# Factoring Cubic Polynomials

A **cubic polynomial** is a polynomial of the form $f(x)=ax^3+bx^2+cx+d,$ where $a\ne 0.$ If the coefficients are real numbers, the polynomial must factor as the product of a linear polynomial and a quadratic polynomial.

## Existence of a Linear Factor

The fundamental theorem of algebra implies that every irreducible polynomial with real coefficients is linear or quadratic, so a cubic polynomial must split as a product of two lower-degree factors. One of these must be linear and the other quadratic (the quadratic might be irreducible or might itself split into a product of two linear polynomials).

A more down-to-earth way to see that every cubic polynomial has a real root (and hence a linear factor) is to notice that for large $x,$ the lead term $ax^3$ dominates, so the sign of $f(x)$ for large positive $x$ is the sign of $a,$ and the sign of $f(x)$ for large negative $x$ is the sign of $-a.$ So $f(x)$ is negative for some $x$ and positive for others; by the intermediate value theorem there must be at least one point where the graph of $y=f(x)$ crosses the $x$-axis.

## Factoring in Practice

If a given cubic polynomial has rational coefficients and a rational root, it can be found using the rational root theorem.

Factor the polynomial $3x^3 + 4x^2+6x-35$ over the real numbers.

Any rational root of the polynomial has numerator dividing $35$ and denominator dividing $3.$ The possibilities are $\pm 1, \pm 5, \pm 7, \pm 35, \pm \frac13, \pm \frac53, \pm \frac73, \pm \frac{35}3.$ Inspection of these sixteen possibilities yields the unique root $x=\frac53.$ By the remainder factor theorem, $x-\frac53$ (or $3x-5)$ can be factored out of the original polynomial.

Polynomial division of $3x^3+4x^2+6x-35$ by $3x-5$ yields $x^2+3x+7,$ so the polynomial factors as $3x^3+4x^2+6x+35 = (3x-5)(x^2+3x+7).$ Note that the quadratic factor is irreducible over the real numbers, since its discriminant is $3^2-4\cdot 7 = - 19 <0.$ $_\square$

## Factoring - General Case

Unfortunately, "most" polynomials with real coefficients do not have rational roots. So the hardest part of factoring a cubic polynomial in general is finding a real root. Once a root $r$ is found, the polynomial factors as $f(x) = (x-r)g(x),$ where $g(x)$ is quadratic, and quadratic polynomials can be factored easily via the quadratic formula.

Techniques for finding a real root of a cubic polynomial date back to the 16$^\text{th}$ century. See the Cardano Method wiki for an account of the original "cubic formula," and see also the Lagrange Resolvent wiki for details on an alternate approach.

**Cite as:**Factoring Cubic Polynomials.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/factor-polynomials-cubics/