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Chapter 5: Problem 31

Write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}+6 x+3}{(x+1)\left(x^{2}+2 x+2\right)}$$

Short Answer

Expert verified
The partial fraction decomposition of the given expression \( \frac{5x^2 + 6x + 3}{(x+1)(x^2 + 2x + 2)} \) is obtained by finding constants A, B, and C from a set of linear equations. The resulting decomposition can then be written in the form \( \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2x + 2} \) .

Step by step solution

01

Set Up the Decomposition

The first step in Partial Fraction Decomposition is to set up the expression according to the factors in the denominator. For every linear factor \( (x + a) \), the form will be \( \frac{A}{x + a} \), and for every quadratic factor \( (x^2 + bx + c) \), the form will be \( \frac{Ax + B}{x^2 + bx + c} \). Applying these rules, the expression can be set up as: \[\frac{5x^2 + 6x + 3}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 2x + 2}\]
02

Equate Coefficients

The next step is to multiply through by the denominator on left side to get a polynomial equation. This will give:\(5x^2 + 6x + 3 = A(x^2 + 2x + 2) + (Bx + C)(x + 1)\)Now, by equating coefficients of like terms on both sides, we can establish three distinct equations to find the values of A, B, and C.
03

Solve for Constants

Expanding and equating the coefficients, we get:For \(x^2\): \(5 = A + B\)For \(x\): \(6 = 2A + B + C\)And for constants: \(3 = 2A + C\)Solve these equations to find the values of A, B, and C.
04

Substitute the Values back

After finding the values of A, B, and C, substitute them back into the partial fraction set up in Step 1 to achieve the final decomposition.

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Most popular questions from this chapter

In Exercises \(31-42,\) solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution sets. \(x=9-2 y\) \(x+2 y=13\)

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