Question 199208
{{{x=4+i}}} or {{{x=4-i}}} Start with the given solutions.



{{{x-4=i}}} or {{{x-4=-i}}} Subtract 4 from both sides (for each equation).



{{{(x-4)^2=i^2}}} or {{{(x-4)^2=(-i)^2}}} Square both sides



{{{(x-4)^2=-1}}} or {{{(x-4)^2=-1}}} Square i to get -1 and square -i to get -1



{{{(x-4)^2+1=0}}} or {{{(x-4)^2+1=0}}} Add 1 to both sides.



Since the equations are the same, we can focus on one equation:



{{{(x-4)^2+1=0}}}



{{{x^2-8x+16+1=0}}} FOIL



{{{x^2-8x+17=0}}} Combine like terms.



So the quadratic with the roots 4+i and 4-i is {{{y=x^2-8x+17}}}



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# 2



First, let's find the <i>possible</i> rational zeros



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 1 (the last coefficient):


*[Tex \LARGE p=\pm1]


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{-1}{1}]







Now simplify


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


*[Tex \LARGE  1, -1]



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



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

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


Take note that the first three values in the bottom row are 1, -3, and -1. So this means that



{{{x^3-4x^2+2x+1=(x-1)(x^2-3x-1)}}}



Now all we need to do is solve {{{x^2-3x-1=0}}} to find the next two zeros:



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(1)(-1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=-1}}}



{{{x = (3 +- sqrt( (-3)^2-4(1)(-1) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{x = (3 +- sqrt( 9-4(1)(-1) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--4 ))/(2(1))}}} Multiply {{{4(1)(-1)}}} to get {{{-4}}}



{{{x = (3 +- sqrt( 9+4 ))/(2(1))}}} Rewrite {{{sqrt(9--4)}}} as {{{sqrt(9+4)}}}



{{{x = (3 +- sqrt( 13 ))/(2(1))}}} Add {{{9}}} to {{{4}}} to get {{{13}}}



{{{x = (3 +- sqrt( 13 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3+sqrt(13))/(2)}}} or {{{x = (3-sqrt(13))/(2)}}} Break up the expression.  



So the next two zeros are {{{x = (3+sqrt(13))/(2)}}} or {{{x = (3-sqrt(13))/(2)}}} 



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Answer:


So the three roots are {{{x=1}}}, {{{x = (3+sqrt(13))/(2)}}} or {{{x = (3-sqrt(13))/(2)}}}



where the irrational roots are {{{x = (3+sqrt(13))/(2)}}} and {{{x = (3-sqrt(13))/(2)}}}




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# 3


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 -3 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm3]


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{3}{1}, \frac{-1}{1}, \frac{-3}{1}]







Now simplify


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


*[Tex \LARGE  1, 3, -1, -3]