91Ó°ÊÓ

Factoring quadratics is a core skill in algebra, particularly for solving equations and simplifying expressions. The quadratic expression in this exercise is ewline \[ 2x^2 - 5x - 3 \]To factor it, identify two numbers that multiply to the product of the leading coefficient (2) and the constant term (-3), which is -6, and add to the middle coefficient (-5). These numbers are -6 and +1. ewline Next, rewrite the middle term, splitting it into two terms that use these numbers: ewline \[ 2x^2 - 6x + x - 3 &\] Recollapse by grouping to get: ewline \[ 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3) &\]Now, the quadratic is fully factored into \'2x + 1\' and \'x - 3\'. This simplifies the problem and makes further calculations easier.

A rational expression is a fraction where both the numerator and the denominator are polynomials. In this problem, the rational expression is: ewline \[ \frac{4}{2x^2 - 5x - 3} \]Rational expressions can often be simplified or decomposed into simpler parts, especially by factoring the denominator. For the given example, after factoring the denominator, it becomes: ewline \[ \frac{4}{(2x + 1)(x - 3)} \] Next, this expression can be split into partial fractions.
Partial fraction decomposition rewrites a complex fraction into a sum of simpler fractions. The goal is to express the fraction in a form like \[ \frac{A}{2x + 1} + \frac{B}{x - 3} \] which can be easier to work with in various mathematical operations.

Solving the coefficients of partial fractions often involves setting up and solving a system of equations. Here, we decompose: \[ \frac{4}{(2x + 1)(x - 3)} = \frac{A}{2x + 1} + \frac{B}{x - 3} \] Multiply through by the common denominator to clear fractions:\[ 4 = A(x - 3) + B(2x + 1) \] Expand and combine like terms: \[ 4 = Ax - 3A + 2Bx + B \] Set up the system of equations:\[ A + 2B = 0 \] And \[ -3A + B = 4 \] Now solve for A and B. From the first equation, \'A = -2B\'. Substitute into the second equation: \[ -3(-2B) + B = 4 \] Simplify and solve for B: \[6B + B = 4 \Rightarrow 7B = 4 \Rightarrow B = \frac{4}{7} \] Thus, \'A = -2(\frac{4}{7}) = -\frac{8}{7}\'. Finally, we substitute these values back into the partial fractions to get: \[ \frac{4}{2x^2 - 5x - 3} = -\frac{8}{7(2x + 1)} + \frac{4}{7(x - 3)} \]