Algebra 2 help!

[trytej]

So I'm doing some shit that I don't understand at all right know, and I'm stumped on one of the problems...

Let P(x) = x^4-2x^3+17x^2-50x-200. find all roots of p(x), given that -5i is a root.

 

[corrosivefrost]

It's been a long time since I did any of this, but I think you might find some of the stuff at http://oakroadsystems.com/math/polysol.htm helpful.

 

[Icen]

x = -2, 4, -5i, 5i

 

[Oktoberfest]

Don't you use like the rational roots theorem to solve for a couple of the roots, then since imaginary roots always come in pairs, you know that 5i is also a root

 

[Logain8955]

All this has done is remind me how much from high school I've forgotten.



Damn. It feels weird knowing that I used to be *good* at this stuff too . Now I look at it "Uh....3.50?"

 

[vietgurl]

What Icen said. If you take the roots of the constant (200), you get +/- 1, 2, 4, 5, 10, 20, 40, 50, 100. The factor(s) of the leading coefficient (x^4) is 1. So you divide the roots of the constant by the factor(s) of the leading coefficient. Then you use...what do you call that kind of division? Well, here's an example:

For -2
1 -2 17 -50 -200
-2 8 -50 200
-------------------------------
1 -4 25 -100 0

To do this, you take the coefficients in the equation starting with the leading coefficient. If there is a missing power of x (it doesn't go down chronologically), you substitute 0 for it. You take the number you're testing and multiply it by the first coefficient. Take that number and add it to the second coefficient. Repeat this. If you end up with a remainder of 0 at the end, then the number is a root.

When you have an unreal root (like the -5i that is given), then its conjugate is also a root.

It's 5 AM so sorry if my explanations are unclear or if I used the wrong terms, hehe.

 

[shieryda]

I put my root down.

 

[wubb]

[quote name='Logain8955']All this has done is remind me how much from high school I've forgotten.[/QUOTE]

I'm right there with ya.

 

[sblymnlcrymnl]

Damn, it's been ... 8 years since I took Algebra 2. :lol:[quote name='shieryda']I put my root down.[/QUOTE]

 

[afedock]

You are supposed to used the Rational Roots theorem to find a root, then use synthetic division with that root. You essentially factor out the root you found from the polynomial, and should be left with a more managable polynomial that can be factored easier, or use quadratic equation to get the rest of the roots.

 

[paynus]

[quote name='shieryda']I put my root down.[/quote]

that's pretty funny.

 

[Sleepkyng]

[quote name='shieryda']I put my root down.[/quote]

yeah, but how you gonna kick it?

that's always the question...

 

[shieryda]

[quote name='Sleepkyng']yeah, but how you gonna kick it?

that's always the question...[/quote]

Gonna kick it root down, of course!

 

[Squee]

what do you call that kind of division? Well, here's an example:

For -2
1 -2 17 -50 -200
-2 8 -50 200
-------------------------------
1 -4 25 -100 0

I've heard it called synthetic division before.

 

[alonzomourning23]

It's been 5 years since I've taken a math course, but I sucked at algebra so I probably would have given you the wrong answer anyway.

 

[vietgurl]

[quote name='Squee']I've heard it called synthetic division before.[/QUOTE]

THAT'S IT! Haha, I was half asleep when I was writing that and I barely remember any Algebra 2, hehe. Makes me sad knowing I've forgotten almost everything I've learned in less than few years.