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Magazine Chapter 7: Problem 32
Find the decomposition of the partial fraction for the irreducible non repeating quadratic factor. \(\frac{4 x^{2}+9 x+23}{(x-1)\left(x^{2}+6 x+11\right)}\)
Short Answer
Expert verified
Fraction decomposed as \(\frac{2}{x-1} + \frac{2x-1}{x^{2}+6x+11}\).
Step by step solution
01
Understand the Problem
We need to decompose the given fraction \(\frac{4x^{2}+9x+23}{(x-1)(x^{2}+6x+11)}\). The denominator is a product of a linear factor \((x-1)\) and an irreducible quadratic factor \((x^{2} + 6x + 11)\), which are not repeating.
02
Set Up the Partial Fraction Decomposition
The partial fraction decomposition will have the form: \(\frac{A}{x-1} + \frac{Bx+C}{x^{2}+6x+11}\), where A, B, and C are constants to be determined.
03
Clear the Denominator
Multiply both sides by the denominator \((x-1)(x^{2} + 6x + 11)\) to clear the fractions. This gives: \[4x^{2} + 9x + 23 = A(x^{2} + 6x + 11) + (Bx + C)(x-1)\]
04
Expand and Combine Terms
Expand both sides:1. \(A(x^2 + 6x + 11) = Ax^2 + 6Ax + 11A\)2. \((Bx + C)(x-1) = Bx^2 + (C-B)x - C\)Combine the expressions: \[4x^2 + 9x + 23 = Ax^2 + 6Ax + 11A + Bx^2 + (C-B)x - C\]
05
Match Coefficients
Equate the coefficients of corresponding powers of \(x\) on both sides:1. \(A + B = 4\) (for \(x^2\))2. \(6A + C - B = 9\) (for \(x\))3. \(11A - C = 23\) (constant term).
06
Solve the System of Equations
Solve the system of equations derived from the coefficients:1. \(A + B = 4\)2. \(6A + C - B = 9\)3. \(11A - C = 23\)By substituting and solving, we find:- From Equation 3, solve for \(C\): \(C = 11A - 23\).- Substitute \(C\) in Equation 2: \(6A + (11A - 23) - B = 9\).- Simplify and solve for \(B\): \(17A - 23 - B = 9\), leading to \(B = 17A - 32\).- Substitute \(B\) in Equation 1: \(A + (17A - 32) = 4\), solve for \(A\): \(18A = 36\) gives \(A = 2\).- Substitute \(A = 2\) back to find \(B\) and \(C\): \(B = 17(2) - 32 = 2\), \(C = 11(2) - 23 = -1\).
07
Write the Final Solution
Plug the solved values of \(A\), \(B\), and \(C\) back into the partial fraction form:\[\frac{2}{x-1} + \frac{2x-1}{x^{2}+6x+11}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Irreducible Quadratic Factor
An irreducible quadratic factor is a quadratic polynomial that cannot be factored further over the real numbers. For example, in the fraction we are decomposing, the term \(x^2 + 6x + 11\) is irreducible.
This means there are no real numbers \(a\) and \(b\) such that \(x^2 + 6x + 11\) can be factored as \((x+a)(x+b)\). The irreducibility is confirmed by calculating the discriminant \(b^2 - 4ac\). If the result is negative, the quadratic has no real roots and is thus irreducible.
Knowing this helps in decomposing fractions since you can determine the form of the partial fractions based on the type of factors present in the denominator.
This means there are no real numbers \(a\) and \(b\) such that \(x^2 + 6x + 11\) can be factored as \((x+a)(x+b)\). The irreducibility is confirmed by calculating the discriminant \(b^2 - 4ac\). If the result is negative, the quadratic has no real roots and is thus irreducible.
Knowing this helps in decomposing fractions since you can determine the form of the partial fractions based on the type of factors present in the denominator.
Linear Factor
A linear factor is a simple polynomial of the form \(x + k\) or \(x - k\), where \(k\) is a constant. These factors exhibit the simplest behavior and are straightforward to decompose in a partial fraction.
In our given problem, \(x-1\) is the linear factor. This is important because it affects how we set up our partial fraction, usually contributing a term like \(\frac{A}{x-1}\) in the decomposition.
Linear factors are often easier to recognize and handle within polynomial expressions and play a critical role when simplifying complex fractions using partial fraction decomposition.
In our given problem, \(x-1\) is the linear factor. This is important because it affects how we set up our partial fraction, usually contributing a term like \(\frac{A}{x-1}\) in the decomposition.
Linear factors are often easier to recognize and handle within polynomial expressions and play a critical role when simplifying complex fractions using partial fraction decomposition.
Partial Fractions
Partial fraction decomposition is a technique to express a rational function as a sum of simpler fractions.
In our exercise, the fraction \(\frac{4x^2+9x+23}{(x-1)(x^2+6x+11)}\) is expressed as partial fractions of the form:
Being able to decompose complex polynomials into partial fractions is a foundational skill in mathematics, assisting in simplifying computations and enhancing problem-solving techniques.
In our exercise, the fraction \(\frac{4x^2+9x+23}{(x-1)(x^2+6x+11)}\) is expressed as partial fractions of the form:
- \(\frac{A}{x-1}\)
- \(\frac{Bx+C}{x^2+6x+11}\)
Being able to decompose complex polynomials into partial fractions is a foundational skill in mathematics, assisting in simplifying computations and enhancing problem-solving techniques.
Polynomial Equation System
To find the constants \(A\), \(B\), and \(C\) in the partial fraction decomposition, we need to solve a system of polynomial equations.
This involves equating the coefficients of corresponding powers of \(x\) and constant terms from both the expanded form of the equation and the left-hand side polynomial. In our problem, the system is:
Working through a system of equations essentially shows how each factor contributes to creating the original polynomial, and mastering this skill can significantly aid in understanding complex polynomial relationships.
This involves equating the coefficients of corresponding powers of \(x\) and constant terms from both the expanded form of the equation and the left-hand side polynomial. In our problem, the system is:
- \(A + B = 4\)
- \(6A + C - B = 9\)
- \(11A - C = 23\)
Working through a system of equations essentially shows how each factor contributes to creating the original polynomial, and mastering this skill can significantly aid in understanding complex polynomial relationships.
