Question
\( 1 \leftarrow \quad \) Factor completely. \( v^{2}-2 v+1-36 f^{2} \) \( v^{2}-2 v+1-36 f^{2}=\square \) (Factor completely.)
Ask by Dickson Garrett. in the United States
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( v^{2}-2v+1-36f^{2} = (v+6f-1)(v-6f-1) \)
Solution
Factor the expression \( v^{2}-2v+1-36f^{2} \).
Factor the expression by following steps:
- step0: Factor:
\(v^{2}-2v+1-36f^{2}\)
- step1: Calculate:
\(v^{2}-6vf-v+6fv-36f^{2}-6f-v+6f+1\)
- step2: Rewrite the expression:
\(v\times v-v\times 6f-v+6fv-6f\times 6f-6f-v+6f+1\)
- step3: Factor the expression:
\(v\left(v-6f-1\right)+6f\left(v-6f-1\right)-\left(v-6f-1\right)\)
- step4: Factor the expression:
\(\left(v+6f-1\right)\left(v-6f-1\right)\)
The expression \(v^{2}-2v+1-36f^{2}\) can be factored completely as \((v+6f-1)(v-6f-1)\).
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Beyond the Answer
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!
