Question 140329
First graph the function {{{f(x)=2x^5-5x^4+3x^3-3x^2+x+2}}}



{{{ graph( 500, 500, -10, 10, -10, 10,2x^5-5x^4+3x^3-3x^2+x+2   ) }}}



From the graph, we can see that there is a zero at {{{x=2}}}. So our test zero is 2




Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the function to the right of the test zero.<TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)

<TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by 2 and place the product (which is 4)  right underneath the second  coefficient (which is -5)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 4 and -5 to get -1. Place the sum right underneath 4.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by -1 and place the product (which is -2)  right underneath the third  coefficient (which is 3)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -2 and 3 to get 1. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by 1 and place the product (which is 2)  right underneath the fourth  coefficient (which is -3)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 2 and -3 to get -1. Place the sum right underneath 2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 2 by -1 and place the product (which is -2)  right underneath the fifth  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD>-2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD></TD><TD></TD></TR></TABLE>

    Add -2 and 1 to get -1. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD>-2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD><TD></TD></TR></TABLE>

    Multiply 2 by -1 and place the product (which is -2)  right underneath the sixth  coefficient (which is 2)

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD>-2</TD><TD>-2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD><TD></TD></TR></TABLE>

    Add -2 and 2 to get 0. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>2</TD><TD>|</TD><TD>2</TD><TD>-5</TD><TD>3</TD><TD>-3</TD><TD>1</TD><TD>2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>4</TD><TD>-2</TD><TD>2</TD><TD>-2</TD><TD>-2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x-2}}} is a factor of  {{{2x^5 - 5x^4 + 3x^3 - 3x^2 + x + 2}}}


Now lets look at the bottom row of coefficients:


The first 5 coefficients (2,-1,1,-1,-1) form the quotient


{{{2x^4 - x^3 + x^2 - x - 1}}}



So {{{(2x^5 - 5x^4 + 3x^3 - 3x^2 + x + 2)/(x-2)=2x^4 - x^3 + x^2 - x - 1}}}


You can use this <a href=http://calc101.com/webMathematica/long-divide.jsp>online polynomial division calculator</a> to check your work


Basically  {{{2x^5 - 5x^4 + 3x^3 - 3x^2 + x + 2}}} factors to {{{(x-2)(2x^4 - x^3 + x^2 - x - 1)}}}


Now lets break  {{{2x^4 - x^3 + x^2 - x - 1}}} down further





Now let's graph the function {{{f(x)=2x^4 - x^3 + x^2 - x - 1}}}



{{{ graph( 500, 500, -10, 10, -10, 10,2x^4 - x^3 + x^2 - x - 1   ) }}}



From the graph, we can see that there is a zero at {{{x=1}}}. So our test zero is 1. So this time our test zero is 1




Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the function to the right of the test zero.<TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)

<TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 1 by 2 and place the product (which is 2)  right underneath the second  coefficient (which is -1)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 2 and -1 to get 1. Place the sum right underneath 2.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 1 by 1 and place the product (which is 1)  right underneath the third  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add 1 and 1 to get 2. Place the sum right underneath 1.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply 1 by 2 and place the product (which is 2)  right underneath the fourth  coefficient (which is -1)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR></TABLE>

    Add 2 and -1 to get 1. Place the sum right underneath 2.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD></TD></TR></TABLE>

    Multiply 1 by 1 and place the product (which is 1)  right underneath the fifth  coefficient (which is -1)

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD></TD></TR></TABLE>

    Add 1 and -1 to get 0. Place the sum right underneath 1.

    <TABLE cellpadding=10><TR><TD>1</TD><TD>|</TD><TD>2</TD><TD>-1</TD><TD>1</TD><TD>-1</TD><TD>-1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{x-1}}} is a factor of  {{{2x^4 - x^3 + x^2 - x - 1}}}


Now lets look at the bottom row of coefficients:


The first 4 coefficients (2,1,2,1) form the quotient


{{{2x^3 + x^2 + 2x + 1}}}



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


You can use this <a href=http://calc101.com/webMathematica/long-divide.jsp>online polynomial division calculator</a> to check your work


Basically  {{{2x^4 - x^3 + x^2 - x - 1}}} factors to {{{(x-1)(2x^3 + x^2 + 2x + 1)}}}


Now lets break  {{{2x^3 + x^2 + 2x + 1}}} down further






Now let's graph the function {{{f(x)=2x^3 + x^2 + 2x + 1}}}



{{{ graph( 500, 500, -10, 10, -10, 10,2x^3 + x^2 + 2x + 1   ) }}}



From the graph, we can see that there is a zero at {{{x=-1/2}}}. So our test zero is {{{-1/2}}}. So this time our test zero is 1






Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the function to the right of the test zero.<TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)

<TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1/2 by 2 and place the product (which is -1)  right underneath the second  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD></TD><TD></TD><TD></TD></TR></TABLE>

    Add -1 and 1 to get 0. Place the sum right underneath -1.

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>0</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -1/2 by 0 and place the product (which is 0)  right underneath the third  coefficient (which is 2)

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>0</TD><TD></TD><TD></TD></TR></TABLE>

    Add 0 and 2 to get 2. Place the sum right underneath 0.

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>0</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>0</TD><TD>2</TD><TD></TD></TR></TABLE>

    Multiply -1/2 by 2 and place the product (which is -1)  right underneath the fourth  coefficient (which is 1)

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>0</TD><TD>-1</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>0</TD><TD>2</TD><TD></TD></TR></TABLE>

    Add -1 and 1 to get 0. Place the sum right underneath -1.

    <TABLE cellpadding=10><TR><TD>-1/2</TD><TD>|</TD><TD>2</TD><TD>1</TD><TD>2</TD><TD>1</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>0</TD><TD>-1</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>2</TD><TD>0</TD><TD>2</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. This means {{{2x+1}}} is a factor of  {{{2x^3 + x^2 + 2x + 1}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (2,0,2) form the quotient


{{{2x^2 + 2}}}



Notice in the denominator {{{2x+1}}}, the x term has a coefficient of 2, so we need to divide the quotient by 2 like this:

{{{(2x^2 + 2)/2=x^2 + 1}}}

  

So {{{(2x^3 + x^2 + 2x + 1)/(2x+1)=x^2 + 1}}}


You can use this <a href=http://calc101.com/webMathematica/long-divide.jsp>online polynomial division calculator</a> to check your work


Basically  {{{2x^3 + x^2 + 2x + 1}}} factors to {{{(2x+1)(x^2 + 1)}}}


Now lets break  {{{x^2 + 1}}} down further




{{{x^2 + 1=0}}} Set the factor equal to zero



{{{x^2=-1}}} Subtract 1 from both sides



{{{x=0+-sqrt(-1)}}} Take the square root of both sides



Simplify


{{{x=i}}} or {{{x=-i}}}





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Answer:



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



{{{x=-1/2}}}, {{{x=1}}}, {{{x=2}}}, {{{x=i}}}, or {{{x=-i}}}