Find the partial fraction decomposition for each rational expression. See Examples 1–4. 4/(x(1 - x))

Verified step by step guidance

Identify the denominator and factor it completely. Here, the denominator is $x(1 - x)$, which is already factored into linear factors $x$ and $(1 - x)$.

Set up the partial fraction decomposition form. Since both factors are linear and distinct, write the expression as $\frac{4}{x(1 - x)} = \frac{A}{x} + \frac{B}{1 - x}$, where $A$ and $B$ are constants to be determined.

Multiply both sides of the equation by the common denominator $x(1 - x)$ to clear the fractions: \$4 = A(1 - x) + Bx$.

Expand the right side: \$4 = A - Ax + Bx$. Group like terms to get $4 = A + (B - A)x$.

Equate the coefficients of corresponding powers of $x$ on both sides. For the constant term: $4 = A$. For the coefficient of $x$: $0 = B - A$. Use these equations to solve for $A$ and $B$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as a sum of simpler fractions. This technique is especially useful for integrating rational functions or solving equations. It involves breaking down a fraction into parts with simpler denominators, typically linear or quadratic factors.

Factoring the Denominator

Factoring the denominator is essential to identify the simpler components for decomposition. In the given expression, the denominator x(1 - x) is already factored into linear terms. Recognizing these factors helps set up the correct form of partial fractions, where each factor corresponds to a separate term in the decomposition.

Setting Up and Solving Equations for Coefficients

After expressing the rational function as a sum of partial fractions with unknown coefficients, you multiply both sides by the common denominator to clear fractions. Then, equate coefficients of corresponding powers of x or substitute convenient values to form a system of equations. Solving this system yields the values of the unknown coefficients.

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