How do you find all the zeros of #f(x)=x^5+4x^4+5x^3#?

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Answer 1 · 2016-02-26T07:13:45

#f(x) = x^5+4x^4+5x^3 =x^3(x+2-i)(x+2+i)#

Hence the zeros of #f(x)# are #x=0# and #x=-2+-i#.

First separate out the common factor #x^3#.

Then complete the square and use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=x+2# and #b=i#, as follows:

#f(x) = x^5+4x^4+5x^3#

#=x^3(x^2+4x+5)#

#=x^3(x^2+4x+4+1)#

#=x^3((x+2)^2-i^2)#

#=x^3((x+2)-i)((x+2)+i)#

#=x^3(x+2-i)(x+2+i)#

Hence the zeros of #f(x)# are #x=0# and #x=-2+-i#.