Question 90246
When we graph, we find a zero at x=-1


So 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 numerator to the right of the test zero.<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>1</TD><TD>-3</TD><TD>-5</TD><TD>-2</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 1)

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

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

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

    Multiply -1 by 0 and place the product (which is 0)  right underneath the third  coefficient (which is -3)

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

    Add 0 and -3 to get -3. Place the sum right underneath 0.

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

    Multiply -1 by -3 and place the product (which is 3)  right underneath the fourth  coefficient (which is -5)

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

    Add 3 and -5 to get -2. Place the sum right underneath 3.

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

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

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

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

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>1</TD><TD>-3</TD><TD>-5</TD><TD>-2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>0</TD><TD>3</TD><TD>2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>0</TD><TD>-3</TD><TD>-2</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  {{{x^4 + x^3 - 3x^2 - 5x - 2}}}


Now lets look at the bottom row of coefficients:


The first 4 coefficients (1,0,-3,-2) form the quotient


{{{x^3 - 3x - 2}}}



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



-------------------------------------------------------------------

Now use the same test zero (which is -1) and perform synthetic division on {{{x^3 - 3x - 2}}}



Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from {{{1x^3}}} to {{{-3x^1}}} there is a zero coefficient for {{{x^2}}}. This is simply because {{{x^3 - 3x - 2}}} really looks like {{{1x^3+0x^2+-3x^1+-2x^0}}}<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-3</TD><TD>-2</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 1)

<TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-3</TD><TD>-2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD></TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></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 second  coefficient (which is 0)

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

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

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

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

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

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

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

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

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

    <TABLE cellpadding=10><TR><TD>-1</TD><TD>|</TD><TD>1</TD><TD>0</TD><TD>-3</TD><TD>-2</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-1</TD><TD>1</TD><TD>2</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>-1</TD><TD>-2</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  {{{x^3 - 3x - 2}}}


Now lets look at the bottom row of coefficients:


The first 3 coefficients (1,-1,-2) form the quotient


{{{x^2 - x - 2}}}



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


Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{x^2-x-2=0}}} ( notice {{{a=1}}}, {{{b=-1}}}, and {{{c=-2}}})


{{{x = (--1 +- sqrt( (-1)^2-4*1*-2 ))/(2*1)}}} Plug in a=1, b=-1, and c=-2




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




{{{x = (1 +- sqrt( 1-4*1*-2 ))/(2*1)}}} Square -1 to get 1  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)




{{{x = (1 +- sqrt( 1+8 ))/(2*1)}}} Multiply {{{-4*-2*1}}} to get {{{8}}}




{{{x = (1 +- sqrt( 9 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (1 +- 3)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




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


So now the expression breaks down into two parts


{{{x = (1 + 3)/2}}} or {{{x = (1 - 3)/2}}}


Lets look at the first part:


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


{{{x=4/2}}} Add the terms in the numerator

{{{x=2}}} Divide


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(1 - 3)/2}}}


{{{x=-2/2}}} Subtract the terms in the numerator

{{{x=-1}}} Divide


So another answer is

{{{x=-1}}}


So our solutions are:

{{{x=2}}} or {{{x=-1}}}


===================================


Answer:


So our zeros are: 

{{{x=-1}}} (with a multiplicity of 3), {{{x=2}}}