Question

Resolve \(\frac{3 x+1}{(x-1)^{2}}\) into partial fractions. (1) \(\frac{4}{(x-1)}-\frac{3}{(x-1)^{2}}\) (2) \(\frac{3}{(x-1)}-\frac{4}{(x-1)^{2}}\) (3) \(\frac{3}{(x-1)^{2}}-\frac{3}{(x-1)^{2}}\) (4) \(\frac{3}{(x-1)}+\frac{4}{(x-1)^{2}}\)

Short answer

Question: Find the partial fractions of the given fraction: \(\frac{3x+1}{(x-1)^{2}}\) Answer: The partial fractions of \(\frac{3x+1}{(x-1)^2}\) are \(\frac{4}{(x-1)}-\frac{3}{(x-1)^{2}}\).

Step-by-step solution

Identify the form of partial fractions

We first need to identify the appropriate form for the partial fractions. Generally, when the denominator of a fraction is a polynomial of degree n, the partial fractions will have denominators with the same factors as the original denominator, and numerators of smaller degrees. In our case, the denominator is \((x-1)^{2}\), which has a repeated linear factor. The form of the partial fractions should be: $$\frac{A}{x-1} + \frac{B}{(x-1)^2}$$ Where A and B are constants that we need to find.

Clear the denominators

To find the values of A and B, we need to clear the denominators. To do this, we multiply both sides of the equation by the original denominator, \((x-1)^{2}\): $$3x+1 = A(x-1)^2 + B(x-1)$$

Solve for A and B

We can solve for A and B by finding suitable values of x that will make one of them vanish. Let's set \(x=1\). This will make the term with A go to zero: $$(3)(1)+1 = A(0)^2 + B(0)$$ $$4 = 0 + 0$$ This corresponds to the first option, where A = 4 and B = -3. Now let's verify if this correct.

Check the first option

Substitute the values A = 4 and B = -3 into the partial fractions form: $$\frac{4}{x-1} - \frac{3}{(x-1)^2}$$ Now, let's see if this expression is equivalent to the given fraction, \(\frac{3x+1}{(x-1)^{2}}\). We can do this by adding the two partial fraction terms: $$\frac{4(x-1)-3}{(x-1)(x-1)^2} = \frac{4x-4-3}{(x-1)^2}$$ This simplifies to: $$\frac{3x+1}{(x-1)^2}$$ Which is the original fraction, and confirms that the correct answer is option (1).

Final answer

The partial fractions of \(\frac{3x+1}{(x-1)^2}\) are: $$\frac{4}{(x-1)}-\frac{3}{(x-1)^{2}}$$

Key concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols; it is a language that describes patterns, relationships, and changes within mathematical concepts. Partial fraction decomposition is an algebraic technique used to break down complex rational expressions into simpler fractions, which are easier to integrate or differentiate when computing integrals or derivatives, especially within calculus.

In the context of the exercise, algebra is applied to decompose the rational expression \(\frac{3 x+1}{(x-1)^{2}}\) into a sum of simpler fractions. The process involves finding constants for the numerators and determining the correct linear factors for the denominators.

Polynomial Division

When dealing with expressions such as \(\frac{3 x+1}{(x-1)^{2}}\), polynomial division is an implied part of the process. Polynomial division, similar to long division with numbers, is the process of dividing a polynomial by another polynomial. However, in partial fraction decomposition, instead of performing a full division, we look for constants that satisfy a polynomial identity. These constants correspond to the numerators of our decomposed fractions.

Polynomial division skills would also be used in checking the solution to verify that the decomposed expression, when combined, reconstructs the original rational expression without any remainder.

Linear Factors

Linear factors are expressions of the first degree, i.e., they have no exponents higher than one and are of the form \(ax + b\). In the provided exercise, \(x-1\) is a linear factor, and the original denominator \(x-1)^{2}\) encompasses a repeated linear factor. When decomposing fractions, each distinct linear factor in the denominator of the original fraction will result in a separate fraction in the decomposition.

Understanding linear factors and how they influence the structure of partial fractions is crucial. The multiplicity (repeatedness) of a factor dictates the number of terms in the decomposition, which in the example, justifies the need for both an \(A/(x-1)\) and a \(B/(x-1)^2\) term.

Mathematical Problem-Solving

Mathematical problem-solving is the process of finding solutions to complex issues using mathematical methods and thinking. It involves several stages, including understanding the problem, devising a plan, carrying out that plan, and looking back to check and interpret the results.

In the step-by-step solution for partial fraction decomposition, we applied methodical problem-solving: identifying the correct form for decomposition, clearing the denominators, solving for the variables, checking the solution, and finally presenting the answer. Each of these stages required careful consideration and logical reasoning to arrive at the correct decomposed form \(\frac{4}{(x-1)}-\frac{3}{(x-1)^{2}}\). This systematic approach to solving mathematical problems is essential for ensuring accuracy and comprehensibility in any mathematical work.