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Third order Characteristic Equation
- Thread starter lilshai
- Start date
Hello, lilshai!
Are you familiar with the Rational Roots Theorem and the Factor Theorem?
For this equation, if there are any rational roots,
. . they must be factors of 6 (the constant term).
The only choices are: ±1, ±2, ±3, ±6
We're in luck . . . We find immediately that r = 1 works.
. . Hence, (r - 1) is a factor of the polynomial.
Using long division, we have: .(r - 1)(r<sup>2</sup> - 4r + 3) .= .(r - 1)(r - 1)(r - 3)
Are you familiar with the Rational Roots Theorem and the Factor Theorem?
You have: .r<sup>3</sup> - 6r<sup>2</sup> - 11y - 6 .= .0y"' - 6y" + 11y' - 6y = 0. How do I factor this?
For this equation, if there are any rational roots,
. . they must be factors of 6 (the constant term).
The only choices are: ±1, ±2, ±3, ±6
We're in luck . . . We find immediately that r = 1 works.
. . Hence, (r - 1) is a factor of the polynomial.
Using long division, we have: .(r - 1)(r<sup>2</sup> - 4r + 3) .= .(r - 1)(r - 1)(r - 3)
Hello,
Thanks for the help. Now I think I remember how to do it. I think you copied the equation incorrectly though. It should be r^3-6r^2+11y-6=0. (r-1) is still a factor but you get r=1, 2, and 3 as roots instead after long division. Thanks for the refresh on Rational Roots/Factor Theorem. :wink:
Thanks for the help. Now I think I remember how to do it. I think you copied the equation incorrectly though. It should be r^3-6r^2+11y-6=0. (r-1) is still a factor but you get r=1, 2, and 3 as roots instead after long division. Thanks for the refresh on Rational Roots/Factor Theorem. :wink: