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In order to do this, we need to recognize that factors of the equation always take the form:
`p(x-a)(x-b)(x-c)... = 0`
Where `p` is the coefficient of the term of highest degree (the highest power of `x`). In our case, for example, `p=1` so we're going to have it pretty easy.
Going back to the form of our equation that'll give you easy roots:
`p(x-a)(x-b)(x-c) = 0`
You can FOIL each term and get the following:
`p(x^2-(a+b)x + ab)(x-c) = 0`
`p(x^3-(a+b+c)x^2 + (ab+bc+ac)x - abc)= 0`
Since our `p=1` we can just get rid of that part to get our final full cutbic form:
`x^3-(a+b+c)x^2+(ab+bc+ac)x - abc = 0`
If you'll look at the last term you'll notice that it is the product of each root! Therefore, if you factor the constant term, you'll find the possible rational roots of the function!
So, our constant term is 2, so our factors will be:
Other than that, there will be no rational roots! It's actually quite beautiful in that respect.
Notice, though, that this is by no means the general case. If we had a factor in front of the `x^3` term, we would be dealing with something completely different, where we have each factor of the constant term divided by factors of `p`.
However, it looks like we're good!
Your possible roots will be: `+-1, +-2`
I hope that helps!
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