Factoring Cubic Polynomials | Brilliant Math & Science Wiki

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Brody Acquilano, Patrick Corn, Ahmed Kamel, and 1 other

• Jimin Khim

contributed

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.

Contents

• Existence of a Linear Factor
• Factoring in Practice
• Factoring - General Case

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.

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 \( 3^2-4\cdot 7 = - 19 <0.\) \(_\square\)

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.

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