Question

Resolve \(\frac{3 x+5}{x^{2}+8 x-20}\) into partial fractions. (1) \(\frac{11}{24(x-2)}+\frac{25}{24(x+10)}\) (2) \(\frac{11}{6(x-2)}+\frac{25}{6(x+10)}\) (3) \(\frac{11}{12(x+2)}+\frac{25}{12(x-10)}\) (4) \(\frac{11}{12(\mathrm{x}-2)}+\frac{25}{12(\mathrm{x}+10)}\)

Short answer

Answer: $\frac{11}{12(x-2)}-\frac{25}{12(x+10)}$.

Step-by-step solution

Factor the denominator

First, we need to factor the denominator, \(x^2 + 8x - 20\). This can be factored as \((x - 2)(x + 10)\).

Determine the form of the partial fractions

Since the factors of the denominator are linear and distinct, the partial fraction decomposition will have the form: $$\frac{A}{x-2}+\frac{B}{x+10}$$ Our goal is to determine the constants A and B.

Set up an equation to find the unknown coefficients

To find the unknown coefficients A and B, we'll add the partial fractions to get: $$\frac{3x+5}{(x - 2)(x + 10)} = \frac{A}{x-2}+\frac{B}{x+10}$$ Now, clear the denominators by multiplying each side of the equation by \((x - 2)(x + 10)\): $$(x - 2)(x + 10) \cdot \frac{3x+5}{(x - 2)(x + 10)} = (x - 2)(x + 10) \cdot \left(\frac{A}{x-2}+\frac{B}{x+10}\right)$$ Since the denominators cancel out, the equation becomes: $$3x+5=A(x+10)+B(x-2)$$

Solve for the unknown coefficients A and B

We evaluate the left side of the equation in two scenarios: When \(x = 2\), \(A(2 + 10) + B(2 - 2) = 3\cdot2 + 5 \Rightarrow 12A = 11 \Rightarrow A = \frac{11}{12}\). When \(x = -10\), \(A(-10+10) + B(-10-2) = 3\cdot(-10) + 5 \Rightarrow -12B = 25 \Rightarrow B = \frac{-25}{12}\). Now we substitute the found values of A and B in the partial fractions: $$\frac{3x+5}{(x - 2)(x + 10)}=\frac{\frac{11}{12}}{(x-2)}+\frac{\frac{-25}{12}}{(x+10)}$$ Finally, we can simplify the partial fractions: $$\frac{11}{12(x-2)}-\frac{25}{12(x+10)}$$ So the correct answer is (4) \(\frac{11}{12(\mathrm{x}-2)}-\frac{25}{12(\mathrm{x}+10)}\).

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 numbers.

When dealing with algebraic expressions like \(\frac{3x+5}{x^2+8x-20}\), the goal is often to simplify or to solve for an unknown quantity. In our exercise, the focus is on simplification using a method called partial fraction decomposition, which is particularly useful when integrating rational expressions in calculus.

Understanding algebra is crucial for mastering this technique, as it requires identifying the structure of algebraic expressions, factoring polynomials, and solving linear equations—all fundamental skills in algebra that allow us to break down complex expressions into simpler parts.

Mathematical Factorization

Mathematical factorization is the process of breaking down a complex expression into a product of simpler factors. In the context of partial fraction decomposition, factorization is the first step and is essential in simplifying rational expressions.

In the given exercise, the denominator \(x^2+8x-20\) is factored into \(x-2)(x+10)\), revealing the roots of the polynomial: x=2 and x=-10. Recognizing these roots helps us to construct the form \(\frac{A}{x-2}+\frac{B}{x+10}\), where A and B are constants to be determined. Once the polynomial is factored, solving for these constants becomes straightforward, allowing for the expression to be decomposed into partial fractions that are easier to work with, particularly for integration and finding limits.

Rational Expressions

Rational expressions are fractions in which both the numerator and the denominator are polynomials. The expression \(\frac{3x+5}{x^2+8x-20}\) is an example of a rational expression.

Simplifying rational expressions, like the one in our exercise, often involves partial fraction decomposition, which is the process of breaking a complex rational expression into a sum of simpler fractions. The method requires that the denominator be factored first. Once this is done, we set up partial fractions where each fraction contains a part of the factored denominator with an undetermined numerator, which we solve for. This technique simplifies integration and differentiation of rational expressions in calculus, and understanding it requires a solid grasp of algebraic operations involving polynomials.