Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 10 - Systems of Equations and Inequalities - Section 10.5 Partial Fraction Decomposition - 10.5 Assess Your Understanding - Page 788: 19

Answer

$\displaystyle \dfrac{x^2}{(x-1)^2(x+1)}=\dfrac{\frac{3}{4}}{x-1}+\dfrac{\frac{1}{2}}{(x-1)^2}+\dfrac{\frac{1}{4}}{x+1} $

Work Step by Step

We can notice from the given rational expression $\dfrac{A(x)}{B(x)}$ that $B(x)$ has repeated irreducible quadratic factor. So, we will write the given rational expression into partial fraction form as: $$\displaystyle \frac{x^2}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\dfrac{C}{x+1} ~~~(1)$$ Now, write the RHS with a common denominator using the LCD $(x-1)^2(x+1)$:. $$\displaystyle \frac{x^2}{(x-1)^2(x+1)}=\dfrac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+1)}\\ \displaystyle \frac{x^2}{(x-1)^2(x+1)}=\dfrac{(Ax-A) (x+1)+Bx+B+C(x^2+1-2x)}{(x-1)^2(x+1)}\\ \displaystyle \frac{x^2}{(x-1)^2(x+1)}=\dfrac{Ax^2+Ax-Ax-A+Bx+B+Cx^2+C-2Cx}{(x-1)^2(x+1)}\\ \displaystyle \frac{x^2}{(x-1)^2(x+1)}=\dfrac{(A+C)x^2+(B-2C)x+(-A+B+C)}{(x-1)^2(x+1)}$$ Next, equate the numerators to obtain: $$(A+C)x^2+(B-2C)x+(-A+B+C)=x$$ Equate the like coefficients of the polynomials on the LHS and RHS to obtain: $$A+C=1 ~~~(2) \\ B-2C=0 ~~~(3)\\-A+B+C=0~~~(4)$$ Add equations (2) and (3). $$A+C+B-2C=1 \\ A+B-C=1 ~~~(5)$$ Then add equations (4) and (5) to obtain tve value of $B$: $$-A+B+C +A+B-C=0+1 \\ 2 B=1 \\ B=\dfrac{1}{2}$$ Plug $B=\dfrac{1}{2}$ into equation (3) to solve for $C$. $$B-2C=0 \\ \dfrac{1}{2}-2C=0 \\ -2C=0-\dfrac{1}{2} \\ C=\dfrac{1}{4}$$ Finally, plug $C=\dfrac{1}{4}$ into equation (2) to solve for $A$ $$A+\dfrac{1}{4}=1 \\ A=1-\dfrac{1}{4}\\ A=\dfrac{3}{4}$$ Therefore, the equation (1) becomes: $\displaystyle \frac{x^2}{(x-1)^2(x+1)}=\dfrac{\frac{3}{4}}{x-1}+\dfrac{\frac{1}{2}}{(x-1)^2}+\dfrac{\frac{1}{4}}{x+1} $
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