How do you solve #x^5-5x^3+4x=0#?

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Answer 1 · 2015-10-23T19:02:14

#x=0#, #x=1#, #x=2#, #x=-1#, and #x=-2#

Start off by factoring out an #x# as follows

#x(x^4-5x^2+4)=0#

We can factor #x^4-5x^2+4#

Doing so we have

#x(x^2-4)(x^2-1)=0#

So in order for this to be equal to zero

#x=0#

OR

#x^2-4=0#

OR

#x^2-1=0#

Well #x=0#

#x^2=4#

#x=+-sqrt(4)=+-2#

#x^2=1#

#x=+-sqrt(1)=+-1#