I am wondering if there is a methodical, algorithmic, brain-dead way to factor polynomials. For example:
$x^6 – 14x^{5} + 73x^{4} – 188x^{3} + 256x^{2} – 176x^{1} + 48$
can be written as
$(x-1)^2 (x-2)^3 (x-6)$
But how do you actually get there?
[Math] How to factor polynomials
algebra-precalculusfactoringpolynomials
Related Solutions
If all the terms have even powers, you have a polynomial of half the degree in the square of the variable. $5x^4+24x^2-72=6((x^2)^2+4(x^2)-12)$, which is a quadratic that you should know how to factor. It might be easier to see if you define $y=x^2$ and write the polynomial as $6(y^2+4y-12)$. Once you have factored that, plug the $x^2$s back in and you will have two quadratic factors that you can factor.
Generally and here it is not useful to factor part of an expression. In your first example, you can write it as $(x^2+3x)(x^2-3x)+14$ (you have extra factors of $2$) but you are right that you don't know what to do with the $14$. You could complete the square to get $x^4-9x^2+14=(x^2-\frac 92)^2-\frac {25}4$ and then use the difference of squares.
This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 7 x + 3 \right) $$ $$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( 2 x^{2} - 7 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } + \left( \frac{ 15 x - 45 }{ 4 } \right) $$ $$ \left( 2 x^{2} - 7 x + 3 \right) = \left( \frac{ 15 x - 45 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 5 }{ 4 } \right) }{ \left( \frac{ 2 x - 9 }{ 4 } \right) } $$ $$ \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 12 x + 20 }{ 15 } \right) }{ \left( \frac{ 4 x^{2} - 20 x + 24 }{ 15 } \right) } $$ $$ \left( x^{2} - 3 x + 5 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{2} - 5 x + 6 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( -1 \right) $$ $$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{2} - 3 x + 5 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( x^{2} - 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$ $$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$ $$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( - x + 3 \right) $$
............
Best Answer
The problem of factoring polynomials is in general difficult and/or labor-intensive, but for the sake of concreteness, I'll indicate one way to factor the example polynomial that uses a few methods (caution: these methods do not apply to all polynomials!).
The Rational Root Theorem says that any (reduced form) rational root $\frac{s}{t}$ of a polynomial $$p(x) := a_n x^n + \cdots + a_1 x + a_0$$ satisfies $s \mid a_0$ and $t \mid a_n$. In particular, if the polynomial is monic (i.e., $a_n = 1$) any rational root is in fact an integer, and all such roots are factors of $a_0$. In our case, the possible integer roots of $$q(x) := x^6 - 14x^{5} + 73x^{4} - 188x^{3} + 256x^{2} - 176x + 48$$ are just the factors of $48$.
There are, unfortunately, 20 of these (counting sign), but we can simplify our search considerably by observing that the signs of the coefficients of $q(x)$ alternate with degree, and hence the signs of $q(-x)$ are all the same (in this case positive). So, $q(-0) > 0$ and $q(-x)$ is increasing on $[0, \infty)$; hence all of the roots of $q(-x)$ are negative, or equivalently, all of the roots of $q(x)$ (rational or otherwise) are positive, reducing the list of integers to check by one-half.
If we start with the largest factors of $a_0 = 48$ and work down, we find that the largest root of $q$ is $6$, and so $(x - 6)$ is a factor of $q(x)$. Polynomial long division gives that $$q(x) = (x - 6) r(x) ,$$ where $$r(x) := x^5-8 x^4+25 x^3-38 x^2+28 x-8,$$ and we can see that by starting with the largest factors of $a_0$ we produce the smallest constant terms after polynomial long division, reducing more the list of possibilities to check at the next step. Indeed, $8$ only has four positive factors to check. (Of course, if you're checking this without calculator assistance, it is generally more computationally intensive to evaluate $p(x)$ for integers $x$ large in magnitude than ones small in magnitude, in which case working from largest to smallest need not be optimal.)
Checking shows that $r(8), r(4) \neq 0$, and so any integer roots of $r$ are $1$, $2$. On the other hand, for any polynomial $p$, the product of the roots (counting multiplicity) is $(-1)^{\deg p} a_0$. So if we know in advance that all of the roots of $q$ are integers, so are all the roots $b_j$ of $r$, and if the multiplicity of $2$ as a root of $r$ is $k$, the multiplicity of $1$ is $5 - k$, and by the above we have that $-(-8) = 1^{5 - k} 2^k = 2^k$. Thus, $k = 3$, and so we conclude that $$q(x) = (x - 6) (x - 2)^3 (x - 1)^2 $$ as desired. Note that this method only required evaluating a polynomial $8$ times and a carrying out a single polynomial long division (and again, knowing in advance that all of the roots of $p$ were integers).
Caution Several of the techniques here can only be applied when the given polynomial has certain properties. Most importantly, (1) not all roots of a real polynomial need be rational (indeed, none need be, as is the case for $x^2 - 2$), and (2) not all roots of a real polynomial need even be real (as in the case of $x^2 + 1$), and correspondingly, a polynomial need not be factorable over $\Bbb R$. In general, factoring a polynomial over $\Bbb Q$ or $\Bbb R$ is difficult and/or labor-intensive (and in the latter case there's a question of what form one wants). See the Wikipedia article, Factorization of Polynomials, for more information.