[SPLIT] Solve equation using De Moivre's theorem

[anil86]

Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

2. x^7 - x^4 + x^3 - 1 = 0


I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.

 

[DrPhil]

Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

2. x^7 - x^4 + x^3 - 1 = 0


I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.

For an odd-order polynomial, there must be at least one real root. For (1), a real root is \(\displaystyle x=-1\), and for (2) a real root is \(\displaystyle x=+1\). You may divide out those roots if you wish.

For complex roots, \(\displaystyle x = |x|\ (\cos\theta \pm \mathrm i\ \sin\theta)\).

EDIT: I highly recommend dividing out the real root of each equation, and then factoring the resulting 6th-order polynomial into irreducible quartic and quadratic factors.

 

[anil86]

For an odd-order polynomial, there must be at least one real root. For (1), a real root is \(\displaystyle x=-1\), and for (2) a real root is \(\displaystyle x=+1\). You may divide out those roots if you wish.

For complex roots, \(\displaystyle x = |x|\ (\cos\theta \pm \mathrm i\ \sin\theta)\).

EDIT: I highly recommend dividing out the real root of each equation, and then factoring the resulting 6th-order polynomial into irreducible quartic and quadratic factors.

Hi DrPhil,

I divided equation by (x + 1) to get 6th order polynomial. How to simplify that?

Thanks

 

[anil86]

Solve 7th degree equation using complex number

Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

2. x^7 - x^4 + x^3 - 1 = 0


I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.

 

[stapel]

Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

x^7 + x^4 + x^3 + 1 = x^4(x^3 + 1) + 1(x^3 + 1) = (x^4 + 1)(x^3 + 1) = 0, so x^4 = -1 or x^3 = -1.

Where did you go from here?

 

[Romsek]

Solve equations using De Moivre's theorem:

1. x^7 + x^4 + x^3 + 1 = 0

2. x^7 - x^4 + x^3 - 1 = 0


I tried multiplying with (x - 1). Also tried putting x^3 = y; didn't work.

Ok, This one is sort of a trick question in a way because the answer requires you to notice something not totally obvious.
It took a bit of toying with this to see what they are after.

Take

x^7 + x^4 + x^3 + 1 = x^4(x^3 + 1) + (x^3 + 1) = (x^4 + 1)(x^3 + 1) = 0

I took the 2 leftmost terms of the sum and factored those separately and a factor just happens to be the same as the remaining terms in the sum so it can all be expressed by the product of those two terms above. Clearly the problem was set up this way.

Now in order for the product to be zero one or the other or both of those two factors must be 0, i.e.

(x^4 + 1) = 0 or (x^3 + 1) = 0, or both

This is where D'Moivre's theorem comes into play as you use it to immediately pick out the roots of each of these two terms.

(x^4 + 1) = 0, e^(i 4 theta) = e^(i (2k+1) pi) k an integer (do you understand why? because -1 = e^(i (2k+1) pi) )

theta = (2k+1) pi/4 for k=0,1,2,3 they just repeat after k=3

similarly

(x^3 + 1) = 0, e^(i 3 theta) = e^(i (2k+1) pi)

theta = (2k+1) pi/3 for k=0,1,2

take e^(i theta) for each of those theta's above and those are your 7 roots of the original polynomial.

You give the 2nd one a shot now that you've got an outline of how to proceed.

 

[anil86]

Ok, This one is sort of a trick question in a way because the answer requires you to notice something not totally obvious.
It took a bit of toying with this to see what they are after.

Take

x^7 + x^4 + x^3 + 1 = x^4(x^3 + 1) + (x^3 + 1) = (x^4 + 1)(x^3 + 1) = 0

I took the 2 leftmost terms of the sum and factored those separately and a factor just happens to be the same as the remaining terms in the sum so it can all be expressed by the product of those two terms above. Clearly the problem was set up this way.

Now in order for the product to be zero one or the other or both of those two factors must be 0, i.e.

(x^4 + 1) = 0 or (x^3 + 1) = 0, or both

This is where D'Moivre's theorem comes into play as you use it to immediately pick out the roots of each of these two terms.

(x^4 + 1) = 0, e^(i 4 theta) = e^(i (2k+1) pi) k an integer (do you understand why? because -1 = e^(i (2k+1) pi) )

theta = (2k+1) pi/4 for k=0,1,2,3 they just repeat after k=3

similarly

(x^3 + 1) = 0, e^(i 3 theta) = e^(i (2k+1) pi)

theta = (2k+1) pi/3 for k=0,1,2

take e^(i theta) for each of those theta's above and those are your 7 roots of the original polynomial.

You give the 2nd one a shot now that you've got an outline of how to proceed.

Thank you Romsek

 

[Romsek]

were you able to solve the second one?

 

[Deleted member 4993]

Interesting:
Add the equations: x^7 = x^3

Subtract them: x^4 = -1

Interesting - but those are two different problems - there is no apparent reason that we should be able add or subtract.

 

[anil86]

were you able to solve the second one?

Yes