$$ \begin{array}{l} ext { vercise } \ ext { actions with repeated linear factors }\end{array} $$ $$ \frac{4 x-3}{(x+1)^{2}} $$

Question

$$ \begin{array}{l}	ext { vercise } \ 	ext { actions with repeated linear factors }\end{array} $$ $$ \frac{4 x-3}{(x+1)^{2}} $$

UpStudy AI · Accepted Answer

We wish to decompose
$$
\frac{4x-3}{(x+1)^{2}}
$$
into partial fractions. Since the denominator has a repeated linear factor, we write:
$$
\frac{4x-3}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}.
$$

**Step 1. Multiply both sides by $$(x+1)^2$$:**
$$
4x-3 = A(x+1) + B.
$$

**Step 2. Expand and collect like terms:**
$$
4x-3 = Ax + A + B.
$$

**Step 3. Equate the coefficients on both sides:**

- For the $$x$$-term:
$$
A = 4.
$$

- For the constant term:
$$
A+B = -3.
$$
Substitute $$A = 4$$:
$$
4+B = -3 \quad \Longrightarrow \quad B = -7.
$$

**Step 4. Write the final partial fraction decomposition:**
$$
\frac{4x-3}{(x+1)^2} = \frac{4}{x+1} - \frac{7}{(x+1)^2}.
$$
$$ \frac{4x-3}{(x+1)^2} = \frac{4}{x+1} - \frac{7}{(x+1)^2} $$

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