Determine all of the roots of the given equation:
x^3-2x^2-3x+10=0

- anonymous

Determine all of the roots of the given equation:
x^3-2x^2-3x+10=0

- schrodinger

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- anonymous

Complex, too?
@mmbuckaroos?

- anonymous

x=-2
x=2-i
x=2+i

- anonymous

Could you explaint o me how you got those so I can try my other problems please?

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- anonymous

Sorry, I'm a little low on time... if you bump your question, i'm sure someone else can help you. sorry!

- anonymous

ok thanks

- Mertsj

The possible rational roots are the factors of the constant term which is 10 divided by the factors of the coefficient of the term with the highest exponent which is the x^3 term. It's coefficient is 1. So the possible rational roots are:
\[\frac{\pm10}{1}, \frac{\pm5}{1}, \pm\frac{2}{1}, \frac{\pm1}{1}\]

- Mertsj

Use synthetic division to see if any of those are actually roots.

- Mertsj

Since we have reason to believe that -2 is a root, let's try it.

- Mertsj

|dw:1335664232988:dw|

- Mertsj

So we see that -2 is a root and now we must solve the quadratic equation x^2-4x+5=0 to find the other roots.

- anonymous

ok this is the part that confuses me this division. My book here makes it so confusing

- anonymous

like where do you get all of those numbers from? Sorry I just don't get this at all.

- Mertsj

|dw:1335664387666:dw|

- Mertsj

So you need help with synthetic division?

- Mertsj

Do you understand the part about the possible rational roots?

- anonymous

I actually think I figured out what you did to get those numbers now so nvm on that ? I understand the possible rational roots

- Mertsj

So you need help with the synthetic division process?

- anonymous

do i have to do the synthetic division for every possible rational root?

- Mertsj

No. Just until you find one where the remainder is 0. That indicates that you have found a factor. Then you can start working on the quotient to find more roots.

- anonymous

so it is a rootif it comes out equaling zero. ok so now that im following you we can keep going sorry about that

- Mertsj

|dw:1335664711981:dw|

- Mertsj

So since the second factor is a quadratic and it won't factor, we just plug it into the formula to find the roots.

- Mertsj

Do you have another problem? Perhaps if we start another one from scratch, it will help you understand. I can guide you.

- anonymous

yes one sec sorry

- anonymous

x^3-x^2-4x-6=0

- Mertsj

Can you list the possible rational roots?

- Mertsj

It is so hard to help someone who won't respond.

- anonymous

Sorry my computer shut down. Ok:
+/-1, +/-1/6
?

- anonymous

He's not online anymore.... btw. I don't think I can help you, though, sorry:/ bump your question i guess?

- anonymous

It's alright. I think I got it figured out. Feel bad I left him hanging there darn computer for some reason it doesn't ever like this site

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