Multiply out the trinomials:
x4-2x3-7x2+8x+7<-5,
x4-2x3-7x2+8x+12<0.
Now let's see if we can factorise. The factors of 12 are 2, 3, 4, 6 as well as 12 and 1.
It's always worth trying x=1 or -1, that is x=±1, so we have:
1-2-7+8+12=12 or 1+2-7-8+12=0. So x=-1 is a zero. We can use synthetic division to find the cubic:
-1 | 1 -2 -7 8 12
1 -1 3 4 | -12
1 -3 -4 12 | 0 = x3-3x2-4x+12.
This time x=±1 won't give us zero, so let's move on to the next factor: 2.
x=-2: -8-12+8+12=0, so x=-2 is a zero.
x=2: 8-12-8+12=0, so x=2 ia also a zero.
We can again use synthetic division. Let's use 2.
2 | 1 -3 -4 12
1 2 -2 | -12
1 -1 -6 | 0 = x2-x-6=(x-3)(x+2).
So we have four zeroes: -2, -1, 2, 3.
We have five regions to investigate:
(1) x<-2
(2) -2<x<-1
(3) -1<x<2
(4) 2<x<3
(5) x>3
The polynomial we're testing is factorised: (x+2)(x+1)(x-2)(x-3)<0.
Region (1) makes this >0.
Region (2) (+)(-)(-)(-)<0.
Region (3) (put x=0) (+)(+)(-)(-)>0.
Region (4) (+)(+)(+)(-)<0.
Region (5) (+)(+)(+)(+)>0.
Therefore regions (2) and (4) are both negative so the solution is -2<x<-1 or 2<x<3.