How do you evaluate #v^ { - 2} \cdot v \cdot 2u ^ { 2} v ^ { - 2}#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Barney V. Mar 8, 2017 #(2u^2)/v^3# Explanation: #v^-2*v*2u^2v^-2# #:.v^-2*v^1*2u^2v^-2# #:.v^(-2+1-2)*2u^2# #:.v^-3*2u^2# #:.1/v^3*(2u^2)/1# #:.(2u^2)/v^3# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 1278 views around the world You can reuse this answer Creative Commons License