Help Finding Roots of Polynomial

  • Thread starter Thread starter theintarnets
  • Start date Start date
  • Tags Tags
    Polynomial Roots
AI Thread Summary
The discussion focuses on finding the roots of the polynomial f(x) = x^3 + 5x^2 - 8x + 2 by first identifying rational zeros and then using synthetic division. Possible rational zeros identified include 2, -2, 1, and -1, with synthetic division revealing a quotient of x^2 + 6x - 2. The correct roots are determined to be x = 1 and x = -3 ± √(11), clarifying the source of the additional root. Participants express differing opinions on the utility of synthetic division versus polynomial long division, with some preferring the former for its speed. Overall, the conversation emphasizes the importance of understanding the division process in polynomial root-finding.
theintarnets
Messages
64
Reaction score
0

Homework Statement


First find all rational zeros of f, then use the depressed equation to find all roots of the equation f(x) = 0.
f(x) = x^3 + 5x^2 - 8x + 2

Homework Equations


The Attempt at a Solution


Possible rational zeros: 2, -2, 1, -1
Synthetic division:

1 | 1 5 -8 2
_____1 6 -2
=============
1 6 -2 0

Quotient: x^2 + 6x - 2
Factored: (x + 3)^2 - 11

I would think that the answer would just be x = -3 ± √(11) but the answer in the book says: {1, -3 ± √(11)}
Where'd the 1 come from?
 
Physics news on Phys.org
Remember, you divided by (x-1), so your equation become (x-1)(x^2+6x-2)=0 -> x= 1 in addition to the other roots you found.
 
Ohhhhhhh! I see now, thank you!
 
synthetic division, oh lord
if you want a nicer division algorithm try this

or proper polynomial long division
... I really hated synthetic division :p
 
Last edited by a moderator:
I believe we've had this discussion before, but there's nothing wrong with synthetic division. It's quick and pretty straightforward, IMO. Of course, if you're trying to divide a polynomial by a quadratic or a higher degree polynomial, then long division is the way to go.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top