Find x in x^3+x^2-x-1=0 Equation

  • Thread starter Thread starter superelf83
  • Start date Start date
AI Thread Summary
To solve the equation x^3 + x^2 - x - 1 = 0, it is established that x = 1 is a solution, making (x - 1) a factor. By factoring out (x - 1), the remaining quadratic factor can be determined using methods such as Horner's rule or by equating coefficients. This involves setting up the equation (x - 1)(x^2 + Ax + B) and solving for A and B based on the coefficients of the original polynomial. The quadratic equation derived from this process can then be solved to find the other roots. The discussion highlights the complexity of the general cubic formula, which is not necessary for this specific problem.
superelf83
Messages
5
Reaction score
0
simple "find the x" question

how do you solve for x in this equation?

x^3+x^2-x-1=0

i know one of them is 1. but the other one...?
 
Physics news on Phys.org
If you know that x = 1 is a solution, then (x-1) is a factor of the polynomial. Factor it out and determine the remaining (quadratic) factor, e.g. using Horner's rule.
 
Another way to get the quadratic that is left is to write:
(x- 1)(x2+ Ax+ B)= x3+ Ax2+ Bx- x2- Ax- B= x3+ (A- 1)x2+ (B-A)x- B= x3+ x2- x- 1. In order for those to be equal for all x, corresponding coefficients must be the same: A- 1= 1, B- A= -1, -B= -1.
Solve for A and B and then solve the quadratic equation.

There is a general "cubic" formula but it is very compliciated.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

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
View full post »

Thread 'Greatest possible value of a constant in polynomial'

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
View full post »

Similar threads

Find integer points on this equation
Replies
8
Views
1K
Solve ##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##
Replies
24
Views
2K
Solve the simultaneous equations involving x, y, and their squares
Replies
2
Views
1K
Show that if ##x>1##, ##\log_e\sqrt{x^2-1}=\log_ex-\dfrac{1}{2x^2}-##
Replies
8
Views
2K
Solve the given equation that involves fractional exponent
Replies
4
Views
2K
Find all possible solutions of x^3 + 2 = 0
Replies
9
Views
2K
Solve the given simultaneous equations in x^2, x and y
Replies
1
Views
1K
Solve the problem involving the cubic function
Replies
12
Views
2K
What Are the Values of A, B, and C for the Given Expression?
Replies
19
Views
3K
Express ##x^3+y^3## in terms of ##m## and ##n##
Replies
17
Views
3K

Hot Threads

Recent Insights

Back
Top