1} = 2\left(1 + 2x + (2x)^2 + \cdots\right) = 2\left(1 + 2x + 4x^2 + \cdots\right)
   \]
   Up to the term in \( x^2 \), this becomes:
   \[
   \frac{2}{1 - 2x} = 2 + 4x + 8x^2
   \]

2. **Expand** \( \frac{x + 1}{1 + x^2} \):
   \[
   \frac{x + 1}{1 + x^2} = (x + 1)\left(1 - x^2 + x^4 - \cdots\right)
   \]
   Multiply \( (x + 1) \) by the series:
   \[
   (x + 1)(1 - x^2) = x + 1 - x^2 - x^3
   \]
   Up to the term in \( x^2 \), this becomes:
   \[
   \frac{x + 1}{1 + x^2} = x + 1 - x^2
   \]

3. **Combine the two expansions**:
   \[
   f(x) = \left(2 + 4x + 8x^2\right) + \left(x + 1 - x^2\right)
   \]
   Simplify:
   \[
   f(x) = (2 + 1) + (4x + x) + (8x^2 - x^2)
   \]
   \[
   f(x) = 3 + 5x + 7x^2
   \]

Thus, the expansion of \( f(x) \) up to the term in \( x^2 \) is:

\[
f(x) = 3 + 5x + 7x^2
\]

---

Would you like more details or clarification on any part of this solution?

### Related Questions:
1. How does the binomial expansion work for terms like \( (1 - 2x)^{-1} \)?
2. Can partial fraction decomposition always be used for rational functions?
3. How would the expansion differ if higher powers of \( x \) were included?
4. What happens to the expansion if \( f(x) \) has higher-degree polynomials in the denominator?
5. How can you check if the partial fraction decomposition is correct?

### Tip:
When expanding functions using series, it helps to know the radius of convergence to understand where the expansion is valid.

Let f(x) = (3 - x) / [(1 - 2x)(1 + x²)]

a) Express f(x) in partial fractions.

b) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x².

Learn how to express f(x) = (3 - x) / [(1 - 2x)(1 + x²)] in partial fractions and obtain its series expansion up to the x² term. This detailed guide covers key algebraic concepts like partial fraction decomposition and series expansion. Ideal for students in Grades 11-12.

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