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

Revision as of 16:33, 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 ${\frac {3x-37}{(x+1)(x-4)}}$ ?
2) What is the form of the partial fraction decomposition of ${\frac {4x^{2}}{(x-1){(x-2)}^{2}}}$ ?
Answer:
1) ${\frac {A}{x+1}}+{\frac {B}{x-4}}$
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 $(x+3)^{2}(x-1)$ , we get
$6x^{2}+27x+31=A(x+3)^{2}+B(x+3)(x-1)+C(x-1)$ .

Final Answer:

@@ Line 3: / Line 3: @@
 ! Foundations
 |-
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+|1) What is the form of the partial fraction decomposition of <math>\frac{3x-37}{(x+1)(x-4)}</math>?
+|-
+|2) What is the form of the partial fraction decomposition of <math>\frac{4x^2}{(x-1){(x-2)}^2}</math>?
 |-
 |Answer:
 |-
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+|1) <math>\frac{A}{x+1}+\frac{B}{x-4}</math>
+|-
+|2)<math>\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{{(x-2)}^2}</math>
 |}
@@ Line 16: / Line 20: @@
 ! Step 1:
 |-
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+|We set <math>\frac{6x^2 + 27x + 31}{(x + 3)^2(x-1)}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C}{{(x+3)}^2}</math>.
-|-
-|
 |}
@@ Line 24: / Line 26: @@
 ! Step 2:
 |-
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+|Multiplying both sides of the equation by <math>(x + 3)^2(x-1)</math>, we get
+|-
+|<math>6x^2+27x+31=A(x+3)^2+B(x+3)(x-1)+C(x-1)</math>.
 |}