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
