Question 199774
First, let's find the possible rational zeros of P(x):


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







Now simplify


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


*[Tex \LARGE  1, 2, 4, -1, -2, -4]


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Now let's see which possible roots are actually roots.



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



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

<table cellpadding=10><tr><td>1</td><td>|</td><td>1</td><td>1</td><td>-5</td><td>-4</td><td>4</td></tr><tr><td></td><td>|</td><td> </td><td>1</td><td>2</td><td>-3</td><td>-7</td></tr><tr><td></td><td></td><td>1</td><td>2</td><td>-3</td><td>-7</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 {{{x^4+x^3-5x^2-4x+4}}}



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



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

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



So this means that {{{x^4+x^3-5x^2-4x+4=(x-2)(x^3+3x^2+x-2)}}}



Note: the term {{{x^3+3x^2+x-2}}} was formed by the first four values in the bottom row.



Now that you have {{{x^3+3x^2+x-2}}}, you simply find the possible rational zeros for {{{x^3+3x^2+x-2}}} and test to see which ones are really zeros (ie repeat the first two steps).


It turns out that the possible roots for {{{x^3+3x^2+x-2}}} are: 1, 2, -1, -2


and that -2 is a root of {{{x^3+3x^2+x-2}}}


Here's the synthetic division to prove it:


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


Looking at the bottom row of values (everything but the remainder), we get {{{x^2+x-1}}}. So this means that


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



Note: this consequently means that {{{x^4+x^3-5x^2-4x+4=(x-2)(x+2)(x^2+x-1)}}}



Now we just need to solve {{{x^2+x-1=0}}} to find the last remaining zeros.




{{{x^2+x-1=0}}} Start with the given equation.



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



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



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



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



{{{x = (-1 +- sqrt( 5 ))/(2(1))}}} Add {{{1}}} to {{{4}}} to get {{{5}}}



{{{x = (-1 +- sqrt( 5 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1+sqrt(5))/(2)}}} or {{{x = (-1-sqrt(5))/(2)}}} Break up the expression.  



So the last two roots are {{{x = (-1+sqrt(5))/(2)}}} or {{{x = (-1-sqrt(5))/(2)}}} 



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Answer:



So the four zeros of {{{P(x) = x^4 + x^3 - 5x^2 - 4x + 4 }}}are:


{{{x=2}}}, {{{x=-2}}}, {{{x = (-1+sqrt(5))/(2)}}} or {{{x = (-1-sqrt(5))/(2)}}}



Note: if you wanted to, you could compactly write the zeros as:


{{{x=""+- 2}}}, {{{x = (-1+-sqrt(5))/(2)}}}


just remember that there are 4 zeros.




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