How do you simplify #(x-3)^2-4(x-3)+3#? Algebra Polynomials and Factoring Multiplication of Polynomials by Binomials 1 Answer Lucy Jun 17, 2018 #(x-6)(x-4)# Explanation: Let #u=x-3# Hence, #(x-3)^2-4(x-3)+3# becomes #u^2-4u+3# Solving for u: #(u^2-4u+3=(u-3)(u-1)# Therefore, subbing #u=x-3# back into the equation #(x-3-3)(x-3-1)=(x-6)(x-4)# Answer link Related questions What is FOIL? How do you use the distributive property when you multiply polynomials? How do you multiply #(x-2)(x+3)#? How do you simplify #(-4xy)(2x^4 yz^3 -y^4 z^9)#? How do you multiply #(3m+1)(m-4)(m+5)#? How do you find the volume of a prism if the width is x, height is #2x-1# and the length if #3x+4#? How do you multiply #(a^2+2)(3a^2-4)#? How do you simplify #(x – 8)(x + 5)#? How do you simplify #(p-1)^2#? How do you simplify #(3x+2y)^2#? See all questions in Multiplication of Polynomials by Binomials Impact of this question 1401 views around the world You can reuse this answer Creative Commons License