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Chapter 5: Problem 29

Write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}-6 x+7}{(x-1)\left(x^{2}+1\right)}$$

Short Answer

Expert verified
The partial fraction decomposition of the given rational expression is \(\frac{5}{x-1} - \frac{6x+7}{x^2+1}\)

Step by step solution

01

Expressing the denominator as product of its factors

The expression has already been provided in a factorised form. Therefore, the denominator of the rational function can be written as the product of the linear factor \(x-1\) and the irreducible quadratic factor \(x^2+1\)
02

Form Partial Fraction Decomposition

The form of the partial fraction decomposition for a rational expression with non-repeating linear and quadratic factors in its denominator is \(\frac{A}{x-a} + \frac{Bx+C}{x^2+b}\). In our case, 'a' can be replaced by '1' (from the factor \(x-1\)) and 'b' replaced by '-1' (from the factor \(x^2+1\)). Hence, the expression becomes \(\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\).
03

Set up and solve the equation

The next step is to equate the original fraction to the partial fraction decomposition and then multiply through by the denominator of the left side to clear the fractions. Doing so, we get the equation: \(5x^2-6x+7 = A(x^2+1) + (Bx+C)(x-1)\). Expanding and grouping like terms gives us the equation in the form of \(Ax^2 + Bx + C + Dx + E = 5x^2 - 6x + 7\). By comparing coefficients, we can get a set of equations that can be solved simultaneously to get the values of A, B and C.
04

Find the coefficients

The terms on both sides of the equation should match. In the expanded form 'Ax^2 + Bx + C = 5x^2 - 6x + 7', the coefficients of 'x^2', 'x', and the constant terms on both sides should match. Equating coefficients gives:For \(x^2\) term: A = 5,For \(x\) term: B = -6,For constant term: C = 7Substitute the above values in the partial fraction decomposition.
05

Final Step

Finally, substitute the values of coefficients A, B and C in the partial fraction decomposition, i.e., \(\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\). This gives the solution as: \(\frac{5}{x-1} - \frac{6x+7}{x^2+1}\).

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Most popular questions from this chapter

Compare the graphs of \(3 x-2 y>6\) and \(3 x-2 y \leq 6\). Discuss similarities and differences between the graphs.

In Exercises \(19-30,\) solve each system by the addition method. \(4 x+3 y=15\) \(2 x-5 y=1\)

Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \begin{aligned}&2 x+y \leq 6\\\&x+y \geq 2\\\&1 \leq x \leq 2\\\&y \leq 3\end{aligned} $$

In Exercises \(5-18\), solve each system by the substitution method. $$ \begin{aligned} &2 x-3 y=-13\\\ &y=2 x+7 \end{aligned} $$

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. \(3 x-2 y=-5\) \(4 x+y=8\)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. \(3 x-2 y=-5\) \(4 x+y=8\)
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