Difference between revisions of "004 Sample Final A, Problem 8"
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Kayla Murray (talk | contribs) (Created page with "<span class="exam"> a) List all the possible zeros of the function.<br> b) Find all the zeros, that is, solve <math>f(x) = 0</math>") |
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− | <span class="exam"> a) List all the possible zeros of the function.<br> | + | <span class="exam"> a) List all the possible rational zeros of the function <math>f(x)=x^4-4x^3-7x^2+34x-24.</math> <br> |
b) Find all the zeros, that is, solve <math>f(x) = 0</math> | b) Find all the zeros, that is, solve <math>f(x) = 0</math> | ||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Foundations | ||
+ | |- | ||
+ | |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: | ||
+ | |- | ||
+ | |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>. | ||
+ | |} | ||
+ | |||
+ | |||
+ | Solution: | ||
+ | |||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 1: | ||
+ | |- | ||
+ | |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>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 2: | ||
+ | |- | ||
+ | |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>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! 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>. | ||
+ | |- | ||
+ | |By synthetic division, <math>x^3-3x^2-10x+24=(x-2)(x^2-x-12)</math>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 4: | ||
+ | |- | ||
+ | |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>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Final Answer: | ||
+ | |- | ||
+ | |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>''']] |
Latest revision as of 17:11, 4 May 2015
a) List all the possible rational zeros of the function
b) Find all the zeros, that is, solve
Foundations |
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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: |
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By the rational roots test, the possible roots of are . |
Step 2: |
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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: |
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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: |
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Thus, . |
The zeros of are . |
Final Answer: |
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The possible roots of are . |
The zeros of are |