How do you find all zeros of #y=x^3+x^2-9x-9#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Gió May 12, 2015 Set #y=0# and solve for #x#: #x^3+x^2-9x-9=0# Collecting #x# from the first and third term: #x(x^2-9)+x^2-9=0# #(x^2-9)(x+1)=0# So: #x^2-9=0# #x=+-sqrt(9)=+-3# and #x+1=0# Giving: #x_1=3# #x_2=-3# #x_3=-1# Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 4398 views around the world You can reuse this answer Creative Commons License