find the real number solutions 3x^7-243x^3=0
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write3,001 answers
starTop subjects are Math, Science, and Business
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:
Related Questions
- Solve x^3 - 3x + 8 = 0 for real solutions.
- 1 Educator Answer
- Solve 3x^7-243x^3=0 for real solutions.
- 1 Educator Answer
- Solve equation for real solution x^3-1=0
- 1 Educator Answer
- First, find the maximum number of real zeros. Then, use Decartes' Rule of Signs to determine the...
- 1 Educator Answer
- Find all solutions of x^4 - 3x^2 + 2 = 0?
- 1 Educator Answer
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.