How do you solve #x^2-56=(-x)# by factoring? Algebra Polynomials and Factoring Zero Product Principle 1 Answer Alan P. Oct 23, 2015 #x=-8# or #x=7# Explanation: #x^2-56 = (-x)# #rArr x^2+x-56 = 0# Factoring: #rArr (x+8)(x-7) = 0# Either #color(white)("XXX")(x+8) = 0 rArr x=-8# or #color(white)("XXX")(x-7)=0 rArr x=7# Answer link Related questions What is the Zero Product Principle? How to use the zero product principle to find the value of x? How do you solve the polynomial #10x^3-5x^2=0#? Can you apply the zero product property in the problem #(x+6)+(3x-1)=0#? How do you solve the polynomial #24x^2-4x=0#? How do you use the zero product property to solve #(x-5)(2x+7)(3x-4)=0#? How do you factor and solve #b^2-\frac{5}{3b}=0#? Why does the zero product property work? How do you solve #(x - 12)(5x - 13) = 0#? How do you solve #(2u+7)(3u-1)=0#? See all questions in Zero Product Principle Impact of this question 2424 views around the world You can reuse this answer Creative Commons License