The problem asks to factorize the quadratic expression $3x^2 - 2x - 1$.

1. Problem Description

The problem asks to factorize the quadratic expression $3x^2 - 2x - 1$.

2. Solution Steps

We are looking for two binomials of the form $(ax+b)(cx+d)$ such that their product is $3x^2 - 2x - 1$.

Since the coefficient of $x^2$ is 3, we can assume $a=3$ and $c=1$ (or vice versa).

So we have $(3x+b)(x+d)$.

We need to find $b$ and $d$ such that $bd = -1$ and $3d + b = -2$.

Since $bd = -1$, one of $b$ and $d$ must be 1 and the other must be -

1.

Case 1: $b = 1$ and $d = -1$

Then $3d + b = 3(-1) + 1 = -3 + 1 = -2$. This works.

So the factorization is $(3x+1)(x-1)$.

Case 2: $b = -1$ and $d = 1$

Then $3d + b = 3(1) + (-1) = 3 - 1 = 2$. This does not work.

Therefore, the correct factorization is $(3x+1)(x-1)$.

We can check our result by expanding the expression:

$(3x+1)(x-1) = 3x(x-1) + 1(x-1) = 3x^2 - 3x + x - 1 = 3x^2 - 2x - 1$.

3. Final Answer

$(3x+1)(x-1)$