\( 1 \leftarrow \quad \) Factor completely. \( v^{2}-2 v+1-36 f^{2} \) \( v^{2}-2 v+1-36 f^{2}=\square \) (Factor completely.)

To factor the expression \( v^{2}-2v+1-36f^{2} \), first recognize that \( v^{2}-2v+1 \) is a perfect square trinomial. It can be rewritten as \( (v-1)^{2} \). Thus, the expression becomes: \[ (v-1)^{2} - 36f^{2} \] Next, notice that this is a difference of squares, which can be factored using the identity \( a^{2} - b^{2} = (a-b)(a+b) \): \[ ((v-1) - 6f)((v-1) + 6f) \] So, the completely factored form is: \[ (v - 1 - 6f)(v - 1 + 6f) \] Let’s check what this looks like when expanded to ensure we haven’t missed anything. When you distribute and combine like terms, you’ll confirm that the formulation is indeed equivalent to the initial expression provided!