a=3.
Here’s why:
(2x+1)(2x²-x+2)=4x³-2x²+4x+2x²-x+2=4x³+3x+2.
4x³+ax+2=(2x+1)(2x²+qx+2), where q is another constant, because we know that 4x³/2x=2x², and to get the constant 2 we need a constant 2/1=2. Let’s expand the parentheses:
4x³+2qx²+4x+2x²+qx+2. There’s no x² term so q=-1 to cancel out the x² term.
Therefore qx=-x. Add this to 4x and we get 4x-x=3x. This gives us a=3.
b) When x=-½, p(x)=0, when x>-½, p(x)>0. Note that 2x²-x+2>0 always.