Difference between revisions of "004 Sample Final A, Problem 10"

Latest revision as of 16:43, 4 May 2015

Decompose into separate partial fractions. ${\frac {6x^{2}+27x+31}{(x+3)^{2}(x-1)}}$

Foundations
1) What is the form of the partial fraction decomposition of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {3x-37}{(x+1)(x-4)}}} ?
2) What is the form of the partial fraction decomposition of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {4x^{2}}{(x-1){(x-2)}^{2}}}} ?
Answer:
1) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {A}{x+1}}+{\frac {B}{x-4}}}
2) ${\frac {A}{x-1}}+{\frac {B}{x-2}}+{\frac {C}{{(x-2)}^{2}}}$

Solution:

Step 1:
We set ${\frac {6x^{2}+27x+31}{(x+3)^{2}(x-1)}}={\frac {A}{x-1}}+{\frac {B}{x+3}}+{\frac {C}{{(x+3)}^{2}}}$ .

Step 2:
Multiplying both sides of the equation by Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (x+3)^{2}(x-1)} , we get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)} .

Step 3:
If we set $x=1$ in the above equation, we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 16A=64} and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A=4} .
If we set $x=-3$ in the above equation, we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -4C=4} and $C=-1$ .

Step 4:
In the equation Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)} , we compare the constant terms of both sides. We must have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 9A-3B-C=31} . Substituting Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A=4} and $C=-1$ , we get $B=2$ .
Thus, the partial fraction decomposition is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}}

Final Answer:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {4}{x-1}}+{\frac {2}{x+3}}+{\frac {-1}{{(x+3)}^{2}}}}

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 ! Step 3:
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+|If we set <math>x=1</math> in the above equation, we get <math>16A=64</math> and <math>A=4</math>.
 |-
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+|If we set <math>x=-3</math> in the above equation, we get <math>-4C=4</math> and <math>C=-1</math>.
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 |}
@@ Line 46: / Line 42: @@
 ! Step 4:
 |-
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+|In the equation <math>6x^2+27x+31=A(x+3)^2+B(x+3)(x-1)+C(x-1)</math>, we compare the constant terms of both sides. We must have
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+|<math>9A-3B-C=31</math>. Substituting <math>A=4</math> and <math>C=-1</math>, we get <math>B=2</math>.
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+|Thus, the partial fraction decomposition is <math>\frac{4}{x-1}+\frac{2}{x+3}+\frac{-1}{{(x+3)}^2}</math>
 |}
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 ! Final Answer:
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+|<math>\frac{4}{x-1}+\frac{2}{x+3}+\frac{-1}{{(x+3)}^2}</math>
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 [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]