find the real number solutions 3x^7-243x^3=0
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Eric Bizzell
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Find the real number solutions to `3x^7-243x^3=0` :
Factor out the greatest common factor:
`3x^3(x^4-81)=0`
Recognize `x^4-81` as a quadratic in `x^2` , and as a difference of two squares:
`3x^3(x^2+9)(x^2-9)=0`
Recognize `x^2-9` as the difference of two squares:
`3x^3(x^2+9)(x+3)(x-3)=0`
Setting each factor on the left side equal to zero (applying the zero product property) we get:
`3x^3=0 ==>x=0` is a root (of multiplicity 3)
`x^2+9=0==>x=+-3i` but these are not real
`x+3=0 ==>x=-3`
`x-3=0 ==>x=3`
Thus the real roots are 0,3,-3
The real roots correspond to the x-intercepts of the graph:
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