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

2 Answers
Sep 26, 2016

Looks like you're finding zeros, so factor

Explanation:

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

Sep 26, 2016

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

Explanation:

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