How to Factorize a Quadratic

This breaks $ax^2 + bx + c$ into a product of linear factors. The quadratic formula gives the two roots, and the factors are built from them.

$x_1$ and $x_2$ are the roots of $ax^2 + bx + c = 0$. When a root is a fraction $\dfrac{p}{q}$, the factor $\left(x - \dfrac{p}{q}\right)$ can be multiplied by $q$ and written as $(qx - p)$, which keeps every coefficient whole.

When whole numbers are possible

If the discriminant $D = b^2 - 4ac$ is a perfect square then $\sqrt{D}$ in the quadratic formula is a whole number, so the roots are rational. With whole-number coefficients as well, the factorization comes out in whole numbers. Otherwise the roots are irrational, no whole-number factorization exists, and the roots are shown as decimals.

Example

Take $6x^2 + 5x - 6$. Here $D = 25 + 144 = 169 = 13^2$, a perfect square. The roots are $\dfrac{-5 \pm 13}{12}$, which is $\dfrac{2}{3}$ and $-\dfrac{3}{2}$. Rewriting them as $(3x - 2)$ and $(2x + 3)$ gives $6x^2 + 5x - 6 = (3x - 2)(2x + 3)$.

Notes

A negative discriminant means the expression cannot be split into linear factors over the real numbers.

With $a$ equal to 0 this is not a quadratic, so nothing is calculated.