Simplest Way of finding zeros of cubic eq when rational roots test fails

[aeromat]

Homework Statement

Find the zeros of the cubic equation:
y = x^3 -9x^2 + 15x + 30


How do we find the zeros of this? In this case, subbing in x-values that will make it equal 0 does not work.

 

[Char. Limit]

Well, for this one, you're either going to have to solve the (extremely complicated) cubic equation or use Newton's Method, because there isn't an integer or rational number that will solve this equation for y=0.

 

[vela]

Plotting the function is a good idea as well. At the least, you can obtain initial guesses to use with Newton's method.

 

[Lavabug]

Ruffini's method:
http://en.wikipedia.org/wiki/Ruffini's_rule

 

[gb7nash]

I would probably suggest using Newton's method (or bisection method if you can find a positive function value and a negative function value) Once you've found the first root, take f(x)/(x-r), where r is the root. This will roughly yield a quadratic and you can use the quadratic formula to find the remaining two roots.

 

[Char. Limit]

I would probably suggest using Newton's method (or bisection method if you can find a positive function value and a negative function value) Once you've found the first root, take f(x)/(x-r), where r is the root. This will yield a quadratic and you can use the quadratic formula to find the remaining two roots.

If he's restricting himself to real numbers, there ARE no other roots. Just the one.

 

[gb7nash]

I'm missing something here. How do we know there's only one real root?

 

[Char. Limit]

Oh, sorry. I graphed it first, to get an impression of where the roots were.

 

[gb7nash]

Cheater.

 

[Char. Limit]

Real mathematicians use pictures!

 

[Dick]

I'm missing something here. How do we know there's only one real root?

You can also conclude there is only one root by looking at the derivative and finding the extreme values plus knowing the behavior as x->+/-infinity. Which is basically 'graphing it' without a calculator. Hence, not cheating.

 

[gb7nash]

You can also conclude there is only one root by looking at the derivative and finding the extreme values plus knowing the behavior as x->+/-infinity. Which is basically 'graphing it' without a calculator. Hence, not cheating.

Good call. This is a good method.

 

[aeromat]

Uh.. That was the first time I ever heard any of those methods, so could anyone give me a few pointers just to show how I start off with one of the methods above?

 

[Char. Limit]

Well, Newton's method is given as such: given a guess x₀ as to where the root of a function f(x) is, then...

x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}

x₁ is a better guess, and x₂ will be a better guess than x₁, and so on.

NOTE: With an initial guess close to the answer, you can get a very good approximation after only two or three iterations of this. For example, in this problem, an initial guess of x₀=-1 will get you a very good approximation in three iterations.

 

[Dick]

If you don't know numerical techniques and you are good with a calculator or computer program there is nothing wrong with bisection either. If f(x)=x^3−9x^2+15x+30 then f(-2) is negative and f(0) is positive. That means there is a root in between. Check the mean at (-2+0)/2=(-1). Depending on the sign of f(-1) keep splitting the interval in half until you get whatever precision you want. You'll never get it exact, but do you need to?