Find all zeros of the polynomial 6x4+17x3-2x2+x-6 = 0

We can find the zeros of a polynomial equation by using
synthetic division.

First check to see that the equation is
written in descending powers of the variable.

Then look at
the highest power in the equation.  This will tell how many solutions (zeros) there are
for the equation.

6x^4   +  17x^3  -  2x^2  +  x  -  6   =
  0

The degree of the equation is 4; therefore, we will
have 4 solutions.

Synthetic
Division:

Step 1:    Determine possible zeros by using
(P)/(Q).  The values for (P) will be from the factors of the last term which is -6.  The
values for (Q) will be from the factors of the coefficient of the first term, which is
6.

Possible zeros:  
±1,±2,±3,±6,±½,±⅓,±1/6

Step 2:    Write the
coefficients of the equation.  The goal of testing a possible zero value is to arrive at
a remainder of zero in the last column of your computations.   It will help to go in
order of the possible zeros beginning with 1 and continuing until you have found a zero
by the following process. By trial and error, I have already determined that -3 and 2/3
are zeros for this polynomial.  The following computations were used to identify the two
solutions.

Step 1:    Write the coefficients of the
terms.

Step 2:    Write the test zero to the far left of
the second column.

Step 3:    “Bring down” the first
coefficient

Step 4:   Multiply the first coefficient by the
test value (6 * -3) and place the product (-18) in the second column as shown.  Now,
find the sum of the two numbers (-1) and write it directly underneath the -18.  Multiply
the -1 by the -3 and write the product (3)  under the -2 in the third column.  Find the
sum (1).  Continue the pattern until the sum of the two digits in the last column is 0. 
This verifies that -3 is a remainder.

6        17       
-2        1        -6

-3*                                 
-18          3        -3        6

6         -1          
1        -2        0

Using the third row as new
coefficients and dropping the degree of the equation by one our new “depressed” equation
is

6x^3 -  x^2  +  x   -   2   = 
0

Use the coefficients from the new equation to continue
with the next test value as shown below.

6   
      -1          1        
-2

2/3*                            4           2         
2

6           3          3         
0

We have now depressed the equation twice.  Our new
equation is a quadratic.

6x^2  +  3x  +  3   = 
0

We can simplify the equation by factoring out 
3.

3(2x^2  +  x  +  1) =
0

Dividing by 3 on both sides, our simplified equation
is

2x^2  +  x  +  1  =  0

We
will use the quadratic formula to find the other 2
roots.

-b±sqr rt (b²-4ac) /
2a

From the coefficients of our quadratic equation we know
the following:  a=2; b=1; c=1

-1±sqr rt(1²-4(2)(1)) /
2(2)

After simplifying, our answer is  (
-1±√-7)/4

However, since we cannot have a negative square
root we extract the imaginary “i”

The two roots from the
quadratic  are    (-1±i√7)/4

The  four  zeros for our
original polynomial equation are as follows:  (-3,   2/3,  (-1±i√7)/4
)