I'm learning how to factor polynomials, but I'm having a hard time understanding the approach when the 2nd degree coefficient is greater than $1$.
For example, when I begin to factor $12k^4 + 22k^3 – 70k^2$, I first break it down to $2k^2(6k^2 + 11k – 35)$.
I would think that I'd want to find two numbers that sum up to $11$ and have a product of $-35$, but instead I'm told we need to multiply $-35$ by $6$ so that I now have to find two numbers that sum up to $11$ and have a product of $-210$.
Can anyone help me understand why $-35$ is multiplied by the coefficient $6$? Why isn't $11k$ also multiplied by $6$?
Best Answer
That AC-method reduces to factoring a polynomial that is $\,\rm\color{#c00}{monic}\,$ (lead coeff $\color{#c00}{=1})$ as follows
$$\begin{eqnarray} \rm\: a\:f(x)\:\! \,=\,\:\! a\:(a\:x^2 + b\:x + c) &\,=\,&\!\!\rm\: \color{#c00}{X^2} + b\:X + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\overbrace{ac,}^{\rm\qquad\ \ \ \ \ {\bf\large\ \ AC-method}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \ X = a\:x \\ \end{eqnarray}$$
In your case
$$ {\begin{eqnarray} f \, &\,=\,& \ \ \, 6 x^2+\ 11\ x\,\ -\ \ 35\\ \Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +11(6x)-210\\ &\,=\,& \ \ \ \color{#c00}{X^2}+\, 11\ X\,\ -\ 210,\,\ \ X\, =\, 6x\\ &\,=\,& \ \ (X+21)\ (X-\,10)\\ &\,=\,& \ (6x+21)\,(6x-10)\\ \Rightarrow\ \ f\:=\: \color{#0a0}{6^{-1}}\,(6f)\, &\,=\,& \ (2x+\,\ 7)\ (3x\,-5)\\ \end{eqnarray}}$$
In the final step we cancel $\,\color{#0a0}6\,$ by cancelling $\,3\,$ from the first factor, and $\,2\,$ from the second.
If we denote our factoring algorithm by $\,\cal F,\,$ then the above transformation is simply
$$\cal F f\, = a^{-1}\cal F\, a\,f\quad\,$$
Thus we've transformed by $ $ conjugation $\,\ \cal F = a^{-1} \cal F\, a\ \,$ the problem of factoring non-monic polynomials into the simpler problem of factoring monic polynomials. The same idea also works for higher degree polynomials, see this answer, which also gives links to closely-related ring-theoretic topics.