Question 89872
Lets use the rational root theorem to find any possible roots


Rational Root Theorem

*[Tex \LARGE Roots=\frac{p}{q}] where p and q are factors to the last and first coefficients


So lets list the factors of 8


*[Tex \LARGE p=\pm1, \pm2, \pm4, \pm8, ]


Now let's list the factors of 1


*[Tex \LARGE q=\pm1, ]


Now lets divide them



*[Tex \LARGE \frac{1}{1}, \frac{2}{1}, \frac{4}{1}, \frac{8}{1}, \frac{-1}{1}, \frac{-2}{1}, \frac{-4}{1}, \frac{-8}{1},]


Now simplify


*[Tex \LARGE 1, 2, 4, 8, -1, -2, -4, -8, ]



So these are possible zeros. To find out which possible zero is actually a zero, you need to perform synthetic division on each one.



Lets test -2 as a zero




So our test zero is -2



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>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</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>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</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 -2 by 1 and place the product (which is -2)  right underneath the second  coefficient (which is 6)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</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 -2 and 6 to get 4. Place the sum right underneath -2.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD></TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Multiply -2 by 4 and place the product (which is -8)  right underneath the third  coefficient (which is 12)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD></TD><TD></TD></TR></TABLE>

    Add -8 and 12 to get 4. Place the sum right underneath -8.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD></TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>4</TD><TD></TD></TR></TABLE>

    Multiply -2 by 4 and place the product (which is -8)  right underneath the fourth  coefficient (which is 8)

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD>-8</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>4</TD><TD></TD></TR></TABLE>

    Add -8 and 8 to get 0. Place the sum right underneath -8.

    <TABLE cellpadding=10><TR><TD>-2</TD><TD>|</TD><TD>1</TD><TD>6</TD><TD>12</TD><TD>8</TD></TR><TR><TD></TD><TD>|</TD><TD></TD><TD>-2</TD><TD>-8</TD><TD>-8</TD><TD></TD></TR><TR><TD></TD><TD></TD><TD>1</TD><TD>4</TD><TD>4</TD><TD>0</TD></TR></TABLE>

Since the last column adds to zero, we have a remainder of zero. 



So {{{x=-2}}} is a zero. This means {{{x+2}}} is a factor of  {{{x^3 + 6x^2 + 12x + 8}}}


Now lets look at the bottom row of coefficients:


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


{{{x^2 + 4x + 4}}}



So {{{(x^3 + 6x^2 + 12x + 8)/(x+2)=x^2 + 4x + 4}}}



So we basically factored the expression


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



Now factor further



{{{(x+2)(x+2)(x+2)=0}}}


So the zeros are 


{{{x=-2}}} with a multiplicity of 3