Partial Fractions Quadruple
Find constants $ A, $ $ B, $ $ C, $ and $ D $ such that
\[\frac{4x^3 - 20x^2 + 37x -25}{(x-2)^3(x-1)} = \frac{A}{x - 1} + \frac{B}{(x -2)^3} + \frac{C}{(x-2)^2}+\frac{D}{x-2}.\]Enter the order quadruple $ (A,B,C,D) $.
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
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- $\infty$
Solution
Multiplying both sides by $ (x-2)^3(x-1) $ gives us
$$4x^3 - 20x^2 + 37x -25 = A(x -2)^3+B(x-1)+C(x -1)(x -2)+D(x-1)(x -2)^2.$$Setting $ x=2 $ gives $ 4(8)-20(4)+74-25=B $. Evaluating the expression on the left gives us $ B=1 $.
Setting $ x=1 $ gives $ 4-20+37-25=A(-1)^3 $, and so $ A=4 $.
We still need to find $ C $ and $ D $. By choosing 2 new values for $ x $, we can get two equations which we can solve for $ C $ and $ D $. We can pick convenient values to make our work easier.
When $ x=0 $, we get
$$-25=4(-2)^3+(-1)+C(-1)(-2)+D(-1)(-2)^2$$which simplifies to
$$2C-D=8.$$When $ x=-1 $, we get
$$4(-1)^3-20(-1)^2+37(-1)-25=4(-3)^3+(-2)+C(-2)(-3)+D(-2)(-3)^2$$which simplifies to
$$C-3D=4.$$We can multiply this equation by $ 2 $ and subtract it from the previous one to get $ -D+6D=8-2\cdot4=0 $ and hence $ D=0 $. Then $ 2C=8 $ and $ C=4 $. Therefore $ (A,B,C,D)=\boxed{(4,1,4,0)} $.
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