Solving degree 7 polynomials

• May 18th 2008, 11:27 AM
sean.1986
Solving degree 7 polynomials
How would one go about solving the following...

ax^7 + bx^6 + cx^5 + dx^4 + fx^3 + gx^2 + hx + j = 0

I'm looking for a general answer, i.e. no specific values of the coefficients. Thanks.
• May 18th 2008, 11:28 AM
Moo
Hello,

Quote:

Originally Posted by sean.1986
How would one go about solving the following...

ax^7 + bx^6 + cx^5 + dx^4 + fx^3 + gx^2 + hx + j = 0

I'm looking for a general answer, i.e. no specific values of the coefficients. Thanks.

Do you know that it's even not possible to find a general solution to a polynomial of degree 5 ?
Or at least, the people who said they found were being laughed at... :/
• May 18th 2008, 11:31 AM
sean.1986
Quote:

Originally Posted by Moo
Hello,

Do you know that it's even not possible to find a general solution to a polynomial of degree 5 ?
Or at least, the people who said they found were being laughed at... :/

I had read that, yes, so let me rephrase the question... how would one go about approaching this sort of problem?

Thanks.
• May 18th 2008, 11:49 AM
ThePerfectHacker
Quote:

Originally Posted by sean.1986
How would one go about solving the following...

ax^7 + bx^6 + cx^5 + dx^4 + fx^3 + gx^2 + hx + j = 0

I'm looking for a general answer, i.e. no specific values of the coefficients. Thanks.

It is not possible. It is impossible to find a forumla. Are you shocked? (Surprised)

Quote:

how would one go about approaching this sort of problem?
You can approximate zeros.
• May 18th 2008, 11:54 AM
sean.1986
Quote:

Originally Posted by ThePerfectHacker
Are you shocked? (Surprised)

No, because I knew there were no formulae for degree 5 and above. I was looking for a general approach to the problem, not a general formula.
• May 18th 2008, 12:06 PM
Moo
Quote:

Originally Posted by sean.1986
No, because I knew there were no formulae for degree 5 and above. I was looking for a general approach to the problem, not a general formula.

It also depends on the shape of the polynomial :D
If there is an obvious root..

Also, if you want to factorise (which is almost equivalent to finding roots), the only irreducible polynomials in $\mathbb{R}$ are of degree 1 and 2.

So a polynomial of degree 7 would be written into the product of 1-2-2-2 (at least, because it can be 1-1-1-2-2).

Have to check for it because it has been a long time since I didn't use it..
• May 18th 2008, 03:58 PM
Reckoner
Quote:

Originally Posted by sean.1986
No, because I knew there were no formulae for degree 5 and above. I was looking for a general approach to the problem, not a general formula.

Hi, sean. As mentioned, it is not possible to find a general formula for finding roots of polynomials of degree 5 or higher (at least, not in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions), so it is not possible to create a general method of solving this problem that will work in every situation. However, there are some factoring techniques you can try. In my examples, I am going to use lower degree polynomials for convenience, but you can extend these methods to solving septic equations.

First, you could try finding the rational roots. The rational zero theorem, or rational roots theorem, states that if a polynomial has integer coefficients, then any rational root will have a factor of the constant term as its numerator and a factor of the leading coefficient as its denominator.

Using the rational zero theorem, you can determine possible rational zeros, which you can then check by seeing if you can divide the polynomial evenly by $(x - p)$, where $p$ is the potential root you are testing.

For example:

$\text{Factor }8x^4-22x^3+2x^2-22x-6$

The constant term is 6, and the leading coefficient is 8, so our possible rational roots are:

$\frac{\text{possible numerators}}{\text{possible denominators}} = \frac{\text{factors of 6}}{\text{factors of 8}} = \frac{\pm1,\ \pm2,\ \pm3,\ \pm6}{\pm1,\ \pm2,\ \pm4,\ \pm8}$

$= \pm1,\ \pm\frac12,\ \pm\frac14,\ \pm\frac18,\ \pm2,\ \pm3,\ \pm\frac32,\ \pm\frac34,\ \pm\frac38,\ \pm6$

Now, systematically check each of these values until you determine that one of them is a root (but it is possible that none of them are, in which case the polynomial has no rational roots). To do this, take the potential root $p$, and divide the polynomial by $x - p$. If there is no remainder, then $p$ is a root of the polynomial. You can use polynomial long division for this, but I recommend you look up the method of "synthetic division" as it is much faster and easier.

This will take a while, but eventually you would find that

$\frac{8x^4-22x^3+2x^2-22x-6}{x + \frac14} = 8x^3-24x^2+8x-24$

$\Rightarrow8x^4-22x^3+2x^2-22x-6 = \left(x + \frac14\right)\left(8x^3-24x^2+8x-24\right)$

If you continue testing, you will also see that

$\frac{8x^3-24x^2+8x-24}{x - 3} = 8x^2 + 8$

and so, we find that

$8x^4-22x^3+2x^2-22x-6 = \left(x + \frac14\right)\left(x - 3\right)\left(8x^2 + 8\right) = 2\left(4x + 1\right)\left(x - 3\right)\left(x^2 + 1\right)$

And the roots are $x = -\frac14,\ 3,\ \pm\,\text{i}$.

This is a very nice method, because it works for most polynomials that "factor nicely." However, it can be very tedious, and there are often simpler ways to factor such expressions, but such techniques are highly dependent on the nature of the polynomial you are working with.

One technique you can try is factoring by grouping. Here, you place similar terms together in groups, and factor each group separately to form new terms that have a common factor. For example, here is how to factor the polynomial given earlier by grouping:

$8x^4-22x^3+2x^2-22x-6$

$= \left(8x^4 + 2x^2 - 6\right) - \left(22x^3 + 22x\right)$

$= 2\left(4x^4 + x^2 - 3\right) - 22x\left(x^2 + 1\right)$

$= 2\left(4\left(x^2\right)^2 + \left(x^2\right) - 3\right) - 22x\left(x^2 + 1\right)$

$= 2\left(x^2 + 1\right)\left(4x^2 - 3\right) - 22x\left(x^2 + 1\right)$

$= {\color{blue}2}{\color{red}\left(x^2 + 1\right)}{\color{blue}\left(4x^2 - 3\right)} - {\color{blue}22x}{\color{red}\left(x^2 + 1\right)}$

$= {\color{red}\left(x^2 + 1\right)}{\color{blue}\left(8x^2 - 6 - 22x\right)}$

$= \left(x^2 + 1\right)\left(8x^2 - 22x - 6\right)$

$= 2\left(x^2 + 1\right)\left(x - 3\right)\left(4x + 1\right)$

Another trick you might be able to use: Notice above, on line 4, how I treated the quartic expression in $x$, $4x^4 + x^2 - 3$, as a quadratic expression in $x^2$, $4\left(x^2\right)^2 + \left(x^2\right) - 3$. This technique is useful for factoring higher-degree polynomials that are missing certain terms. For example:

$\text{Factor }x^6 - 2x^3 + 1$

Although this is a 6th degree polynomial in $x$, you may notice that it is also a 2nd degree polynomial in $x^3$. To see this, substitute $u = x^3$:

$x^6 - 2x^3 + 1$

$= x^{3\cdot2} - 2x^3 + 1$

$= \left(x^3\right)^2 - 2\left(x^3\right) + 1$

$=u^2 - 2u + 1$

$=(u - 1)^2$

$=\left(x^3 - 1\right)^2$

$=\left[\left(x - 1\right)\left(x^2 + x + 1\right)\right]^2$

$=\left(x - 1\right)^2\left(x^2 + x + 1\right)^2$

Here, we would find the only real root to be $x = 1$.

Finally, some higher-degree polynomial equations are easily solved due to their form. For example:

$x^8 - 256 = 0$

$\Rightarrow x^8 = 256$

$\Rightarrow x = \pm\sqrt[8]{256} = \pm2$

Does that give you some ideas?
• May 18th 2008, 04:00 PM
KGene
Quote:

Originally Posted by sean.1986
No, because I knew there were no formulae for degree 5 and above. I was looking for a general approach to the problem, not a general formula.

There is no general method to find an exact root of a general polynomial, all you can do is just approximation.
• May 19th 2008, 07:49 AM
sean.1986
OK, thank for all your help. I think I will try out some of the things mentioned in this thread.