Question 188143
Any possible 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 -6 (the last coefficient):


*[Tex \LARGE p=\pm1, \pm2, \pm3, \pm6]


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{3}{1}, \frac{6}{1}, \frac{-1}{1}, \frac{-2}{1}, \frac{-3}{1}, \frac{-6}{1}]







Now simplify


These are all the distinct <i>possible</i> rational zeros of the function.


*[Tex \LARGE  1, 2, 3, 6, -1, -2, -3, -6]



Note: these are the possible zeros. The function may not even have rational zeros (they may be irrational or complex).



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Now simply use synthetic division to find the real rational zeros



Let's see if the possible zero {{{1}}} is really a root for the function {{{x^3-2x^2+3x-6}}}



So let's make the synthetic division table for the function {{{x^3-2x^2+3x-6}}} given the possible zero {{{1}}}:

<table cellpadding=10><tr><td>1</td><td>|</td><td>1</td><td>-2</td><td>3</td><td>-6</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>-1</td><td>2</td></tr><tr><td></td><td></td><td>1</td><td>-1</td><td>2</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 {{{1}}} is <font size=4><b>not</b></font> a zero of {{{x^3-2x^2+3x-6}}}



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Let's see if the possible zero {{{2}}} is really a root for the function {{{x^3-2x^2+3x-6}}}



So let's make the synthetic division table for the function {{{x^3-2x^2+3x-6}}} given the possible zero {{{2}}}:

<table cellpadding=10><tr><td>2</td><td>|</td><td>1</td><td>-2</td><td>3</td><td>-6</td></tr><tr><td></td><td>|</td><td> </td><td>2</td><td>0</td><td>6</td></tr><tr><td></td><td></td><td>1</td><td>0</td><td>3</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^3-2x^2+3x-6}}}




Because {{{x=2}}} is a zero, this means that {{{x-2}}} is a factor of {{{x^3-2x^2+3x-6}}}


The first three numbers in the last row 1, 0, and 3 form the coefficients to the polynomial {{{x^2+3}}}. So this consequently means that {{{x^3-2x^2+3x-6=(x-2)(x^2+3)}}}



{{{(x-2)(x^2+3)=0}}} Set the right side equal to zero



{{{x-2=0}}} or {{{x^2+3=0}}} Set each factor equal to zero



Since we know that {{{x=2}}} is already a zero, we can ignore the first equation. So simply solve the quadratic equation {{{x^2+3=0}}} to find the remaining solutions:



{{{x^2+3=0}}} Start with the given equation



{{{x^2=-3}}} Subtract 3 from both sides



{{{x=sqrt(-3)}}} or {{{x=-sqrt(-3)}}} Take the square root of both sides (don't forget the "plus/minus")



{{{x=sqrt(3)*i}}} or {{{x=-sqrt(3)*i}}} Simplify. Note: {{{i=sqrt(-1)}}}





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Answer:



So the solutions of {{{x^3-2x^2+3x-6}}} are {{{x=2}}}, {{{x=sqrt(3)*i}}} or {{{x=-sqrt(3)*i}}}