The problem asks us to factor the polynomial $16a^2 - 8ab - 3b^2$.

1. Problem Description

The problem asks us to factor the polynomial $16a^2 - 8ab - 3b^2$.

2. Solution Steps

We are looking for two binomials of the form $(xa + yb)(za + wb)$ such that when multiplied, they yield the given polynomial.

We need to find $x, y, z, w$ such that

$(xa + yb)(za + wb) = xza^2 + (xw + yz)ab + ywb^2 = 16a^2 - 8ab - 3b^2$.

This means $xz = 16$, $xw + yz = -8$, and $yw = -3$.

Let's try $x = 4$ and $z = 4$. Then we have $4w + 4y = -8$, which simplifies to $w + y = -2$.

We also need $yw = -3$. We can try $y = 1$ and $w = -3$.

Then $yw = (1)(-3) = -3$ and $w + y = -3 + 1 = -2$, which satisfy our conditions.

So, $(4a + b)(4a - 3b) = 16a^2 - 12ab + 4ab - 3b^2 = 16a^2 - 8ab - 3b^2$.

Thus, the factorization is $(4a + b)(4a - 3b)$.

3. Final Answer

(4a+b)(4a-3b)