For part (a), you might find it easier to take out the factor of $4$ at the very beginning and work with

For part (b), the number we want is found by letting $x=99$ - if we put this into the factorised form that we found in part (a), this is much easier to work out.

1. First, we expand and simplify

   $$\begin{array}{rl}& 4(x+2{)}^2+8(x+1)-4\\ =& 4[(x+2{)}^2+2(x+1)-1]\\ =& 4[x^2+4x+4+2x+2-1]\\ =& 4[x^2+6x+5]\\ =& 4(x+5)(x+1)\end{array}$$

2. By inspection, we notice that $4\times {101}^2+796$ is given by $4(x+2{)}^2+8(x+1)-4$ when $x=99$. This is much easier to caclulate by hand in the factorised form we calculated in part (a).

   $$4\times (104)\times (100)=41600$$