Problem 2
Question
Write the form of the partial fraction decomposition of the function (as in Example 4). Do not determine the numerical values of the coefficients. \(\frac{x}{x^{2}+3 x-4}\)
Step-by-Step Solution
Verified Answer
Decomposition: \(\frac{A}{x+4} + \frac{B}{x-1}\).
1Step 1: Identify the Type of Denominator
The denominator of the given function is \(x^2 + 3x - 4\). It is a quadratic polynomial, since it is a polynomial of degree 2. For partial fraction decomposition, we first need to identify and factor the denominator if possible.
2Step 2: Factor the Denominator
The quadratic \(x^2 + 3x - 4\) can be factored into two linear factors. By solving \(x^2 + 3x - 4 = 0\), we find the roots by using the quadratic formula or factoring directly. Here, it can be factored as \((x + 4)(x - 1)\).
3Step 3: Set Up the Partial Fraction Decomposition
Since \(x^2 + 3x - 4\) factors into two distinct linear factors \((x + 4)\) and \((x - 1)\), the partial fraction decomposition will take the form: \[ \frac{x}{x^2 + 3x - 4} = \frac{A}{x+4} + \frac{B}{x-1} \] where \(A\) and \(B\) are constants that need to be determined.
Key Concepts
Quadratic PolynomialFactoringLinear Factors
Quadratic Polynomial
A quadratic polynomial is a type of polynomial with a degree of 2. This means its highest power of the variable, usually denoted by \( x \), is 2. The general form of a quadratic polynomial is \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Quadratic polynomials are important in many mathematical and real-world scenarios because they can describe parabolic shapes in graphing. Additionally, they are key components in algebra and calculus, particularly in solving equations and analyzing functions.
- The term "quadratic" comes from "quad," meaning square, hinting at the degree of the polynomial.
- Every quadratic polynomial graphically represents a parabola.
Factoring
Factoring is the process of breaking down a polynomial into simpler components called factors. For a quadratic polynomial, this means expressing it as a product of two binomials, each of which is zero at the polynomial's roots. Factoring is a powerful algebraic tool because it simplifies complex expressions and helps solve equations.In our given exercise, the quadratic polynomial \( x^2 + 3x - 4 \) is factored as \((x + 4)(x - 1)\). This was achieved by solving the equation \( x^2 + 3x - 4 = 0 \) to find the roots, \( x = -4 \) and \( x = 1 \). These roots are used to rewrite the polynomial as a product of binomials:
- For quadratic polynomials, the factorization depends on the specific roots derived from setting the polynomial equal to zero.
- This method streamlines solving equations because it breaks the problem into manageable parts.
Linear Factors
Linear factors are the simplest form of polynomial factors. In the context of quadratic polynomials, once the expression has been factored, each binomial is considered a linear factor. Linear factors are expressions of the form \( ax + b \), where \( a \) and \( b \) are constants.In the partial fraction decomposition, linear factors are key because they help to decompose the original function into a sum of fractions, making complex integrals or algebraic processes simpler. For the expression \( \frac{x}{x^2 + 3x - 4} \),
- The linear factors are \( (x + 4) \) and \( (x - 1) \).
- By expressing the function as a sum, \( \frac{A}{x+4} + \frac{B}{x-1} \), the problem is simplified to solving for the constants \( A \) and \( B \).
Other exercises in this chapter
Problem 1
Use the substitution method to find all solutions of the system of equations. \(\left\\{\begin{aligned} x-y &=2 \\ 2 x+3 y &=9 \end{aligned}\right.\)
View solution Problem 2
1–14 Graph the inequality. $$y \geq-2$$
View solution Problem 2
\(1-8\) Find the determinant of the matrix, if it exists. $$ \left[\begin{array}{rr}{0} & {-1} \\ {2} & {0}\end{array}\right] $$
View solution Problem 2
Calculate the products \(A B\) and \(B A\) to verify that \(B\) is the inverse of \(A .\) \(A=\left[\begin{array}{ll}{2} & {-3} \\ {4} & {-7}\end{array}\right],
View solution