Question 166084
Any rational zero can be found through this equation


*[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 2 (the first coefficient):


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


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



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







Now simplify


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


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




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Now let's test all of the possible zeros:




Let's see if the possible zero {{{1}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{1}}}:

<table cellpadding=10><tr><td>1</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>2</td><td>-5</td></tr><tr><td></td><td></td><td>2</td><td>2</td><td>-5</td><td>-3</td></tr></tr></table>

Since the remainder {{{-3}}} (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 {{{2x^3-7x+2}}}



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Let's see if the possible zero {{{1/2}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{1/2}}}:

<table cellpadding=10><tr><td>1/2</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>1/2</td><td>-13/4</td></tr><tr><td></td><td></td><td>2</td><td>1</td><td>-13/2</td><td>-5/4</td></tr></tr></table>

Since the remainder {{{-5/4}}} (the right most entry in the last row) is <font size=4><b>NOT</b></font> equal to zero, this means that {{{1/2}}} is <font size=4><b>NOT</b></font> a zero of {{{2x^3-7x+2}}}



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Let's see if the possible zero {{{2}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{2}}}:

<table cellpadding=10><tr><td>2</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>4</td><td>8</td><td>2</td></tr><tr><td></td><td></td><td>2</td><td>4</td><td>1</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 {{{2x^3-7x+2}}}



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Let's see if the possible zero {{{-1}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{-1}}}:

<table cellpadding=10><tr><td>-1</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>-2</td><td>2</td><td>5</td></tr><tr><td></td><td></td><td>2</td><td>-2</td><td>-5</td><td>7</td></tr></tr></table>

Since the remainder {{{7}}} (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 {{{2x^3-7x+2}}}



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Let's see if the possible zero {{{-1/2}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{-1/2}}}:

<table cellpadding=10><tr><td>-1/2</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>-1</td><td>1/2</td><td>13/4</td></tr><tr><td></td><td></td><td>2</td><td>-1</td><td>-13/2</td><td>21/4</td></tr></tr></table>

Since the remainder {{{21/4}}} (the right most entry in the last row) is <font size=4><b>NOT</b></font> equal to zero, this means that {{{-1/2}}} is <font size=4><b>NOT</b></font> a zero of {{{2x^3-7x+2}}}



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Let's see if the possible zero {{{-2}}} is really a root for the function {{{2x^3-7x+2}}}



So let's make the synthetic division table for the function {{{2x^3-7x+2}}} given the possible zero {{{-2}}}:

<table cellpadding=10><tr><td>-2</td><td>|</td><td>2</td><td>0</td><td>-7</td><td>2</td></tr><tr><td></td><td>|</td><td> </td><td>-4</td><td>8</td><td>-2</td></tr><tr><td></td><td></td><td>2</td><td>-4</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 {{{2x^3-7x+2}}}


Also, it turns out that the numbers 2,-4, and 1 (the bottom row of numbers) form the coefficients of the quotient. So this means that {{{(2x^3-7x+2)/(x+2)=2x^2-4x+1}}}



{{{2x^3-7x+2=(x+2)(2x^2-4x+1)}}} Now multiply both sides by {{{x+2}}}



So this means that {{{2x^3-7x+2}}} factors to {{{(x+2)(2x^2-4x+1)}}}



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Now let's solve {{{2x^2-4x+1=0}}}



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-4}}}, and {{{c=1}}}



Let's use the quadratic formula to solve for x



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



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



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



{{{x = (4 +- sqrt( 16-4(2)(1) ))/(2(2))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16-8 ))/(2(2))}}} Multiply {{{4(2)(1)}}} to get {{{8}}}



{{{x = (4 +- sqrt( 8 ))/(2(2))}}} Subtract {{{8}}} from {{{16}}} to get {{{8}}}



{{{x = (4 +- sqrt( 8 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (4 +- 2*sqrt(2))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (4+2*sqrt(2))/(4)}}} or {{{x = (4-2*sqrt(2))/(4)}}} Break up the expression.  



{{{x = (2+sqrt(2))/(2)}}} or {{{x = (2-sqrt(2))/(2)}}} Reduce



So the last two zeros are {{{x = (2+sqrt(2))/(2)}}} or {{{x = (2-sqrt(2))/(2)}}}



which approximate to {{{x=1.707}}} or {{{x=0.293}}} 




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Answer:


So after all of that, we get the three zeros: {{{x=-2}}}, {{{x=(2+sqrt(2))/(2)}}} or {{{x=(2-sqrt(2))/(2)}}}



Which in decimal form are {{{x=-2}}}, {{{x=1.707}}}, or {{{x=0.293}}}