Mathematicians: How would you find the 6 roots (real or not) from x^6+7x-8=0 using an algebraic method?

Question

I have my own way to solve it, but Id like to see if someone of you can teach me something different.

THIS IS NOT A PIECE OF HOMEWORK. This is a serious question, addressed to Mathematics lovers.

Later!

Ana

Im sorry, I mistyped the equation.
x^6+7x^3-8=0 is the right one

Answer 1

As pointed out in previous messages, x^6+7x-8=(x-1)(x^5+x^4+x^3+x^2+x+8). By "using an algebraic method", I assume you mean "solve by radicals"? That is, you want solutions expressed using only addition, subtraction, multiplication, division, and n-th roots for any integer n? If this is the case, there is a great method that tells you if the above 5-th degree polynomial is solvable by radicals. (As pointed out above, not all such polynomials are so solvable.) This method is very computational and takes no advanced knowledge, once someone tells you what to look at. There is a "magic" polynomial of degree 6 which has a *rational* root if and only if this polynomial of degree 5 is solvable by radicals. Since checking for rational roots is a matter of factoring one or two integers and plugging-in to the polynomial, this is very simple once you know the 6-th degree polynomial.

The more high-brow way to try this problem is to take that 5-th degree polynomial and try to find its Galois group. It is a theorem that the 5-th degree polynomial is solvable by radicals if and only if the Galois group is solvable. In particular, if the Galois group is the whole symmetric group on 5 letters, then the polynomial is *not* solvable by radicals. It turns out that the polynomial x^5+x^4+x^3+x^2+x+8 has exactly this Galois group. This shows that there is *no way* to solve the original equation by radicals.

To show that the Galois group is the symmetric group on 5 letters you have to do some work. Let f(x)=x^5+x^4+x^3+x^2+x+8. First we want to show that f(x) is irreducible over the rational numbers. One way to do this is show that it is irreducible when you reduce it modulo 31. Another way to do this is as follows. f(x) has no rational roots, by the rational root test (any rational root must lie in the set {2, -2, 4, -4, 8, -8}). So f(x) is not a linear times a quartic. The only other possibility is a quadratic times a cubic, say (x^3+ax^2+bx+c)(x^2+dx+e). Matching coefficients gives ce=8, d+a=1, cd+be=1, ae+bd+c=1, e+b+ad=1. Play around with these equations for a little bit and you'll get a contradiction (use the fact that ce=8; in particular, neither c nor e is zero, so you can multiply and divide by them freely). So f(x) is irreducible. This implies that its Galois group is a transitive subgroup of S_5 the symmetric group on 5 letters. It is a theorem that if there is a 5-cycle and a 2-cycle in a subgroup of S_5, then the subgroup is the whole group. Reducing modulo 31, f(x) is an irreducible quintic. Reducing modulo 53, f(x) is an irreducible quadratic times the product of three linears. It is a theorem that these degrees give cycle types in the Galois group of the polynomial. Hence, the Galois group has a 5-cycle and a 2-cycle so that it must be S_5.

Answer 2

There is no general way to find the root of a sixth degree polynomial. In fact, it can be proven that there are such polynomials whose roots cannot be expressed using the algebraic operations (+, -, *, /, powers and roots). General methods only exist up to fourth degree.

It helps that we can find one root fairly easily: x = 1. Dividing the polynomial by (x - 1) gives us
x^5 + x^4 + x^3 + x^2 + x + 8 = 0

This is much harder. I note, first of all, that other solutions cannot be positive. Since there are 5 roots and complex roots come in pairs, there must be at least one negative real root. If this root is -a, then
-a^5 + a^4 - a^3 + a^2 - a + 8 = 0 or
a^5 + a^3 + a = a^4 + a^2 + 8

If a increases, the left side increases faster than the right side. It is easy to see that a lies between 1 and 2. It is no rational number, because the original equation is monic and without fractions. A precise solution is hard to find...

Answer 3

You should first select the numbers that 8 is dividable to them. they are:{1,2,4,8,(-1),(-2),(-4),(-8)}. Then, you should put these number in the formula. If you put a in the formula and the answer was zero, it means that this formula is dividable to (x-a). for example first we try 1:

1^6 + 7*1 - 8 = 0
Great! it means that we can divide the formula to (x-1):

x^6 + 7x - 8 = (x-1)(x^5 + x^4 + x^3 + x^2 + x + 8)
the last number in the new formula is 8. so we should try another number from the above list except 1. However, you know that it can't be a positive number. If so, the answer would be positive, not zero. so we try negative numbers.

(-1): (-1)^5 + (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 8 = 7 ≠ 0
(-2): (-2)^5 + (-2)^4 + (-2)^3 + (-2)^2 + (-2) + 8 = (-6) ≠ 0
(-4): (-4)^5 + (-4)^4 + (-4)^3 + (-4)^2 + (-4) + 8 = (-812) ≠ 0
And the answer about (-8) is not zero neither.
It means that you can't simplify it any more and (x-1)(x^5 + x^4 + x^3 + x^2 + x + 8) is the final result.

Answer 4

There is no general algebraic way to solve a polynomial higher than 4th order. It is a simple matter to generate any number of polynomials by first picking the roots but finding the solution is no more than good guessing.

In a case like this one, since all the coefficients sum to zero and there is a zeroth order coefficent, you can factor the first root of x=1 leaving:

x^5 + x^4 + x^3 + x^2 + x + 8

From here, I don't see a cute guess for the next root. Once you guess the next root you can use the closed form solution for a quartic to get the rest.

Answer 5

There are 6 distinctive roots of this equation. enable y=x^3, to get y^2+7y-8=0. This components as (y+8)(y-one million)=0, so y=-8 or one million. consequently, x^3=-8 or x^3=one million. Rewrite those: x^3+8=0 or x^3-one million=0. ingredient: (x+2)(x^2-2x+4)=0 or (x-one million)(x^2 +x+one million)=0. the 1st supplies x=-2, (2+- sqrt(-12))/2 and the 2nd supplies x=one million, (-one million+-sqrt(-3))/2. -2 and one million are no longer repeated roots!!!!!!!

Answer 6

1st try to find root by inspection x = 1 makes
f(1) = 1+7-8 =0
so split 7x into 8x -x

we get x^6-x+8x- 8 = 0

so x(x^5-1) +8(x-1) =0
so x(x-1)(x^4+x^3+x^2+x+1) +8(x-1) =0
(x-1)(x(x^4+x^3+x^2+x+1)+8) = 0
(x-))(x^5+x^4+x^3+x^2+x^1+8) = 0
x= 1
or x^5+x^4+x^3+x^2+x^1+8= 0

you should be able to proceed

Answer 7

for the equation x^6 + 7x^3 - 8 = 0
WE can put the equation in the formula
(x³ +8 ) (x³ - 1) = 0
so we have
x³ = -8
and this has the roots
-2 , -2ω , -2ω²

AND
x³ = 1
and this has the roots
1 , ω , ω²

ω = (-1 + √3 i )/ 2
ω² = (-1 - √3 i )/ 2
i = √-1
*******************************************

For the real Polynomial:
x ^ 6+7x-8 = 0
Start Newton Iteration. Stage 1=>Stop Condition: f(z)<1.51e-14
Start : z[1]=(0.2+i0) dz=(-2.48e-1+i0.00e+0) f(z)=6.3e+0
Iteration: 1
Newton Step: z[1]=(1+i0) dz=(-8.95e-1+i0.00e+0) f(z)=2.2e+0
Function value decrease=>try multiple steps in that direction
Try Step: z[1]=(2+i0) dz=(-8.95e-1+i0.00e+0) f(z)=7.8e+1
: No improvement=>Discard last try step
Iteration: 2
Newton Step: z[1]=(1+i0) dz=(1.19e-1+i0.00e+0) f(z)=3.1e-1
Enter Stage 2=>New Stop Condition: f(z)<4.66e-14
Iteration: 3
Newton Step: z[1]=(1+i0) dz=(2.28e-2+i0.00e+0) f(z)=8.3e-3
Enter Stage 2=>New Stop Condition: f(z)<4.12e-14
Iteration: 4
Newton Step: z[1]=(1+i0) dz=(6.38e-4+i0.00e+0) f(z)=6.1e-6
Enter Stage 2=>New Stop Condition: f(z)<4.11e-14
Iteration: 5
Newton Step: z[1]=(1+i0) dz=(4.71e-7+i0.00e+0) f(z)=3.3e-12
Enter Stage 2=>New Stop Condition: f(z)<4.11e-14
Iteration: 6
Newton Step: z[1]=(1+i0) dz=(2.56e-13+i0.00e+0) f(z)=0.0e+0
Enter Stage 2=>New Stop Condition: f(z)<4.11e-14
Stop Criteria satisfied after 6 Iterations
Final Newton z[1]=(1+i0) dz=(2.56e-13+i0.00e+0) f(z)=0.0e+0
Deflate the real root z=1
Start Newton Iteration. Stage 1=>Stop Condition: f(z)<1.26e-14
Start : z[1]=(-0.3+i0) dz=(2.60e-1+i0.00e+0) f(z)=7.8e+0
Iteration: 1
dz>5*dz0 =>Alter direction: Old dz=(12+i0.0) New dz=(0.78+i1.0)
Newton Step: z[1]=(-1-i1) dz=(7.80e-1+i1.04e+0) f(z)=1.0e+1
Function value increase=>try shorten the step
Try Step: z[1]=(-0.7-i0.5) dz=(3.90e-1+i5.20e-1) f(z)=7.7e+0
: Improved=>Continue stepping
Try Step: z[2]=(-0.46-i0.26) dz=(1.95e-1+i2.60e-1) f(z)=7.7e+0
: Improved=>Continue stepping
: Probably local saddlepoint=>Alter Direction: z[1]=(-0.2-i0.2) dz=(-4.55e-2+i1.56e-1) f(z)=7.8e+0
Iteration: 2
Newton Step: z[1]=(-1e+1-i3) dz=(1.15e+1+i2.85e+0) f(z)=2.4e+5
Function value increase=>try shorten the step
Try Step: z[1]=(-6-i2) dz=(5.74e+0+i1.42e+0) f(z)=7.6e+3
: Improved=>Continue stepping
Try Step: z[1]=(-3-i0.9) dz=(2.87e+0+i7.12e-1) f(z)=2.6e+2
: Improved=>Continue stepping
: Probably local saddlepoint=>Alter Direction: z[1]=(-0.8-i2) dz=(5.76e-1+i1.36e+0) f(z)=4.9e+0
Iteration: 3
Newton Step: z[2]=(-0.66-i1.4) dz=(-1.28e-1-i7.12e-2) f(z)=8.2e-1
Function value decrease=>try multiple steps in that direction
Try Step: z[2]=(-0.54-i1.4) dz=(-1.28e-1-i7.12e-2) f(z)=2.7e+0
: No improvement=>Discard last try step
Iteration: 4
Newton Step: z[3]=(-0.631-i1.46) dz=(-3.15e-2+i1.02e-2) f(z)=3.6e-2
Enter Stage 2=>New Stop Condition: f(z)<4.44e-14
Iteration: 5
Newton Step: z[4]=(-0.6318-i1.458) dz=(4.74e-4+i1.38e-3) f(z)=7.1e-5
Enter Stage 2=>New Stop Condition: f(z)<4.46e-14
Iteration: 6
Newton Step: z[7]=(-0.6317840-i1.457620) dz=(-8.91e-7-i2.71e-6) f(z)=2.7e-10
Enter Stage 2=>New Stop Condition: f(z)<4.46e-14
Iteration: 7
Newton Step: z[12]=(-0.631783963993-i1.45761996594) dz=(-3.14e-12-i1.04e-11) f(z)=1.8e-15
Enter Stage 2=>New Stop Condition: f(z)<4.46e-14
Stop Criteria satisfied after 7 Iterations
Final Newton z[12]=(-0.631783963993-i1.45761996594) dz=(-3.14e-12-i1.04e-11) f(z)=1.8e-15
Deflate the complex conjugated root z=(-0.6317839639928163-i1.4576199659396556)
Start Newton Iteration. Stage 1=>Stop Condition: f(z)<2.99e-15
Start : z[1]=(0.2+i0) dz=(-2.45e-1+i0.00e+0) f(z)=2.9e+0
Iteration: 1
dz>5*dz0 =>Alter direction: Old dz=(-2.5+i0.0) New dz=(-0.74-i0.98)
Newton Step: z[1]=(1+i1) dz=(-7.36e-1-i9.82e-1) f(z)=2.4e-1
Function value decrease=>try multiple steps in that direction
Try Step: z[1]=(2+i2) dz=(-7.36e-1-i9.82e-1) f(z)=1.5e+1
: No improvement=>Discard last try step
Iteration: 2
Newton Step: z[3]=(0.951+i1.01) dz=(3.10e-2-i3.06e-2) f(z)=7.6e-3
Enter Stage 2=>New Stop Condition: f(z)<1.05e-14
Iteration: 3
Newton Step: z[4]=(0.9520+i1.013) dz=(-1.19e-3-i6.20e-4) f(z)=7.2e-6
Enter Stage 2=>New Stop Condition: f(z)<1.05e-14
Iteration: 4
Newton Step: z[7]=(0.9519675+i1.012985) dz=(1.27e-6-i1.51e-7) f(z)=6.5e-12
Enter Stage 2=>New Stop Condition: f(z)<1.05e-14
Iteration: 5
Newton Step: z[13]=(0.9519674919814+i1.012985269888) dz=(2.91e-13-i1.12e-12) f(z)=0.0e+0
Enter Stage 2=>New Stop Condition: f(z)<1.05e-14
Stop Criteria satisfied after 5 Iterations
Final Newton z[13]=(0.9519674919814+i1.012985269888) dz=(2.91e-13-i1.12e-12) f(z)=0.0e+0
Deflate the complex conjugated root z=(0.9519674919814203+i1.0129852698877781)
The Solutions are:
X1=-1.6403670559772078
X2=(0.9519674919814203-i1.0129852698877781)
X3=(0.9519674919814203+i1.0129852698877781)
X4=(-0.6317839639928163+i1.4576199659396556)
X5=(-0.6317839639928163-i1.4576199659396556)
X6=1

Answer 8

I'd just use an online solver. :)

For the real Polynomial:
+1x^6+7x-8
The Solutions are:
X1=-1.6403670559772078
X2=(0.9519674919814203-i1.0129852698877781)
X3=(0.9519674919814203+i1.0129852698877781)
X4=(-0.6317839639928163+i1.4576199659396556)
X5=(-0.6317839639928163-i1.4576199659396556)
X6=1