91Ó°ÊÓ

In mathematics, polynomial factorization is a process where a polynomial, which is an expression consisting of variables and coefficients, is expressed as a product of its factors. These factors are simpler polynomials or numbers, and they multiply together to give the original polynomial. When we factor polynomials, we’re essentially breaking down a complex polynomial into simpler pieces that we can work with more easily. Consider the polynomial given in the original exercise: the denominator \(2x^2 - x - 1\). We aim to find two binomials that can be multiplied to give us this original expression.

• First, identify two numbers that multiply to the product of the leading coefficient (2) and the constant term (-1), which is -2.
• The numbers must also add up to the middle coefficient (-1). For \(2x^2 - x - 1\), the numbers \(-2\) and \(1\) work.

Using these numbers, the factorization results in \((2x + 1)(x - 1)\), allowing us to decompose the rational expression into simpler fractions.

A rational function is a type of function represented as a ratio of two polynomials. In simpler terms, you have a fraction where the top and bottom are both polynomials. These functions are interesting because they can exhibit a variety of behaviors, such as asymptotes, holes, and intercepts.
In our specific exercise, the rational function is \(\frac{4x^3 + 4x^2 - 4x + 2}{2x^2 - x - 1}\). The goal of partial fraction decomposition is to take this complex rational function and rewrite it as a sum of simpler fractions.

• The denominator \(2x^2 - x - 1\) is already factored, allowing us to write the rational expression as \(\frac{A}{2x+1} + \frac{B}{x-1}\).
• Here, \(A\) and \(B\) are constants that we need to determine by solving a system of equations.

Simplifying a rational function into partial fractions helps in performing operations such as integration and gives insight into the behavior of the function.

When dealing with partial fractions, a system of equations arises naturally during the process of finding the unknown constants in the decomposition. In the exercise, once we write the partial fractions form \(\frac{A}{2x+1} + \frac{B}{x-1}\), we proceed to clear the denominators by multiplying through by \((2x+1)(x-1)\). This results in an equation relating the coefficients of like terms from both sides.

• From the expanded form \((A + 2B)x + (B - A)\), we set up two equations based on the coefficients of \(x\) and the constant term.
• The system of equations \(A + 2B = 4\) and \(B - A = 2\) is derived.

Solving these equations allows us to deduce \(A = 0\) and \(B = 2\). In this system, we often use basic algebraic methods like substitution or elimination to find the values of the unknown constants efficiently.