Been a while since I've done synthetic division.
1a. Let's assume that's supposed to be an equals sign
p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6
possible rational roots have factors of 6 in the numerator and of 2 in the denominator. We'll only worry about negative numerators.
Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6
Factors of 2: 1,2
Possible rational roots:
(dividing by 1:) 1,-1,2,-2,3,-3,6,-6
(dividing by 2:) 1/2, -1/2 (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2
Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2
Synthetic division, trying x=1,
1 | 2 -3 -6 5 6
2 -1 -7 -2
2 -1 -7 -2 4
Got a remainder of 4, so 1 isn't a root;
Trying x=-1
-1 | 2 -3 -6 5 6
-2 5 1 -6
2 -5 -1 6 0
Zero remainder, found a root, x=-1. This division says
(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6
Same set of rational roots on the cubic, we continue with x=2
2 | 2 -5 -1 6
4 -2 -6
2 -1 -3 0
Another zero remainder, x=2 is a root. We're left with 2x² - x - 3 = 0, which factors as
(2x - 3)(x + 1) = 0
That's a second factor of x+1. Our final factorization is
2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)
Fourth degree with a positive leading coefficient so goes to +infinity at both ends. Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.
I'll leave the actually sketching and the other two polynomials to you -- that took some time.