SOLUTION: use synthetic division to show that the given x value is a zero of the polynomial. Then find all other zeros.? P(x)=x^ 3 -2x^ 2 -5x+6; x = 1

Instead of doing your problem for you, I'll do another one that is
EXACTLY like yours, step by step.  Use it as a guide to do yours. The
problem I will do is this one:

We write down the coefficients of P(x) in a row, and the 2 from x=2
to the left



  2 | 1  -3  -10  24
    |               

Bring the 1 down below the bottom line:

 2 | 1   -3 -10   24
   |                
     1

Multiply the 1 on the bottom by the 2 at the left, getting 2.
Then write it ABOVE and TO THE RIGHT of the 1 on the bottom, 
immediately under the -3, like this:

 2 | 1 -3 -10  24
   |    2        
     1

Add the -3 and the 2, getting -1, and write it below the line,
like this:

 2 | 1 -3 -10  24
   |    2        
     1 -1

Multiply the -1 on the bottom by the 2 at the left, getting -2.
Then write it ABOVE and TO THE RIGHT of the -1 on the bottom, 
immediately under the -10, like this: 
 
 2 | 1 -3 -10  24
   |    2  -2    
     1 -1

Add the -10 and the -2 under it, getting -12, and write it below the line,
like this:

 2 | 1 -3 -10  24
   |    2  -2    
     1 -1 -12 

Multiply the -12 on the bottom by the 2 at the left, getting -24.
Then write that -24 ABOVE and TO THE RIGHT of the -12 on the bottom,
under the 24, like this:

 2 | 1 -3 -10  24
   |    2  -2 -24
     1 -1 -12  

Add the 24 and the -24 under it, getting 0, and write the 0 below the line,
under the 24, like this:

 2 | 1 -3 -10  24
   |    2  -2 -24
     1 -1 -12   0

That completes the synthetic division.  Now we must interpret what we have.
The right-most number 0 on the bottom is the REMAINDER.

The fact that this REMAINDER number on the bottom right came out to be 0,
shows that the given x value of 2 is a zero of the polynomial. 

The numbers on the bottom, all except the remainder 0 on the bottom right, indicate the quotient.

We have factored this polynomial:

P(x) = x3 - 3x2 - 10x + 24

like this:

P(x) = (x - 2)(1x2 - 1x - 12)

We erase the understood 1's

P(x) = (x - 2)(x2 - x - 12)

To find all the zeros, we set the factored polynomial equal to 0

       (x - 2)(x2 - x - 12) = 0         

We use the zero-factor property, solve the x-2=0 for the zero 2
that we already know we had, and factor the quadratic trinomial
inside the second parentheses:

        x - 2 = 0;    x2 - x - 12 = 0
            x = 2;    (x - 4)(x + 3) = 0
                      x - 4 = 0;   x + 3 = 0
                          x = 4        x = -3

So the other zeros besides the 2 are:   4 and -3.

Now do your problem the exact same way, step by step.

Edwin