p4 - 6p³ + 7p² + 7p - 3

If there are any rational zeros (roots) they must be ± a factor of 
the absolute value of the contant term, |-3| or 3

All candidates for roots are 1,-1,3,-3

We try 1, using synthetic division:

1|1 -6  7  7 -3
 |   1 -5  2  9
  1 -5  2  9  6

No, 1 is not a zero (root) for that remainder 
is 6, not 0

We try -1

-1|1 -6  7   7 -3
  |  -1  7 -14  7
   1 -7 14  -7  4

No, -1 is not a zero (root) for that remainder 
is 4, not 0

We try 3

3|1 -6  7  7 -3
 |   3 -9 -6  3
  1 -3 -2  1  0

Yes, 3 is a zero (root) for that remainder 
is 0.  And the numbers across the bottom except
for the last one, the 0, are the coefficients
of the quotient,  So we have factored

p4 - 6p³ + 7p² + 7p - 3  as

(p - 3)(p³ - 3p² - 2p + 1)

So the binomial factor is p - 3.

Edwin