How do you solve #5x^{5} - 80x = 0#?

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Answer 1 · 2016-09-26T20:33:15

Looks like you're finding zeros, so factor

Take out a GCF (Greatest Common Factor)
#5x (x^4 - 16) = 0#

Then, set #5x# equal to zero
#5x = 0#

Solve for #x# by dividing the #5#
#x=0#

Set #x^4 - 16# equal to zero and solve
#x^4 - 16 = 0#
#x^4 = 16#
#x = +-2#

Answer:
#x = 0, x = +-2#

Answer 2 · 2016-09-26T20:34:29

#x=0" or " x = +2" or "x =-2#

#5x^5 -80x = 0" "larrdiv 5#

#x^5 - 16x = 0" "larr# take out common factor of #x#

#x(x^4-16)=0" "larr# factor further by diff of squares

#x(x^2+4)(x^2-4) =0" "larr# factor further by diff of squares

#x(x^2+4)(x+2)(x-2)=" "larr# there are 4 factors.

Each factor could be equal to 0

#x = 0#
#x+2=0 " "rarr x = -2#
#x-2=0" "rarr x =+2#
#x^2 +4 =0" "x^2 = -4" "larr# there is no real solution

#x=0" or " x = +2" or "x =-2#