How do you solve #5x^2= 4-19x# by factoring? Algebra Polynomials and Factoring Zero Product Principle 1 Answer Konstantinos Michailidis Sep 29, 2015 See explanation Explanation: We have that #5x^2+19x-4=0=>5x^2+20x-x-4=0=>5x(x+4)-(x+4)=0=>(5x-1)(x+4)=0=>x=-4 or x=1/5# 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 1746 views around the world You can reuse this answer Creative Commons License