Question

find the complex zeros of f(x)=x^3-8x^2+25x-26. I have 2 so far

Answer 1

Which two do you have?

Answer 2

I meant I have 2 as a zero

Answer 3

Do you know synthetic division or "long" division of polynomials?

Answer 4

yeah, through synthetic division I got 2, but not sure how to get the 3+-2i I got stuck there

Answer 5

What are the numbers on the last row of the synthetic division? They are the coefficients of the quadratic equation you can solve to get the remaining pair of conjugate complex roots.

Answer 6

I got 2 as a zero. Please continue.

Answer 7

they are 1, -6, and 13. So solve x^2-6x+13 right? I forgot that it dropped to a quadratic so it can be solved like one

Answer 8

i meant -13

Answer 9

WAIT no its 13 sorry lol

Answer 10

You are correct. So use the QF to solve for the remaining two roots.

Answer 11

I have 3 -2i and then 3 + 2i but I have not checked them.

Answer 12

Questions?

Answer 13

Ok now I got it, 6+-4i/2 which simplifies to 3+-2i. Thanks!

Answer 14

Want to try another one?

Answer 15

y = x^3 - 2x^2 - x - 12

Answer 16

sure the next one's \[x ^{4}+2x^{3}+22x ^{2}+50x-75\] ill start and see how far I get

Answer 17

Okay, that's a quartic. Four roots to be accounted for. I'd try the Rational Root Theorem. Do you know Descartes' Rule of Signs?

Answer 18

Let's hope that the RR Theorem gives up two roots.

Answer 19

yeah im pretty sure, with descartes rule I have 4,2,or 0 real, 2 or 0 imaginary? I'm doing the rr theorem now

Answer 20

I have one sign change with Descartes for a guaranteed positive Real root. Let's hope it is rational. I'll check the negative Real possibilities.

Answer 21

With Descartes, I have 3 negative Real roots OR 1 negatives Real roots.

When you finish running the Rat Root Theorem, I would like to hear how you got your Descartes possibilities. Mine might be wrong.

Answer 22

when I looked at the original equation, all were positive besides -75, wouldn't that mean there are 4 possible positive roots? and when I plugged in a -x, two the x^3 and x changed to negative, so 2 possible negative roots?

Answer 23

There's only one sign change --> from + 50x to -75 for the positve Real root possibility.

Answer 24

When I substituted -x for x in the polynomial, the signs went:

+ - + - -

That's 3 negative Real roots or 1 Real root.

Answer 25

I just realized I was doing it backwards, positive roots are when the sign does change, for some reason I thought they had to stay positive. And I forgot to include the -75, I thought since it didnt have an x it wasn't included. I read my notes wrong lol

Answer 26

Consider -75 to be the term -75 x ^ 0. Don't ignore it. :)

Answer 27

How are the Rat Roots coming along?

Answer 28

Ok now I see where I went wrong with the negatives.. yeah the descartes rule is kind of tricky at first lol ok so now I know my possibilities, time to start finding zeros? I have all the factors of +-75, 12 possibilities

Answer 29

Try -3 first.

Answer 30

Sure is a zero. Now continue with the cubic equation I have left?

Answer 31

-3 could be a double root. You could try it again. I usually just pick one of the others to try. There's a theorem that helps cut down on the number of tries but no guarantees. I am uploading it in case you have not seen it.

Answer 32

http://www.mesacc.edu/~marfv02121/readings/bounds/index.html

Answer 33

You mean how to find the upper and lower bounds?

Answer 34

Yes, using synthetic division. I thought about it because of your "all positives" comment on Descartes' Rule.

Answer 35

Found that root yet? I always enjoy the search.

Answer 36

havent found another zero yet, -3 didnt turn out to be a double root

Answer 37

How many Rat Root candidates do you have left to test?

Answer 38

Are your candidates on this thread somewhere? I don't see them. Would you post them?

Answer 39

i have +-15,+-25,+-75 I had no luck with 5

Answer 40

For the numerators from p: I have +- 75, +- 25, +-,5, +-3, +-1

For the denominators, I have +-1

So, the candidates are : +- 75, +- 25, +-,5, +-3, +-1

Answer 41

how about 15? 15x5=75

Answer 42

And i had the rest too, sorry i didnt put in the ones I already tried

Answer 43

Oh, gosh. I missed that. Okay. Did you try 1 or -1?

Answer 44

Hey unfortunately I don't think I'll have time to finish, I have class at 9 but I will keep working on this until I get it right, I havent looked at the answers yet but I'll make sure I get it right. Thanks you've helped a lot!

Answer 45

Okay. One is a "good" number. Have a great day at school.