SOLUTION: How do i factor {{{x^3+3x^2-18x-40}}}

Since the leading coefficient is 1, if there
are any rational zeros, they will all be
± the integer factor of the absolute value of the
numerical term -40.  So if there are any, they will
be among these:

±1,±2,±4,±5,±8,±10,±20

We begin by trying +1 using synthetic division:

1 | 1  3  -18  -40
  |    1    4  -14
    1  4  -14  -54

So, 1 is not a zero, and so x-1 is not a factor.

We try -1:

-1 | 1  3 -18  -40
   |   -1  -2   20
     1  2 -20  -20

So, -1 is not a zero, and so x+1 is not a factor.

We try +2:

2 | 1  3 -18  -40
  |    2  10  -16
    1  5  -8  -56

So, 2 is not a zero, and so x-2 is not a factor.

We try -2:

-2 | 1  3 -18  -40
   |   -2  -2   40
     1  1 -20    0

Eureka!  We get a 0 remainder, so x+2 is a factor.
An the numbers on the bottom row of the synthetic
division left of the remainder 0 tell us the other
factor.  So we have now factored



as



Now we can factor the trinomial in
the second parentheses and we have the
complete factorization as:



Edwin