Factoring Non-Monic Quadratics

A quadratic is monic when its leading coefficient is $1$ ($x^2+bx+c$) and non-monic when it's anything else ($a{x}^2+bx+c$ with $a\ne 1$). Non-monic quadratics need a more disciplined factoring procedure — random guessing breaks down quickly.

The "ac method" (factoring by grouping)

This is the algorithmic approach that always works. Use it whenever $a\ne 1$.

1. Compute $a\cdot c$.
2. Find two integers that multiply to $a\cdot c$ and add to $b$.
3. Use those integers to split the middle term ($bx$) into two terms.
4. Factor by grouping the resulting four-term polynomial.
5. Factor out the common binomial.

Worked example

Factor: $6x^2+11x-10$.

Step 1 — compute $a\cdot c$:

$a\cdot c=6\cdot (-10)=-60$

Step 2 — find two integers that multiply to $-60$ and add to $11$.

Test systematically: $(15,-4)$ multiply to $-60$ and add to $11$. ✓

Step 3 — split the middle term:

$6x^2+15x-4x-10$

Step 4 — factor by grouping. Group the first two and last two terms:

$3x(2x+5)-2(2x+5)$

Both groups now contain the binomial $(2x+5)$. Factor it out:

$\boxed{(3x-2)(2x+5)}$

Step 5 — verify by FOIL (always do this until factoring becomes automatic):

$(3x-2)(2x+5)=6x^2+15x-4x-10=6x^2+11x-10✓$

The discriminant — diagnostic before factoring

Before you commit to factoring, compute the discriminant $\Delta =b^2-4ac$.

• $\Delta >0$: two distinct real roots. The quadratic factors over the reals.
• $\Delta =0$: one repeated real root. The quadratic is a perfect square.
• $\Delta <0$: two complex conjugate roots. The quadratic doesn't factor over the reals.

For our example: $\Delta =11^2-4(6)(-10)=121+240=361=19^2$. Positive, so two real roots, and a perfect square so the roots are rational.

When factoring fails

If the integers in step 2 don't exist (i.e., the quadratic doesn't factor over the integers), fall back to the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

This always works.

Connection to linear algebra and ML

Finding roots of polynomials is mechanically the same operation as finding eigenvalues of a matrix. To find the eigenvalues of an $n\times n$ matrix $A$, you solve

$\det (A-\lambda I)=0$

which is a polynomial equation in $\lambda$. For $2\times 2$ matrices, it's literally a quadratic. For larger matrices, a numerical eigenvalue routine handles it numerically — but the algebraic intuition you build by factoring quadratics by hand carries directly into understanding what PCA, spectral clustering, and the stability of recurrent neural networks are computing under the hood.