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 also depends on the shape of the polynomial
If there is an obvious root..
Also, if you want to factorise (which is almost equivalent to finding roots), the only irreducible polynomials in $\displaystyle \mathbb{R}$ 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..
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 $\displaystyle (x - p)$, where $\displaystyle p$ is the potential root you are testing.
For example:
$\displaystyle \text{Factor }8x^4-22x^3+2x^2-22x-6$
The constant term is 6, and the leading coefficient is 8, so our possible rational roots are:
$\displaystyle \frac{\text{possible numerators}}{\text{possible denominators}} = \frac{\text{factors of 6}}{\text{factors of 8}} = \frac{\pm1,\ \pm2,\ \pm3,\ \pm6}{\pm1,\ \pm2,\ \pm4,\ \pm8}$
$\displaystyle = \pm1,\ \pm\frac12,\ \pm\frac14,\ \pm\frac18,\ \pm2,\ \pm3,\ \pm\frac32,\ \pm\frac34,\ \pm\frac38,\ \pm6$
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 $\displaystyle p$, and divide the polynomial by $\displaystyle x - p$. If there is no remainder, then $\displaystyle p$ 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
$\displaystyle \frac{8x^4-22x^3+2x^2-22x-6}{x + \frac14} = 8x^3-24x^2+8x-24$
$\displaystyle \Rightarrow8x^4-22x^3+2x^2-22x-6 = \left(x + \frac14\right)\left(8x^3-24x^2+8x-24\right)$
If you continue testing, you will also see that
$\displaystyle \frac{8x^3-24x^2+8x-24}{x - 3} = 8x^2 + 8$
and so, we find that
$\displaystyle 8x^4-22x^3+2x^2-22x-6 = \left(x + \frac14\right)\left(x - 3\right)\left(8x^2 + 8\right) = 2\left(4x + 1\right)\left(x - 3\right)\left(x^2 + 1\right)$
And the roots are $\displaystyle x = -\frac14,\ 3,\ \pm\,\text{i}$.
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:
$\displaystyle 8x^4-22x^3+2x^2-22x-6$
$\displaystyle = \left(8x^4 + 2x^2 - 6\right) - \left(22x^3 + 22x\right)$
$\displaystyle = 2\left(4x^4 + x^2 - 3\right) - 22x\left(x^2 + 1\right)$
$\displaystyle = 2\left(4\left(x^2\right)^2 + \left(x^2\right) - 3\right) - 22x\left(x^2 + 1\right)$
$\displaystyle = 2\left(x^2 + 1\right)\left(4x^2 - 3\right) - 22x\left(x^2 + 1\right)$
$\displaystyle = {\color{blue}2}{\color{red}\left(x^2 + 1\right)}{\color{blue}\left(4x^2 - 3\right)} - {\color{blue}22x}{\color{red}\left(x^2 + 1\right)}$
$\displaystyle = {\color{red}\left(x^2 + 1\right)}{\color{blue}\left(8x^2 - 6 - 22x\right)}$
$\displaystyle = \left(x^2 + 1\right)\left(8x^2 - 22x - 6\right)$
$\displaystyle = 2\left(x^2 + 1\right)\left(x - 3\right)\left(4x + 1\right)$
Another trick you might be able to use: Notice above, on line 4, how I treated the quartic expression in $\displaystyle x$, $\displaystyle 4x^4 + x^2 - 3$, as a quadratic expression in $\displaystyle x^2$, $\displaystyle 4\left(x^2\right)^2 + \left(x^2\right) - 3$. This technique is useful for factoring higher-degree polynomials that are missing certain terms. For example:
$\displaystyle \text{Factor }x^6 - 2x^3 + 1$
Although this is a 6th degree polynomial in $\displaystyle x$, you may notice that it is also a 2nd degree polynomial in $\displaystyle x^3$. To see this, substitute $\displaystyle u = x^3$:
$\displaystyle x^6 - 2x^3 + 1$
$\displaystyle = x^{3\cdot2} - 2x^3 + 1$
$\displaystyle = \left(x^3\right)^2 - 2\left(x^3\right) + 1$
$\displaystyle =u^2 - 2u + 1$
$\displaystyle =(u - 1)^2$
$\displaystyle =\left(x^3 - 1\right)^2$
$\displaystyle =\left[\left(x - 1\right)\left(x^2 + x + 1\right)\right]^2$
$\displaystyle =\left(x - 1\right)^2\left(x^2 + x + 1\right)^2$
Here, we would find the only real root to be $\displaystyle x = 1$.
Finally, some higher-degree polynomial equations are easily solved due to their form. For example:
$\displaystyle x^8 - 256 = 0$
$\displaystyle \Rightarrow x^8 = 256$
$\displaystyle \Rightarrow x = \pm\sqrt[8]{256} = \pm2$
Does that give you some ideas?