Am I missing any?

[moronatmath]

Find all of the zeros of the following polynomial
P(x)=x^4+x^3+2x-4


I got -2, and 1. I dont think -4 is a zero?

Am I missing any? Thank you!

 

[tkhunny]

You're guessing again. How did you find x = 1 and x = -2? Those are correct, but no one knows how you came up with them. After factoring out (x-1) and (x+2), what is left of your polynomial? I expect it is a quadratic polynomial, the roots of which are easily determined by the quadratic formula. do we get to use Complex Numbers or are we staying with Real Numbers?

Give more information. Your salvo of unexplained questions simply is not favorably viewable.

 

[pka]

Am I missing any? Thank you!

Yes you are.
There are two complex roots.
Because the polynomial is of degree 4, there should be four roots.

 

[moronatmath]

TK sorry I seemed to really have pissed you off. I didn't guess on the two I found. I used a graphing calculator.
I really am trying to learn this math. I did not skip any courses I just have a really terrible memory. Sorry.

 

[moronatmath]

When I graph the problem it seems to only pass over X twice. So according to tk I just need to plug that problem in quadratic formula to find the other two (roots)?

 

[pka]

In factored form we have: \(\displaystyle \L
x^4 + x^3 + 2x - 4 = (x - 1)(x + 2)(x^2 + 2)\).

The two factors \(\displaystyle \L
(x - 1)(x + 2)\) give us the two real roots, -2 & 1.

The factor \(\displaystyle \L
(x^2 + 2)\) gives us the two other complex roots \(\displaystyle \L
\sqrt 2 i\quad \& \quad - \sqrt 2 i\).

 

[stapel]

I didn't guess on the two [roots] I found. I used a graphing calculator.

Well, in a way, that is guessing; it's guessing from the pretty picture.

When I graph the problem it seems to only pass over X twice.

Yes...? I'm sorry, but I'm not clear on the point of this. Are you saying that you're only supposed to be finding the real-value roots? That you haven't yet covered complex numbers?

A direct answer to this question would be very helpful. Thank you.

Eliz.

 

[tkhunny]

TK sorry I seemed to

There is no anger, here. I'm just telling you what I think. I want to help, like everyone else, but sometimes it seems like we're not going the right direction. "Wing-and-a-Prayer" isn't my favorite method for learning.

 

[moronatmath]

I have another question that is related.

What if the problem said to find all "rational" zeros? Does that mean do not use the sqrt zeros?

 

[pka]

As you see a root could be real or complex.
Real numbers are either rational, the ratio of two integers, or irrational, ex. √2
So you are to find the roots that are the ratio of two integers.