Question 199776
First, let's find the possible rational roots.



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 -4 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm4]


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{4}{1}, \frac{4}{2}, \frac{-1}{1}, \frac{-1}{2}, \frac{-2}{1}, \frac{-2}{2}, \frac{-4}{1}, \frac{-4}{2}]







Now simplify


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


*[Tex \LARGE  1, \frac{1}{2}, 2, 4, -1, \frac{-1}{2}, -2, -4]


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Now let's test the possible roots to see if they are actually roots.



Let's see if the possible zero {{{1}}} is really a root for the function {{{2x^3+7x^2+4x-4}}}



So let's make the synthetic division table for the function {{{2x^3+7x^2+4x-4}}} given the possible zero {{{1}}}:

<table cellpadding=10><tr><td>1</td><td>|</td><td>2</td><td>7</td><td>4</td><td>-4</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>9</td><td>13</td></tr><tr><td></td><td></td><td>2</td><td>9</td><td>13</td><td>9</td></tr></tr></table>

Since the remainder {{{9}}} (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+4x-4}}}



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Let's see if the possible zero {{{1/2}}} is really a root for the function {{{2x^3+7x^2+4x-4}}}



So let's make the synthetic division table for the function {{{2x^3+7x^2+4x-4}}} given the possible zero {{{1/2}}}:

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




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



Note: the quotient of the division results from taking half of the first three values in the bottom row.



Note: if you didn't find a root, then you would have to keep going until either you find one or you are done with the list.



Now let's solve {{{x^2+4x+4=0}}} to find the next two zeros:




{{{x^2+4x+4=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=4}}}, and {{{C=4}}}



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(1)(4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=4}}}, and {{{C=4}}}



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



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



{{{x = (-4 +- sqrt( 0 ))/(2(1))}}} Subtract {{{16}}} from {{{16}}} to get {{{0}}}



{{{x = (-4 +- sqrt( 0 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-4 +- 0)/(2)}}} Take the square root of {{{0}}} to get {{{0}}}. 



{{{x = (-4 + 0)/(2)}}} or {{{x = (-4 - 0)/(2)}}} Break up the expression. 



{{{x = (-4)/(2)}}} or {{{x =  (-4)/(2)}}} Combine like terms. 



{{{x = -2}}} or {{{x = -2}}} Simplify. 



So the two solutions are {{{x = -2}}} or {{{x = -2}}} 


  
We can simply write this as {{{x=-2}}} with a multiplicity of 2



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Answer:


So the three zeros of {{{P(x) = 2x^3 + 7x^2 + 4x - 4}}} are: {{{x=1/2}}} and {{{x = -2}}} (with a multiplicity of 2)



If you have any questions, email me at jim_thompson5910@hotmail.com