Difference between revisions of "008A Sample Final A, Question 11"

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|1) How many fractions will this decompose into? What are the denominators?
 
|1) How many fractions will this decompose into? What are the denominators?
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|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>
 
|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>
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|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>.
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|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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|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.
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|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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|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>
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|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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|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>
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|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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|<math>\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}</math>
 
|<math>\frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}</math>

Latest revision as of 23:58, 25 May 2015

Question: Decompose into separate partial fractions

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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x -1} , will appear once in the denominator of the decomposition. The other two denominators will be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x + 3 \text{, and } (x + 3)^2} .
2) After writing the equality, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3x^2 +6x + 7}{(x + 3)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}} , clear the denominators, and evaluate both sides at x = 1, -3, and any third value. Each evaluation will yield the value of one of the three unknowns.

Solution:

Step 1:  
From the factored form of the denominator we can observe that there will be three denominators: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x - 1, x + 3} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x + 3)^2} . So the final answer with have the following form: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}}
Step 2:  
Now we have the equality Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3x^2 +6x + 7}{(x + 3)^2(x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{C}{(x + 3)^2}} . We clear the denominators and end up with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(x + 3)^2 + B(x - 1)(x + 3) + C(x - 1) = 3x^2 + 6x + 7} .
Step 3:  
By evaluation both sides by using x = 1, we will zero out the B and C. This leads to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(1 + 3)^2 = 3(1)^2 + 6(1) + 7, 16A = 3 + 6 + 7}  , and finally Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = 1} .
Step 4:  
Now evaluating at x = -3 to zero out both A and B. This yields the following equations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C((-3) - 1) = 3(-3)^2 + 6(-3) + 7} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -4C = 3(9) - 18 + 7} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -4C = 27 - 11} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C = -4}
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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(0 + 3)^2 + B(0 - 1)(0 + 3) + C(0 - 1) = 9A -3B -C = 7} . However, we know the values of both A and C, which are 1 and -4, respectively. So Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 9 -3B - (-4) = 7} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3B + 13 = 7} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -3B = -6} , and finally Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B = 2} .
Final Answer:  
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{x - 1} + \frac{2}{x + 3} - \frac{4}{(x + 3)^2}}

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