roots of a polynomial

xc630

Junior Member
Joined
Sep 1, 2005
Messages
164
Hello. I need help finding all real and imaginary roots as well as multiple roots of P (x) =0. The polynomial is P (x)= 2x^5-3x^4-2x^3+7x^2-2x-6. I would appreciate any help. Thanks.
 
How far have you gotten? You've applied the Rational Roots Test to get the list of potential zeroes, right?

. . . . .±1, 2, 3, 6, 1/2, 3/2

How far have you gotten with synthetic division, testing which potential zeroes are actual zeroes? Have you checked the graph, to help you narrow down which potential zeroes to try?

Eliz.
 
Hi. I tried syntehtic division and none of them equal 0. What should I do?
 
You must be doing something wrong in your division. Try 3/2 and -1 again. If you STILL don't get zero (0), you'll have to show us EXACTLY how you are doing it.
 
This is an odd-degree polynomial. You know that it has to cross the x-axis somewhere. Check your work on the synthetic division, because three of the zeroes can be found quite easily by this method.

Eliz.
 
My mistake I forgot to put the negative sign on four of the numbers. Ok -1 and 3/2 did result in 0.
 
Now look at the graph to find another zero, a second copy of one of the zeroes you've already found. (Hint: Where the graph just touches the x-axis, instead of crossing it, the zero at that spot occurs twice... or four times, or six times, or some other even number of times. So any time the calculator graph shows the line reaching the axis and then turning around, check for a repeated zero.)

Note: Once you've removed three of the zeroes, you'll be left with a quadratic, to which you can apply the Quadratic Formula. (I'd factor the "2" out first, though.)

Eliz.
 
The graph just shows the curve going up to -1, then coming down to about -6 at the y axis, and then going up through 1 1/2. There are no other points where it touches the x-axis.
 
xc630 said:
There are no other points where it touches the x-axis.
Yes, the graph only touches or crosses in two places. Now read my previous post to see which zero you should try again, to find the one that occurs twice.

Eliz.
 
So the zeroes so far are -1, -1, and 3/2. Now I am confiused on how you factor out the 2 and use the quadratic formula. I appreciate your help. :D
 
xc630 said:
I am confiused on how you factor out the 2 and use the quadratic formula. I appreciate your help.
You've taken three factors out already, leaving you with:

. . . . .2x<sup>2</sup> - 4x + 4

Divide the common factor of "2" out front. Then apply the Quadratic Formula you've memorized to find the other two zeroes.

Eliz.

. . .Simple factoring
. . .The Quadratic Formula
 
Additionally, how would I solve P (x) is greater or equal to 0 for the same polynomial? Thanks
 
Use the zeros you discovered already to define regions of the x-axis. Pick the ones that are graeter than zero.
 
The region greater than 1.5 on the x-axis is greater than 0. Is that how I would state the answer?
 
Is that the only region?

Normally, one would use little ">" or "<" symbols. Possibly the ones with the equal signs attached.

0 < x < 12

-3 >= x

x > 2 OR -3 < x < 0

Something like these examples.
 
Well, then I guess you're done. Note: If is said "greater than AND EQUAL TO", there would be one more point.
 
Top