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What is the best method to solving a cubic? watch

1. How do you solve:
X^3 + 2X^2 + X - 100000 = 0.

There are no factors to take out.

What's the best method of solving this? Thank you.

Edit: This is a real one.

2. (Original post by Konnichiwa)
How do you solve:
X^3 + 2X^2 + X - 100000 = 0.

There are no factors to take out.

What's the best method of solving this? Thank you.
Where did you get this from? Can you post the original question?
3. That's one you've created right? Because it doesn't have three real roots. Have you covered complex numbers yet?

Easiest way I do it is using the following steps:

1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root.

3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

You will either be given a solution so you can skip that first step or, like I said, it won't take too long to find your first solution through trial and error.
4. Using Wolfram Alpha :P
5. Cubic formula!

Don't try this one at home!
6. You need to find a value for which that equation is equal to zero. Usually you just do this by trial and error, so if this is from a textbook or something, it's unlikely that you will have to look further than about ±5. Then use synthetic division to turn this into a quadratic, and solve that. You then have values for x from the quadratic and the one you found by trial and error. Check with wolfram alpha
7. (Original post by claret_n_blue)
That's one you've created right? Because it doesn't have three real roots. Have you covered complex numbers yet?

Easiest way I do it is using the following steps:

1) By trial and error, you find a solution such that your function is 0. So let's say you have some function, f(x). You want to find a value for x, such that f(x) = 0.

This is normally just done by guessing, but they tend to make the numbers easy for this. So you would really just try like x = ±1, ±2, ±3 and so on. You wouldn't need to go too far until you find your solution.

2) Let's say that at some point 'a', you get f(a) = 0. What this means is that x = a is solution to your function (i.e a point where it crosses the 'x' axis) and so we can say that (x - a) = 0 is a root.

3) Now you have two options. You can either use the long division way and divide f(x) by (x - a) and this will give you a quadratic. Either that or you can compare coefficients but i don't use this method so don't know how to explain it. You can then factorise this quadratic in the normal way.

You will either be given a solution so you can skip that first step or, like I said, it won't take too long to find your first solution through trial and error.
It's a real one. Thanks.
8. (Original post by Konnichiwa)
It's a real one. Thanks.
What's the original question? You're not going to be expected to solve cubics like that except via numerical approximation!

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