Factorising cubic equations expresses a cubic polynomial as a product of simpler factors, typically one linear factor and one quadratic factor, with solutions obtained from the zeros of the factors. The setting throughout is a cubic with integer coefficients written as \(f(x)=ax^3+bx^2+cx+d\) with \(a \neq 0\).
The root–factor link is the organising principle: \(f(r)=0\) is equivalent to saying \((x-r)\) is a factor of \(f(x)\). When \(f(x)\) is written as a product, the equation \(f(x)=0\) becomes a zero-product statement.
Factor theorem and the zero-product property
- Factor theorem: For a polynomial \(f(x)\), \(f(r)=0\) if and only if \((x-r)\) divides \(f(x)\).
- Zero-product property: If \(uv=0\), then \(u=0\) or \(v=0\). For a factorised polynomial, each factor can be set to zero.
- Typical cubic structure: A real-coefficient cubic can be written as \(a(x-r)(x^2+px+q)\). The quadratic factor may split further over the reals or remain irreducible over the integers.
Rational-root candidates in integer-coefficient cubics
A frequent algebra pathway is a rational root check. For \(f(x)=ax^3+bx^2+cx+d\) with integers \(a,b,c,d\), any rational root has the form \[ r=\frac{p}{q}, \] where \(p\) divides \(d\) and \(q\) divides \(a\) (with \(p\) and \(q\) chosen to be coprime).
Evaluation of \(f(r)\) is a direct verification of a candidate. Once a root \(r\) is confirmed, division by \((x-r)\) reduces the cubic to a quadratic factor. Polynomial long division and synthetic division provide the same quotient.
Common factorisation patterns for cubics
- Common factor extraction: Expressions such as \(x^3+3x^2\) share a factor of \(x^2\).
- Grouping into pairs: Four-term cubics are often rearranged into two binomials with a common binomial factor.
- Special identities: \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and \(a^3+b^3=(a+b)(a^2-ab+b^2)\).
Worked example with three integer roots
Consider the equation \[ x^3 - 6x^2 + 11x - 6 = 0. \]
The constant term is \(-6\), so integer candidates are \(\pm 1,\pm 2,\pm 3,\pm 6\). Direct evaluation shows \[ f(1)=1-6+11-6=0, \] so \((x-1)\) is a factor. Division gives a quadratic quotient: \[ x^3 - 6x^2 + 11x - 6 = (x-1)(x^2 - 5x + 6). \] The quadratic factor splits over the integers: \[ x^2 - 5x + 6 = (x-2)(x-3). \]
The factorised form and solutions are \[ (x-1)(x-2)(x-3)=0 \quad\Rightarrow\quad x\in\{1,2,3\}. \]
Visual interpretation of a factorised cubic
Practice set with a compact answer key
Each equation is presented in the standard form \(f(x)=0\). The factorised form provides the real solutions immediately.
| Cubic equation | Factorised form | Real solutions |
|---|---|---|
| \(x^3-6x^2+11x-6=0\) | \((x-1)(x-2)(x-3)=0\) | \(x=1,2,3\) |
| \(x^3+2x^2-x-2=0\) | \((x+2)(x-1)(x+1)=0\) | \(x=-2,-1,1\) |
| \(2x^3+x^2-8x-4=0\) | \((2x+1)(x-2)(x+2)=0\) | \(x=-\tfrac12,-2,2\) |
| \(x^3-27=0\) | \((x-3)(x^2+3x+9)=0\) | \(x=3\) |
| \(x^3+3x^2=0\) | \(x^2(x+3)=0\) | \(x=0\) (double), \(x=-3\) |
Common pitfalls
- Sign handling: Errors in \(f(r)\) evaluation often arise from distributing minus signs incorrectly in \(bx^2\) or \(d\).
- Candidate list completeness: For non-monic cubics (\(a \neq 1\)), rational candidates include fractions \(\pm \frac{p}{q}\), not only integer factors of \(d\).
- Unfinished factorisation: A confirmed linear factor reduces the problem to a quadratic; completing the quadratic factorisation is essential for a full answer over the reals.
- Multiplicity meaning: Repeated factors correspond to repeated roots; \(x^2(x+3)=0\) contains \(x=0\) with multiplicity \(2\).