Question 697677
<pre>
p<sup>4</sup> - 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
 |<u>   1 -5  2  9</u>
  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
  |<u>  -1  7 -14  7</u>
   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
 |<u>   3 -9 -6  3</u>
  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

p<sup>4</sup> - 6p³ + 7p² + 7p - 3  as

(p - 3)(p³ - 3p² - 2p + 1)

So the binomial factor is p - 3.

Edwin</pre>