Question 1024496
<pre>
Yes it is factorable.  The other tutor thought
because it has no rational zeros (roots) that it
is not factorable.  That is not the case.  A 4th
degree polynomial could factor into the product 
of two quadratic trinomials both which have 
irrational zeros (roots).

Here is the correct approach:

(x-1)(x-2)(x+3)(x+4)+4

If we were going to multiply that out, we would
start by multiplying two pairs of those binomials
by FOIL.  We have a choice of 

1. multiplying the 1st by the 2nd and the 3rd by 
the 4th, or

2. multiplying the 1st by the 3rd and the 2nd by
the 4th, or

3. multiplying the 1st by the 4th and the 2nd by
the 3rd, or


 We investigate the 3 possibilities.

If we multiply the 1st by the 2nd and the 3rd by 
the 4th the middle terms would be -3x and 7x.  

If we multiply the 1st by the 3rd and the 2nd by 
the 4th, the middle terms would be 2x and 2x.
Both the same.

If we multiply the 1st by the 4th and the 2nd by 
the 3rd the middle terms would be 3x and x.

The 2nd way is the best plan for we get the 
same middle term. 

Rearrange the factors so that the ones we multiply
will be side by side:

(x-1)(x+3)(x-2)(x+4)+4

(x²+2x-3)(x²+2x-8)+4

Let u = x²+2x

(u-3)(u-8)+4

u²-11u+24+4

u²-11u+28

(u-7)(u-4)

Replace u by x²+2x

(x²+2x-7)(x²+2x-4)

Edwin</pre>