If (2t+1)is a factor then t=2−1 must be one of the zeroes of the given polynomial. Substituting in the given equation.

q((2−1))=4(2−1)3+4(2−1)2−(2−1)−1=4(8−1)+4(41)+(21)−1=(2−1)+1+(21)−1=0

Hence(2t+1) is a common factor of q(t)