Resolve into partial fractions the expression `((1+x)(1+2x))/((1-x)(1-2x))`.

Answer

To resolve the expression \(\frac{(1+x)(1+2x)}{(1-x)(1-2x)}\) into partial fractions, we will follow these steps:

### Step 1: Set Up the Partial Fraction Decomposition
We start by expressing the given fraction as a sum of partial fractions:
\[ \frac{(1+x)(1+2x)}{(1-x)(1-2x)} = \frac{A}{1-x} + \frac{B}{1-2x} \]
where \(A\) and \(B\) are constants that we need to determine.

### Step 2: Clear the Denominator
To eliminate the denominators, multiply both sides by \((1-x)(1-2x)\):
\[ (1+x)(1+2x) = A(1-2x) + B(1-x) \]

### Step 3: Expand Both Sides
Now, we will expand both sides of the equation:
- Left-hand side:
\[ (1+x)(1+2x) = 1 + 2x + x + 2x^2 = 1 + 3x + 2x^2 \]
- Right-hand side:
\[ A(1-2x) + B(1-x) = A - 2Ax + B - Bx = (A + B) + (-2A - B)x \]

### Step 4: Equate Coefficients
Now, we equate the coefficients of the corresponding powers of \(x\) from both sides:
1. Constant term: \(A + B = 1\)
2. Coefficient of \(x\): \(-2A - B = 3\)
3. Coefficient of \(x^2\): There is no \(x^2\) term on the right side, so we can ignore this for now.

### Step 5: Solve the System of Equations
We have the following system of equations:
1. \(A + B = 1\) (Equation 1)
2. \(-2A - B = 3\) (Equation 2)

From Equation 1, we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]

Substituting \(B\) into Equation 2:
\[ -2A - (1 - A) = 3 \]
Simplifying this gives:
\[ -2A - 1 + A = 3 \implies -A - 1 = 3 \implies -A = 4 \implies A = -4 \]

Now substituting \(A\) back into Equation 1 to find \(B\):
\[ -4 + B = 1 \implies B = 5 \]

### Step 6: Write the Partial Fraction Decomposition
Now we can write the partial fraction decomposition:
\[ \frac{(1+x)(1+2x)}{(1-x)(1-2x)} = \frac{-4}{1-x} + \frac{5}{1-2x} \]

### Final Answer
Thus, the resolved expression into partial fractions is:
\[ \frac{-4}{1-x} + \frac{5}{1-2x} \]

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