8A F11 Q11 - Revision history

Matt Lee at 15:24, 8 April 2015

2015-04-08T15:24:52Z

Matt Lee at 22:36, 6 April 2015

2015-04-06T22:36:54Z

Matt Lee at 23:46, 2 April 2015

2015-04-02T23:46:12Z

Matt Lee: 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..."

2015-03-26T07:56:11Z

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..."

New page

'''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>
|}

@@ Line 45: / Line 45: @@
 |-
 |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>
 |}
 [[8AF11Final|<u>'''Return to Sample Exam</u>''']]

← Older revision		Revision as of 22:36, 6 April 2015
Line 46:		Line 46:
	\|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>		\|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>
	\|}		\|}
		+
		+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

@@ Line 44: / Line 44: @@
 ! 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>
+|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>
 |}

8A F11 Q11 - Revision history

Matt Lee at 15:24, 8 April 2015

Matt Lee at 22:36, 6 April 2015

Matt Lee at 23:46, 2 April 2015

Matt Lee: 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..."

Matt Lee: 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..."