Hello, I have tried solving this using the rational roots theorem and none of the roots seem to be working. I'm trying to figure out where I went wrong.
3x^3 -2x^2 - 7x - 4 = 0
you are right, it does not factor over the rational numbers.
here is a webpage that solves cubics for you, just enter the constants
it gave me 1 irrational root and 2 complex roots.
Thanks for the help. However, I don't see the webpage you were speaking of.
how silly of me
http://www.1728.com/cubic.htm
Thanks!
To solve the equation 3x^3 - 2x^2 - 7x - 4 = 0 using the Rational Roots Theorem, you would first list all the possible rational roots of the equation. The rational roots of a polynomial equation with integer coefficients are given by the fractions p/q, where p is a factor of the constant term (in this case, 4) and q is a factor of the leading coefficient (in this case, 3). However, based on your statement, it appears that none of the rational roots you have tried are solving the equation.
If the Rational Roots Theorem does not yield any rational roots, it's possible that the equation doesn't have any rational solutions. In this case, you can try using other numerical methods, such as graphing the equation or using numerical approximation methods like Newton's method or the bisection method. These methods can help you find approximate solutions for the equation.
To graph the equation, you can plot the points and observe where the curve intersects the x-axis. This can give you an approximate idea of where the roots might lie.
Another approach is to use numerical approximation methods like Newton's method or the bisection method. These methods involve iterative calculations to refine the estimate of the root. Newton's method utilizes calculus to find the root, while the bisection method is a simpler iterative technique.
If you provide information on the specific method you used or any particular steps you took, I can try to help you identify where you may have gone wrong.