Difference between revisions of "004 Sample Final A, Problem 8"
		
		
		
		
		
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Kayla Murray (talk | contribs)  | 
				Kayla Murray (talk | contribs)   | 
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! Foundations  | ! Foundations  | ||
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| − | |    | + | |If <math>f(x)=x^4+bx^3+cx^2+dx+e</math>, what does the rational roots tell us are the possible roots of <math>f(x)</math>?   | 
|-  | |-  | ||
|Answer:  | |Answer:  | ||
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| − | |  | + | |The rational roots tells us that the possible roots of <math>f(x)</math> are <math>\pm k</math> where <math>k</math> is a divisor of <math>e</math>.  | 
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! Step 1:  | ! Step 1:  | ||
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| − | |  | + | |By the rational roots test, the possible roots of <math>f(x)</math> are <math>\pm\{1,2,3,4,6,8,12,24\}</math>.  | 
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! Step 2:  | ! Step 2:  | ||
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| − | |  | + | |Using synthetic division, we test 1 as a root of <math>f(x)</math>. We get a remainder of 0. So, we have that 1 is a root of <math>f(x)</math>.   | 
| + | |-  | ||
| + | |By synthetic division, <math>f(x)=(x-1)(x^3-3x^2-10x+24)</math>.  | ||
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! Step 3:  | ! Step 3:  | ||
|-  | |-  | ||
| − | |  | + | |Using synthetic division on <math>x^3-3x^2-10x+24</math>, we test 2 as a root of this function. We get a remainder of 0. So, we have that 2 is a root of <math> x^3-3x^2-10x+24</math>.  | 
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|-  | |-  | ||
| − | |  | + | |By synthetic division, <math>x^3-3x^2-10x+24=(x-2)(x^2-x-12)</math>.  | 
|}  | |}  | ||
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! Step 4:  | ! Step 4:  | ||
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| − | |  | + | |Thus, <math>f(x)=(x-1)(x-2)(x^2-x-12)=(x-1)(x-2)(x-4)(x+3)</math>.  | 
|-  | |-  | ||
| − | |  | + | |The zeros of <math>f(x)</math> are <math>1,2,4,-3</math>.  | 
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|}  | |}  | ||
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! Final Answer:  | ! Final Answer:  | ||
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| − | |  | + | |The possible roots of <math>f(x)</math> are <math>\pm\{1,2,3,4,6,8,12,24\}</math>.  | 
| + | |-  | ||
| + | |The zeros of <math>f(x)</math> are <math>1,2,4,-3</math>  | ||
|}  | |}  | ||
[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]  | [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]  | ||
Latest revision as of 16:11, 4 May 2015
 a) List all the possible rational zeros of the function  
b) Find all the zeros, that is, solve 
| Foundations | 
|---|
| If , what does the rational roots tell us are the possible roots of ? | 
| Answer: | 
| The rational roots tells us that the possible roots of are where is a divisor of . | 
Solution:
| Step 1: | 
|---|
| By the rational roots test, the possible roots of are . | 
| Step 2: | 
|---|
| Using synthetic division, we test 1 as a root of . We get a remainder of 0. So, we have that 1 is a root of . | 
| By synthetic division, . | 
| Step 3: | 
|---|
| Using synthetic division on , we test 2 as a root of this function. We get a remainder of 0. So, we have that 2 is a root of . | 
| By synthetic division, . | 
| Step 4: | 
|---|
| Thus, . | 
| The zeros of are . | 
| Final Answer: | 
|---|
| The possible roots of are . | 
| The zeros of are |