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

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 start by expressing the fraction in terms of repeated linear factors. Since the denominator is $$(x+1)^2$$, we use the form

$$
\frac{4x-3}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}.
$$

Multiply both sides by $$(x+1)^2$$ to clear the denominators:

$$
4x - 3 = A(x+1) + B.
$$

Expanding the right-hand side:

$$
4x - 3 = Ax + A + B.
$$

Now, we equate the coefficients of like terms from both sides of the equation:

1. Coefficient of $$x$$:

$$
4 = A.
$$

2. Constant term:

$$
-3 = A + B.
$$

Substitute $$A = 4$$ into the constant term equation:

$$
-3 = 4 + B \quad \Rightarrow \quad B = -3 - 4 = -7.
$$

Thus, the values of the constants are $$A = 4$$ and $$B = -7$$.

Now, substitute these back into the partial fractions:

$$
\frac{4x-3}{(x+1)^2} = \frac{4}{x+1} - \frac{7}{(x+1)^2}.
$$

This is the partial-fraction decomposition of the given expression.
$$ \frac{4x-3}{(x+1)^2} = \frac{4}{x+1} - \frac{7}{(x+1)^2} $$

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