How do you find all the zeros of #f(x)=x^4-5x^2-36#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Bdub Mar 10, 2016 #x={-3,-2,2,3}# Explanation: let #u=x^2# then #f(x)=x^4-5x^2-36 ->f(u)=u^2-5u-36# #u^2-5u-36=0# #(u-9)(u+4)=0# #u-9=0,u+4=0# #u=9 or u=-4# #x^2=9,x^2=4# #x=+-3,x=+-2# #x={-3,-2,2,3}# 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 4938 views around the world You can reuse this answer Creative Commons License