How do you find all zeros for #x^3 - 5x^2 + x - 5#? Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Alan P. May 17, 2016 #x =5# or #x=+-i# Explanation: #x^3-5x^2+x-5# can be factored as #color(white)("XXX")color(red)(""(x^3+x))-color(blue)(""(5x^2+5))# #color(white)("XXX")=color(red)(x(x^2+1))-color(blue)(5(x^2+1)# #color(white)("XXX")=(color(red)(x)-color(blue)(5))*(x^2+1)# For the zeroes: either #color(white)("XXX")(x-5)=0color(white)("XX")rarrcolor(white)("XX")x=5# or #color(white)("XXX")(x^2+1)=0color(white)("XX")rarrcolor(white)("XX")x=+-sqrt(-1)=+-i# 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 4128 views around the world You can reuse this answer Creative Commons License