Difference between revisions of "008A Sample Final A, Question 9"

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'''Question: ''' a) List all the possible rational zeros of the function <math>f(x) = x^4 + 5x^3 - 27x^2 +31x -10</math><br>
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'''Question: '''a) List all the possible rational zeros of the function <math>f(x) = x^4 + 5x^3 - 27x^2 +31x -10</math><br>
 
:::&nbsp;b) Find all the zeros, that is, solve f(x) = 0
 
:::&nbsp;b) Find all the zeros, that is, solve f(x) = 0
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
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!Foundations: &nbsp;
 
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|1) What does the Rational Zeros Theorem say about possible zeros?
 
|1) What does the Rational Zeros Theorem say about possible zeros?
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
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!Step 1: &nbsp;
 
|-
 
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|Start by factoring -10, and 1. Then the Rational Zeros Theorem gives us that the possible rational zeros are <math>\pm 1, \pm 2, \pm 5,</math> and <math>\pm 10</math>.
 
|Start by factoring -10, and 1. Then the Rational Zeros Theorem gives us that the possible rational zeros are <math>\pm 1, \pm 2, \pm 5,</math> and <math>\pm 10</math>.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
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!Step 2: &nbsp;
 
|-
 
|-
 
|Start testing zeros with 1 and -1 since they require the least arithmetic. You will also find that 1 is a zero. Applying synthetic division you can reduce the polynomial to <math>(x - 1)(x^3 + 6x^2 - 21x + 10)</math>.  
 
|Start testing zeros with 1 and -1 since they require the least arithmetic. You will also find that 1 is a zero. Applying synthetic division you can reduce the polynomial to <math>(x - 1)(x^3 + 6x^2 - 21x + 10)</math>.  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
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!Step 3: &nbsp;
 
|-
 
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|Now we just need to find the zeros of <math>x^3 + 6x^2 - 21x + 10</math>. Since we are not down to a quadratic polynomial we have to find another zero from the list of rational zeros we found in step 1. You will find 2 is another root, and the polynomial can further be reduced to <math>(x - 1)(x - 2)(x^2 + 8x - 5)</math>
 
|Now we just need to find the zeros of <math>x^3 + 6x^2 - 21x + 10</math>. Since we are not down to a quadratic polynomial we have to find another zero from the list of rational zeros we found in step 1. You will find 2 is another root, and the polynomial can further be reduced to <math>(x - 1)(x - 2)(x^2 + 8x - 5)</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
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!Step 4: &nbsp;
 
|-
 
|-
 
|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4\cdot 21}}{2} = -4 \pm \sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
 
|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4\cdot 21}}{2} = -4 \pm \sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
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!Final Answer: &nbsp;
 
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|The roots are <math> x = 1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
 
|The roots are <math> x = 1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
 
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[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]
 
[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

Latest revision as of 23:55, 25 May 2015

Question: a) List all the possible rational zeros of the function

 b) Find all the zeros, that is, solve f(x) = 0
Foundations:  
1) What does the Rational Zeros Theorem say about possible zeros?
2) How do you check if a possible zero is actually a zero?
3) How do you find the rest of the zeros?
Answer:
1) The possible divisors can be found by finding the factors of -10, in a list, and the factors of 1, in a second list. Then write down all the fractions with numerators from the first list and denominators from the second list.
2) Use synthetic division, or plug a possible zero into the function. If you get 0, you have found a zero.
3) After your reduce the polynomial with synthetic division, try and find another zero from the list you made in part a). Once you reach a degree 2 polynomial you can finish the problem with the quadratic formula.

Solution:

Step 1:  
Start by factoring -10, and 1. Then the Rational Zeros Theorem gives us that the possible rational zeros are and .
Step 2:  
Start testing zeros with 1 and -1 since they require the least arithmetic. You will also find that 1 is a zero. Applying synthetic division you can reduce the polynomial to .
Step 3:  
Now we just need to find the zeros of . Since we are not down to a quadratic polynomial we have to find another zero from the list of rational zeros we found in step 1. You will find 2 is another root, and the polynomial can further be reduced to
Step 4:  
Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are . Thus the zeros of are and
Final Answer:  
The roots are and

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