Question 149763
First, let's list all of the possible rational zeros.


Any rational zero can be found through this formula


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



So let's list the factors of 2 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2]


Now let's list the factors of 1 (the first coefficient):


*[Tex \LARGE q=\pm1]


Now let's divide each factor of the last coefficient by each factor of the first coefficient



*[Tex \LARGE \frac{1}{1}, \frac{2}{1}, \frac{-1}{1}, \frac{-2}{1}]




Now simplify


These are all the distinct rational zeros of the function that could occur


*[Tex \LARGE  1, 2, -1, -2]



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Now let's test each possible rational root with use of synthetic division





Let's see if the possible zero {{{1}}} is really a root for the function {{{x^3-x^2-2x+2}}}



So let's make the synthetic division table for the function {{{x^3-x^2-2x+2}}} given the possible zero {{{1}}}:

<table cellpadding=10><tr><td>1</td><td>|</td><td>1</td><td>-1</td><td>-2</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>0</td><td>-2</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>-2</td><td>0</td></tr></tr></table>

Since the remainder {{{0}}} (the right most entry in the last row) is equal to zero, this means that {{{1}}} is a zero of {{{x^3-x^2-2x+2}}}



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Let's see if the possible zero {{{2}}} is really a root for the function {{{x^3-x^2-2x+2}}}



So let's make the synthetic division table for the function {{{x^3-x^2-2x+2}}} given the possible zero {{{2}}}:

<table cellpadding=10><tr><td>2</td><td>|</td><td>1</td><td>-1</td><td>-2</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>2</td><td>0</td></tr><tr><td></td><td></td><td>1</td><td>1</td><td>0</td><td>2</td></tr></tr></table>

Since the remainder {{{2}}} (the right most entry in the last row) is <font size=4><b>not</b></font> equal to zero, this means that {{{2}}} is <font size=4><b>not</b></font> a zero of {{{x^3-x^2-2x+2}}}



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Let's see if the possible zero {{{-1}}} is really a root for the function {{{x^3-x^2-2x+2}}}



So let's make the synthetic division table for the function {{{x^3-x^2-2x+2}}} given the possible zero {{{-1}}}:

<table cellpadding=10><tr><td>-1</td><td>|</td><td>1</td><td>-1</td><td>-2</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>-1</td><td>2</td><td>0</td></tr><tr><td></td><td></td><td>1</td><td>-2</td><td>0</td><td>2</td></tr></tr></table>

Since the remainder {{{2}}} (the right most entry in the last row) is <font size=4><b>not</b></font> equal to zero, this means that {{{-1}}} is <font size=4><b>not</b></font> a zero of {{{x^3-x^2-2x+2}}}



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Let's see if the possible zero {{{-2}}} is really a root for the function {{{x^3-x^2-2x+2}}}



So let's make the synthetic division table for the function {{{x^3-x^2-2x+2}}} given the possible zero {{{-2}}}:

<table cellpadding=10><tr><td>-2</td><td>|</td><td>1</td><td>-1</td><td>-2</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>-2</td><td>6</td><td>-8</td></tr><tr><td></td><td></td><td>1</td><td>-3</td><td>4</td><td>-6</td></tr></tr></table>

Since the remainder {{{-6}}} (the right most entry in the last row) is <font size=4><b>not</b></font> equal to zero, this means that {{{-2}}} is <font size=4><b>not</b></font> a zero of {{{x^3-x^2-2x+2}}}




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So to recap, we only found one rational zero. So the only rational zero for the function {{{x^3-x^2-2x+2}}} is {{{1}}}


Now if we go back to the corresponding synthetic division table for the test zero {{{1}}}, we get


<table cellpadding=10><tr><td>1</td><td>|</td><td>1</td><td>-1</td><td>-2</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>0</td><td>-2</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>-2</td><td>0</td></tr></tr></table>


Now looking at the bottom row of coefficients, we see the first three numbers: 1, 0 and -2


These coefficients form the quotient {{{x^2+0x-2}}} or just {{{x^2-2}}}


This means that {{{(x^3-x^2-2x+2)/(x-1)=x^2-2}}} or  {{{x^3-x^2-2x+2=(x-1)(x^2-2)}}} 


{{{x^2-2=0}}} Set the quotient equal to zero



{{{x^2=2}}} Add 2 to both sides.



{{{x=0+-sqrt(2)}}} Take the square root of both sides.



{{{x=sqrt(2)}}} or {{{x=-sqrt(2)}}} Break up the expression.



So the remaining two zeros are {{{x=sqrt(2)}}} or {{{x=-sqrt(2)}}}



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Answer:


So the zeros of {{{x^3-x^2-2x+2}}} are:


{{{1}}}, {{{x=sqrt(2)}}} or {{{x=-sqrt(2)}}} where each zero has a multiplicity of 1.