How do you simplify #(2t-2)/(1-t)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Shwetank Mauria Aug 2, 2016 #(2t-2)/(1-t)=-2# Explanation: Observe that when we take #2# common from numerator, we are left with #t-1#, which is exactly negative of denominator i.e. denominator is #(-1)(t-1)#. Hence using this #(2t-2)/(1-t)# = #(2(t-1))/((-1)(1-t))# = #(2cancel((t-1)))/((-1)cancel((1-t)))# = #2/-1=-2# Answer link Related questions What is an example of long division of polynomials? How do you do long division of polynomials with remainders? How do you divide #9x^2-16# by #3x+4#? How do you divide #\frac{x^2+2x-5}{x}#? How do you divide #\frac{x^2+3x+6}{x+1}#? How do you divide #\frac{x^4-2x}{8x+24}#? How do you divide: #(4x^2-10x-24)# divide by (2x+3)? How do you divide: #5a^2+6a-9# into #25a^4#? How do you simplify #(3m^22 + 27 mn - 12)/(3m)#? How do you simplify #(25-a^2) / (a^2 +a -30)#? See all questions in Division of Polynomials Impact of this question 1445 views around the world You can reuse this answer Creative Commons License