How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$

Question

How do I factorize a polynomial $ax^2 + bx + c$ where $a \neq 1$?

E.g. I know that $6x^2 + 5x + 1$ will factor to $(3x + 1)(2x + 1)$, but is there a recipe or an algorithm that will allow me to factorize this?

Answer 1

The roots of a quadratic polynomial $ax^2+bx+c$ are $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$. Then the factorization is just $a(x-x_1)(x-x_2)$.

In your example, $a=6,b=5,c=1$, so $x_1=\frac{-5+\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac13$ and $x_2=\frac{-5-\sqrt{5^2-4\cdot 6\cdot1}}{2\cdot 6}=-\frac12$. The factorization is then $6(x+\frac13)(x+\frac12)$, or $(3x+1)(2x+1)$.

Answer 2

If you know how to factor polynomials for $a=1$, then simply writing $$ax^2+bx+c=a(x^2+\frac bax+\frac ca )$$ makes the task immediate.

Answer 3

Hint: another possibility is to use the vieta's formulas https://en.wikipedia.org/wiki/Vieta%27s_formulas but it's more a "guess and proof strategy"^^

Answer 4

If you are lucky enough to find "by guesswork" values $x_1$ and $x_2$ so that $$x_1 + x_2 = -\frac{b}{a}$$ and $$x_1 x_2 = \frac{c}{a}$$ Then you'll be able to factor $ax^2 + bx + c$ to $a(x-x_1)(x-x_2)$.

Of course it will be easy to guess if $x_1$ and $x_2$ are integers.

Answer 5

I actually learned this one neat trick in middle school. Take your polynomial $$6x^2+5x+1$$ Take out the leading coefficient, and multiply it into the constant term to get $$x^2+5x+6$$ Now factor like usual $$(x+2)(x+3)$$ Divide both of those constant terms by the original leading coefficient, and reduce the fractions $$\left(x+\frac{2}{6}\right)\left(x+\frac{3}{6}\right) = \left(x+\frac{1}{3}\right)\left(x+\frac{1}{2}\right)$$ Now take the denominators and put them in front of the $x$ terms to get $$(3x+1)(2x+1)$$ (this process essentially only works for polynomials with integer coefficients, and it assumes that there are no constant factors in the original polynomial)

Answer 6

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 $$ {\begin{eqnarray} f \, &\,=\,& \ \ \, 6 x^2+\ 5\ x\,\ +\ \ 1\\ \Rightarrow\,\ 6f\, &\,=\,&\!\,\ (6x)^2\! +5(6x)+6\\ &\,=\,& \ \ \ \color{#c00}{X^2}+\, 5\ X\,\ +\ 6,\,\ \ X\, =\, 6x\\ &\,=\,& \ \ (X+2)\ (X+\,3)\\ &\,=\,& \ (6x+2)\,(6x+3)\\ \Rightarrow\ \ f\:=\: \color{#0a0}{6^{-1}}\,(6f)\, &\,=\,& \, (3x+ 1)\ (2x+1)\\ \end{eqnarray}}$$

In the final step we cancel $\,\color{#0a0}6\,$ by cancelling $\,2\,$ from the first factor, and $\,3\,$ 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.