How do you solve #3x^7 - 243x^3=0#?

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Answer 1 · 2015-12-12T04:11:42

#x=0,3#

First thing you want to do is factor. You can factor out #3x^3#

#(3x^3)(x^4-81)=0#

Now set both factors to zero

#3x^3=0#

#x^3=0#

#root(3)(x^3)=root(3)0#

#x=0#

#x^4-81=0#

#x^4=81#

#root(4)(x^4)=root(4)81#

#x=3#

Answer 2 · 2015-12-12T04:33:28

#x=0,+-3#

Factor out a common #3x^3#.

#3x^3(x^4-81)=0#

Recognize a difference of squares.

#3x^3(x^2+9)(x^2-9)=0#

Recognize another difference of squares.

#3x^3(x^2+9)(x+3)(x-3)=0#

Set each part equal to #0#.

#3x^3=0#
#x=0#

#x^2+9=0#
#x^2=-9#
#x=+-3i# (NONREAL ANSWERS)

#x+3=0#
#x=-3#

#x-3=0#
#x=3#

Thus, #x=0,+-3#.

Look at a graph:

graph{3x^7-243x^3 [-12.73, 12.58, -6.17, 6.48]}

The #x#-intercepts (the roots) are when #x=0,+-3#.