College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.3 - Page 550: 30

Answer

$\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{3}{x-4}+\dfrac{2x-1}{x^2+5}$

Work Step by Step

We are given the fraction: $\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}$ As the denominator is already factored, we can write the partial fraction decomposition: $\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{A}{x-4}+\dfrac{Bx+C}{x^2+5}$ Multiply all terms by the least common denominator $(x-4)(x^2+5)$: $(x-4)(x^2+5)\cdot\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=(x-4)(x^2+5)\cdot\dfrac{A}{x-4}+(x-4)(x^2+5)\cdot\dfrac{Bx+C}{x^2+5}$ $5x^2-9x+19=A(x^2+5)+(Bx+C)(x-4)$ $5x^2-9x+19=Ax^2+5A+Bx^2-4Bx+Cx-4C$ $5x^2-9x+19=(A+B)x^2+(-4B+C)x+(5A-4C)$ Identify the coefficients and build the system of equations: $\begin{cases} A+B=5\\ -4B+C=-9\\ 5A-4C=19 \end{cases}$ Solve the system: Multiply Equation 2 by 4 and add it to Equation 3: $\begin{cases} A+B=5\\ -16B+4C=-36\\ 5A-4C=19 \end{cases}$ $\begin{cases} A+B=5\\ -16B+4C+5A-4C=-36+19 \end{cases}$ $\begin{cases} A+B=5\\ 5A-16B=-17 \end{cases}$ $\begin{cases} 16A+16B=16(5)\\ 5A-16B=-17 \end{cases}$ $16A+16B+5A-16B=80-17$ $21A=63$ $A=3$ $A+B=5$ $3+B=5$ $B=2$ $-4B+C=-9$ $-4(2)+C=-9$ $-8+C=-9$ $C=-1$ The partial fraction decomposition is: $\dfrac{5x^2-9x+19}{(x-4)(x^2+5)}=\dfrac{3}{x-4}+\dfrac{2x-1}{x^2+5}$
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