SOLUTION: How do you find all real numbers in this solution?x^3+6x^2+12x +8=0

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Lets use the rational root theorem to find any possible roots

Rational Root Theorem
where p and q are factors to the last and first coefficients

So lets list the factors of 8



Now let's list the factors of 1



Now lets divide them




Now simplify




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.

-2	|	1	6	12	8
	|

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

-2	|	1	6	12	8
	|
		1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 6)

-2	|	1	6	12	8
	|		-2
		1

Add -2 and 6 to get 4. Place the sum right underneath -2.

-2	|	1	6	12	8
	|		-2
		1	4

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

-2	|	1	6	12	8
	|		-2	-8
		1	4

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

-2	|	1	6	12	8
	|		-2	-8
		1	4	4

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

-2	|	1	6	12	8
	|		-2	-8	-8
		1	4	4

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

-2	|	1	6	12	8
	|		-2	-8	-8
		1	4	4	0

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


So is a zero. This means is a factor of

Now lets look at the bottom row of coefficients:

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




So


So we basically factored the expression




Now factor further




So the zeros are

with a multiplicity of 3