Question 149768
First, let's graph the function {{{P(x)=x^4+x^3-3x^2-5x-2}}} to get


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,x^4+x^3-3x^2-5x-2)

)}}}


From the graph, we can see that the graph has the approximate zeros {{{x=-1}}} and {{{x=2}}}


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Now let's use the Rational Root Theorem to list all of the possible rational roots


Rational Root Theorem:


*[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 use synthetic division to test each possible zero





Let's see if the possible zero {{{1}}} is really a root for the function {{{x^4+x^3-3x^2-5x-2}}}



So let's make the synthetic division table for the function {{{x^4+x^3-3x^2-5x-2}}} given the possible zero {{{1}}}:

<table cellpadding=10><tr><td>1</td><td>|</td><td>1</td><td>1</td><td>-3</td><td>-5</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>2</td><td>-1</td><td>-6</td></tr><tr><td></td><td></td><td>1</td><td>2</td><td>-1</td><td>-6</td><td>-8</td></tr></tr></table>

Since the remainder {{{-8}}} (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^4+x^3-3x^2-5x-2}}}



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Let's see if the possible zero {{{2}}} is really a root for the function {{{x^4+x^3-3x^2-5x-2}}}



So let's make the synthetic division table for the function {{{x^4+x^3-3x^2-5x-2}}} given the possible zero {{{2}}}:

<table cellpadding=10><tr><td>2</td><td>|</td><td>1</td><td>1</td><td>-3</td><td>-5</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>6</td><td>6</td><td>2</td></tr><tr><td></td><td></td><td>1</td><td>3</td><td>3</td><td>1</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 {{{2}}} is a zero of {{{x^4+x^3-3x^2-5x-2}}}



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Let's see if the possible zero {{{-1}}} is really a root for the function {{{x^4+x^3-3x^2-5x-2}}}



So let's make the synthetic division table for the function {{{x^4+x^3-3x^2-5x-2}}} given the possible zero {{{-1}}}:

<table cellpadding=10><tr><td>-1</td><td>|</td><td>1</td><td>1</td><td>-3</td><td>-5</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>-1</td><td>0</td><td>3</td><td>2</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>-3</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^4+x^3-3x^2-5x-2}}}



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Let's see if the possible zero {{{-2}}} is really a root for the function {{{x^4+x^3-3x^2-5x-2}}}



So let's make the synthetic division table for the function {{{x^4+x^3-3x^2-5x-2}}} given the possible zero {{{-2}}}:

<table cellpadding=10><tr><td>-2</td><td>|</td><td>1</td><td>1</td><td>-3</td><td>-5</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>-2</td><td>2</td><td>2</td><td>6</td></tr><tr><td></td><td></td><td>1</td><td>-1</td><td>-1</td><td>-3</td><td>4</td></tr></tr></table>

Since the remainder {{{4}}} (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^4+x^3-3x^2-5x-2}}}




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Summary:


So only {{{-1}}} and {{{2}}} are actually rational roots.



Now looking back at the table for the test zero {{{-1}}}, we see

<table cellpadding=10><tr><td>-1</td><td>|</td><td>1</td><td>1</td><td>-3</td><td>-5</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>-1</td><td>0</td><td>3</td><td>2</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>-3</td><td>-2</td><td>0</td></tr></tr></table>


The bottom row of coefficients (minus the last one) form the quotient

{{{x^3-3x-2}}}




Now let's perform synthetic division using the other zero {{{2}}} on the function {{{x^3-3x-2}}}




<table cellpadding=10><tr><td>2</td><td>|</td><td>1</td><td>0</td><td>-3</td><td>-2</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>4</td><td>2</td></tr><tr><td></td><td></td><td>1</td><td>2</td><td>1</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 {{{2}}} is a zero of {{{x^3-3x-2}}}


Once again the first three coefficients in the bottom row form the quotient 


{{{x^2+2x+1}}}



Let's use the quadratic formula to find the zeros of {{{x^2+2x+1}}}



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(2) +- sqrt( (2)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=2}}}, and {{{c=1}}}



{{{x = (-2 +- sqrt( 4-4(1)(1) ))/(2(1))}}} Square {{{2}}} to get {{{4}}}. 



{{{x = (-2 +- sqrt( 4-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{x = (-2 +- sqrt( 0 ))/(2(1))}}} Subtract {{{4}}} from {{{4}}} to get {{{0}}}



{{{x = (-2 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-2 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (-2 + 0)/(2)}}} or {{{x = (-2 - 0)/(2)}}} Break up the expression. 



{{{x = (-2)/(2)}}} or {{{x =  (-2)/(2)}}} Combine like terms. 



{{{x = -1}}} or {{{x = -1}}} Simplify. 



So the zeros of {{{x^2+2x+1}}} are {{{x = -1}}} or {{{x = -1}}} or just {{{x = -1}}} with a multiplicity of 2



Now there are 3 instances where we get a zero of {{{x = -1}}}. So this tells us that the zero {{{x = -1}}} has a multiplicity of 3

  
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Answer:


So the zeros of {{{P(x)=x^4+x^3-3x^2-5x-2}}} are {{{-1}}} (with a multiplicity of 3) and {{{x=2}}}