008A Sample Final A, Question 11 - Revision history

MathAdmin at 06:58, 26 May 2015

2015-05-26T06:58:32Z

MathAdmin at 22:52, 23 May 2015

2015-05-23T22:52:25Z

MathAdmin at 22:44, 23 May 2015

2015-05-23T22:44:56Z

MathAdmin at 22:41, 23 May 2015

2015-05-23T22:41:46Z

MathAdmin: Created page with "'''Question: ''' Decompose into separate partial fractions $\frac{3x^2+6x+7}{(x+3)^2(x-1)}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Fo..."

2015-04-28T21:11:12Z

Created page with "'''Question: ''' Decompose into separate partial fractions <math>\frac{3x^2+6x+7}{(x+3)^2(x-1)}</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Fo..."

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'''Question: ''' Decompose into separate partial fractions <math>\frac{3x^2+6x+7}{(x+3)^2(x-1)}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
|-
|1) How many fractions will this decompose into? What are the denominators?
|-
|2) How do you solve for the numerators?
|-
|Answer:
|-
|1) Since each of the factors are linear, and one has multipliclity 2, there will be three denominators. The linear term, x -1, will appear once in the denominator of the decomposition. The other two denominators will be x + 3, and <math>(x + 3)^2</math>.
|-
|2) After writing the equality, <math>\frac{3x^2 +6x + 7}{(x + 3)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}</math>, clear the denominators, and use the cover up method to solve for A, B, and C. After you clear the denominators, the cover up method is to evaluate both sides at x = 1, -3, and any third value. Each evaluation will yield the value of one of the three unknowns.
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|From the factored form of the denominator we can observe that there will be three denominators: <math>x - 1, x + 3</math>, and <math>(x + 3)^2</math>. So the final answer will be of the form: <math>\frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|Now we have the equality <math>\frac{3x^2 +6x + 7}{(x + 3)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}</math>. Now clearing the denominators we end up with <math>A(x + 3)^2 + B(x - 1)(x + 3) + C(x - 1) = 3x^2 + 6x + 7</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|To proceed we start by evaluating both sides at different x-values. We start with x = 1, since this will zero out the B and C. This leads to <math>A(1 + 3)^2 = 3(1)^2 + 6(1) + 7</math>, <math>16A = 3 + 6 + 7</math>, and finally A = 1.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|Now evaluate at -3 to zero out both A and B. This yields the following equations <math>C((-3) - 1) = 3(-3)^2 + 6(-3) + 7</math>, <math>-4C = 3(9) - 18 + 7</math>, <math>-4C = 27 - 11</math>, and <math>C = -4</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 5:
|-
|To obtain the value of B we can evaluate x at any value except 1, and -3. We do not want to evaluate at 1 and -3 since both of these will zero out the B. Evaluating at x = 0 will make the arithmetic easier, and gives us <math>A(0 + 3)^2 + B(0 - 1)(0 + 3) + C(0 - 1) = 9A -3B -C = 7</math>. However, we know the values of both A and C, which are 1 and -4, respectively. So <math>9 -3B - (-4) = 7</math>, <math>-3B + 13 = 7</math>, <math>-3B = -6</math>, and finally <math>B = 2</math>. This means the final answer is <math>\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|<math>\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}</math>
|}

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations
+!Foundations: &nbsp;
 |-
 |1) How many fractions will this decompose into? What are the denominators?
@@ Line 18: / Line 18: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+!Step 1: &nbsp;
 |-
 |From the factored form of the denominator we can observe that there will be three denominators: <math>x - 1, x + 3</math>, and <math>(x + 3)^2</math>. So the final answer with have the following form: <math>\frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}</math>
@@ Line 24: / Line 24: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+!Step 2: &nbsp;
 |-
 |Now we have the equality <math>\frac{3x^2 +6x + 7}{(x + 3)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}</math>. We clear the denominators and end up with <math>A(x + 3)^2 + B(x - 1)(x + 3) + C(x - 1) = 3x^2 + 6x + 7</math>.
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 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3:
+!Step 3: &nbsp;
 |-
 |By evaluation both sides by using x = 1, we will zero out the B and C. This leads to <math>A(1 + 3)^2 = 3(1)^2 + 6(1) + 7,  16A = 3 + 6 + 7</math>&nbsp;, and finally <math>A = 1</math>.
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 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+!Step 4: &nbsp;
 |-
 |Now evaluating at x = -3 to zero out both A and B. This yields the following equations <math>C((-3) - 1) = 3(-3)^2 + 6(-3) + 7</math>, <math>-4C = 3(9) - 18 + 7</math>, <math>-4C = 27 - 11</math>, and <math>C = -4</math>
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 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 5:
+!Step 5: &nbsp;
 |-
 |To obtain the value of B we  can evaluate x at any value except 1, and -3. We do not want to evaluate at 1 and -3 since both of these will zero out the B. Evaluating at x = 0 will make the arithmetic easier, and gives us <math>A(0 + 3)^2 + B(0 - 1)(0 + 3) + C(0 - 1) = 9A -3B -C = 7</math>. However, we know the values of both A and C, which are 1 and -4, respectively. So <math>9 -3B - (-4) = 7</math>, <math>-3B + 13 =  7</math>, <math>-3B = -6</math>, and finally <math>B = 2</math>.
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 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+!Final Answer: &nbsp;
 |-
 |<math>\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}</math>