How do you simplify # (y²-25) / (3y-7)#? Algebra Rational Equations and Functions Division of Polynomials 1 Answer Meave60 Nov 7, 2015 #(y^2-25)/(3y-7)=((y+5)(y-5))/(3y-7)# Explanation: #(y^2-25)/(3y-7)# Rewrite #25# as #5^2#. #(y^2-5^2)/(3y-7)# #(y^2-5^2)# is an example of the difference of squares, #a^2-b^2=(a+b)(a-b)#, where #a=y and b=5#. #((y+5)(y-5))/(3y-7)# 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 1410 views around the world You can reuse this answer Creative Commons License