Asked by Larry
Do not solve, show possibilities of Descartes's rule of signs for roots.
4x^3-9x^2+6x+4=0
f(x)=4x^3-9x^2+6x=4
- + +
one change
=4(-x)^3-9(-x)^2+6(-x)+4
- + - +
three changes
Positive Negative Imaginary
2 1 0
1 2 0
3 possible roots
4x^3-9x^2+6x+4=0
f(x)=4x^3-9x^2+6x=4
- + +
one change
=4(-x)^3-9(-x)^2+6(-x)+4
- + - +
three changes
Positive Negative Imaginary
2 1 0
1 2 0
3 possible roots
Answers
Answered by
Anonymous
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