Solving a 3rd Degree Polynomial: What are the Options?

[fishturtle1]

Homework Statement

Solve for the roots of the following.
(What do you notice about the complex roots?)

b) x³ + x² + 2x + 1 = 0

Homework Equations

To find roots of a polynomial of degree n > 3, look at the constant and take all its factors. Those are possible roots. Then plug them into see which ones solve the equation.

cubic equations from this link: http://www.mathportal.org/formulas/algebra/solalgebric.php

The Attempt at a Solution

factors of 1: 1, -1

x=1
(1)³ + 1² + 2(1) + 1 = 0

1 + 1 + 2 + 1 = 0
5 = 0, so 1 is not a root.

x=-1
(-1)³ + (-1)² + 2(-1) + 1 = 0

-1 + 1 -2 + 1 = 0 -> -1 = 0
So -1 is not a root.

I also tried to use cubic formulas and did not find the correct answer. I checked my calculations.

So i think there is some method more simple than cubic equations that I cannot think of/find on the web.

Are there any other ways besides testing the constant term's factors, and using the cubic equations, to find the roots of a 3rd degree polynomial?

 

[Ray Vickson]

Homework Statement

Solve for the roots of the following.
(What do you notice about the complex roots?)

b) x³ + x² + 2x + 1 = 0

Homework Equations

To find roots of a polynomial of degree n > 3, look at the constant and take all its factors. Those are possible roots. Then plug them into see which ones solve the equation.

cubic equations from this link: http://www.mathportal.org/formulas/algebra/solalgebric.php

The Attempt at a Solution

factors of 1: 1, -1

x=1
(1)³ + 1² + 2(1) + 1 = 0

1 + 1 + 2 + 1 = 0
5 = 0, so 1 is not a root.

x=-1
(-1)³ + (-1)² + 2(-1) + 1 = 0

-1 + 1 -2 + 1 = 0 -> -1 = 0
So -1 is not a root.

I also tried to use cubic formulas and did not find the correct answer. I checked my calculations.

So i think there is some method more simple than cubic equations that I cannot think of/find on the web.

Are there any other ways besides testing the constant term's factors, and using the cubic equations, to find the roots of a 3rd degree polynomial?

Show us what you got when you used the cubic formulas. Perhaps you made an error, but we cannot tell if you won't show us your work.

 

[FactChecker]

Testing factors of the constant term will almost certainly fail. It's not worth trying unless the problem function has been specially set up for that to work. You should show us how you used the cubic formulas. They should work every time.

 

[fishturtle1]

Sorry, here is my work:

x³ + x² + 2x + 1 = 0
So, a₁ = 1, a₂ = 2, a₃ = 1

Then Q = (3a₂ - a²₁)/9
Q = (3(2) - (1)²)/9
Q = (6-1)/9 = 5/9

R = (9a₁a₂ - 27a₃ - 2a₁³) / 54
R = (9(1)(2) - 27(1) - 2(1)³)/54
R = (18 - 27 - 2) / 54
R = -11/54

S = (R + (Q³ + R²)^(1/2))^(1/3)
S = ( (-11/54) + ( (125/729) + (121/2916) )^(1/2) ) ^(1/3)
S = about 0.63642

T = ( (R - (Q³ - R²)^(1/2) ) ^(1/3)
T = ( (-11/54) - ( (125/729) - (121/2916) )^1/2 ) ^1/3
T = about 0.82632i, actually I just checked T = about 0.74690

So for my first root i did this:

x₁ = S + T + a₁/3 = 0.63642 + 0.82632i - 1/3

x₁ = 0.30309 + 0.82632i
So with the new T, x₁ = 0.30309 + 0.74690i.. but this is still wrong.
at which point I checked with an online calculator and this was not any of the three roots so I'm wondering if I used this formula incorrectly or just made a calculation error.Correct answers to online calculator:

x₁ = -0.56984

x₂ = -0.21508 + 1.30713i

x₃ = -0.21508 - 1.30713i

 

[FactChecker]

For T, you should use the negative real third root of (R - (Q³ + R²)^(1/2)). That is T=-0.872931847948311.
Your values of Q, R, and S are correct.
With the corrected value of T, you should get the correct roots.

 

[Ray Vickson]

Sorry, here is my work:

x³ + x² + 2x + 1 = 0
So, a₁ = 1, a₂ = 2, a₃ = 1

Then Q = (3a₂ - a²₁)/9
Q = (3(2) - (1)²)/9
Q = (6-1)/9 = 5/9

R = (9a₁a₂ - 27a₃ - 2a₁³) / 54
R = (9(1)(2) - 27(1) - 2(1)³)/54
R = (18 - 27 - 2) / 54
R = -11/54

S = (R + (Q³ + R²)^(1/2))^(1/3)
S = ( (-11/54) + ( (125/729) + (121/2916) )^(1/2) ) ^(1/3)
S = about 0.63642

T = ( (R - (Q³ - R²)^(1/2) ) ^(1/3)
T = ( (-11/54) - ( (125/729) - (121/2916) )^1/2 ) ^1/3
T = about 0.82632i, actually I just checked T = about 0.74690

So for my first root i did this:

x₁ = S + T + a₁/3 = 0.63642 + 0.82632i - 1/3

x₁ = 0.30309 + 0.82632i
So with the new T, x₁ = 0.30309 + 0.74690i.. but this is still wrong.
at which point I checked with an online calculator and this was not any of the three roots so I'm wondering if I used this formula incorrectly or just made a calculation error.Correct answers to online calculator:

x₁ = -0.56984

x₂ = -0.21508 + 1.30713i

x₃ = -0.21508 - 1.30713i

For intermediate result you should keep as many digits as your calculator will hold. If you round intermediate quantities to values like 0.74690, etc., you can introduce significant roundoff errors, perhaps even making the final answer useless. (Of course I realize you may have done this, but it is not clear from what you wrote.)

Before embarking on an equation-solving expedition it is useful to first gain some insight into the nature of the problem. First, since all terms are > 0 for positive x, there is no positive root. Descarte's Rule of signs implies that the number of negative roots x < 0 is either three or one, and plotting the function makes it clear that there is only one. Furthermore, the function is > 0 at x = 0 and is < 0 at x =1, so there is a root between -1 and 0.

Almost nobody I know would ever bother using the cubic solution formula, even they know very well what it is; most people would just solve the equation numerically, which you can do to as many digits of accuracy as you could ever want using modern tools.

 

[fishturtle1]

For T, you should use the negative real third root of (R - (Q³ + R²)^(1/2)). That is T=-0.872931847948311.
Your values of Q, R, and S are correct.
With the corrected value of T, you should get the correct roots.

So is that a general rule when using these equations? That you leave the " i " out in the T? and yeah i see that this would help me get the real root. Thank you.

For intermediate result you should keep as many digits as your calculator will hold. If you round intermediate quantities to values like 0.74690, etc., you can introduce significant roundoff errors, perhaps even making the final answer useless. (Of course I realize you may have done this, but it is not clear from what you wrote.)

Before embarking on an equation-solving expedition it is useful to first gain some insight into the nature of the problem. First, since all terms are > 0 for positive x, there is no positive root. Descarte's Rule of signs implies that the number of negative roots x < 0 is either three or one, and plotting the function makes it clear that there is only one. Furthermore, the function is > 0 at x = 0 and is < 0 at x =1, so there is a root between -1 and 0.

Almost nobody I know would ever bother using the cubic solution formula, even they know very well what it is; most people would just solve the equation numerically, which you can do to as many digits of accuracy as you could ever want using modern tools.

I see what you mean. I remember in my physics class they taught us to never round too much in the in between steps and i forgot to use that here.

I feel like i have forgotten something from pre calculus class but what do you mean by "most people would just solve the equation numerically".. is there a method besides the cubic equations and rational roots test?

 

[FactChecker]

Almost nobody I know would ever bother using the cubic solution formula, even they know very well what it is; most people would just solve the equation numerically, which you can do to as many digits of accuracy as you could ever want using modern tools.

If I was writing a program, I would use the formulas. Numerical solutions can have problems converging and can require multiple attempts (with different initial values) to get a solution.

 

[FactChecker]

So is that a general rule when using these equations? That you leave the " i " out in the T? and yeah i see that this would help me get the real root.

Yes. They want T to be a real number. There is always one real cube root of any real number. If you include 'i', that is not a cube root of the number.

 

[Ray Vickson]

If I was writing a program, I would use the formulas. Numerical solutions can have problems converging and can require multiple attempts (with different initial values) to get a solution.

The issue is not at all straightforward, due in part to the effects of roundoff errors. (Even programming the quadratic solution robustly is not as easy as one might think.)

There seems to be a debate among the experts as to whether iterative methods (Newton, etc.) or algebraic methods are better; both suffer from problems in implementation. However, there are newer iterative methods available that do not require "good" starting points, but they are more sophisticated than straight Newton-Raphson.

The OP might like to consult
http://math.stackexchange.com/quest...ng-cubic-and-quartic-equations-by-computer-pr
or
http://stackoverflow.com/questions/2003465/fastest-numerical-solution-of-a-real-cubic-polynomial

Some contributors to the latter recommend solving a cubic by setting up the companion matrix (for which the cubic is the characteristic polynomial) then using modern numerical linear algebra methods to find the eigenvalues of the matrix; they claim that that is how some of the available solvers actually work.

All of this matters only if you want to write your own code for solving the problem; otherwise (under what I call modern methods), just use one of the many, many cubic (or higher) solvers available already. For example, Wolfram Alpha works well and is free.

 

[fishturtle1]

Thanks for all your thoughtful answers. This is a late response because after reading some of those articles I was felt out of my depth and decided to come back to this problem set after practicing some easier problems.

Funny thing is, the instructor said the problem was a typo...it was supposed to read x³ + x² + x + 1 = 0..

which I can see x = -1 is a root

so I can factor out x-1 and rewrite the original equation as: (x+1)(x²+1) = 0

then solve x² + 1 = 0

x² = -1

x = i, x = -i

So the roots are x = -1, -i, and i...now i will mark this as solved..just wanted to add the solution incase someone else wanted it.