Factorisation of Polynomials

You now have all the tools: the Factor Theorem, the algebraic identities, and basic polynomial arithmetic. This lesson is where you put them together into a single workflow for factorising polynomials , writing $p(x)$ as a product of simpler polynomials. Done well, factorisation reveals every zero, simplifies every fraction, and clears the way for everything else.

Definition

To factorise a polynomial $p(x)$ is to write it as a product $p(x) = f_1(x) \cdot f_2(x) \cdots f_k(x),$ where each $f_i(x)$ is a polynomial of strictly smaller degree (and is not just a constant, unless we factor out an overall constant for tidiness). The factors over the rationals or reals are usually linear $(x - a)$ or irreducible quadratics.

Three standard methods

We focus on the three methods you will use over and over.

Method 1: Common factor first. Always pull out the greatest common factor (GCF) before doing anything else. $6x^3 - 9x^2 + 3x = 3x(2x^2 - 3x + 1) = 3x(2x - 1)(x - 1).$ The leading constant $3x$ is a factor of every term; factor it out.

Method 2: Identities. Spot a pattern from the identities and read off the factorisation.

• Difference of squares: $x^2 - 9 = (x-3)(x+3)$.
• Difference/sum of cubes: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$, $x^3 + 27 = (x+3)(x^2 - 3x + 9)$.
• Perfect-square trinomial: $x^2 + 6x + 9 = (x+3)^2$, $4x^2 - 12x + 9 = (2x - 3)^2$.

Method 3: Splitting the middle term (for quadratics). To factorise $ax^2 + bx + c$, look for two numbers $p, q$ such that $p + q = b, \qquad p \cdot q = a \cdot c.$ Then split: $ax^2 + bx + c = ax^2 + px + qx + c$, and factor by grouping.

Worked walk-through. Factorise $6x^2 + 17x + 5$.

• $a \cdot c = 6 \cdot 5 = 30$, $b = 17$.
• Find $p, q$ with $p + q = 17$ and $pq = 30$: try $15$ and $2$. Yes: $15 + 2 = 17$, $15 \cdot 2 = 30$.
• Split: $6x^2 + 15x + 2x + 5$.
• Group: $3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5)$. Done.

Method 4: Factor Theorem (for cubics and higher). Try integer values to find a zero, then divide. We covered this in detail in the Factor Theorem lesson.

Worked walk-through. Factorise $p(x) = x^3 - 6x^2 + 11x - 6$.

• $p(1) = 1 - 6 + 11 - 6 = 0$, so $(x - 1)$ is a factor.
• Divide $p(x) \div (x - 1)$: quotient $x^2 - 5x + 6$.
• $x^2 - 5x + 6 = (x - 2)(x - 3)$.
• Final: $(x - 1)(x - 2)(x - 3)$.

Quadratics , a complete recipe

To factorise $ax^2 + bx + c$ over the integers:

1. Pull out any common factor.
2. Compute $a \cdot c$. List factor pairs of $|a \cdot c|$.
3. Among them, find a pair $\{p, q\}$ whose sum (with appropriate signs) is $b$.
4. Split the middle term as $px + qx$.
5. Group and pull common binomials.

If no such pair exists, the quadratic is irreducible over the integers. (You can then use the quadratic formula , coming in Class X , or simply leave it as is.)

A useful sanity check. For $ax^2 + bx + c$, the discriminant is $\Delta = b^2 - 4ac$. The quadratic factorises nicely over the rationals iff $\Delta$ is a perfect square. We will explore this fully next year; for now, knowing the test exists is enough.

Worked examples

Example 1. Factorise $x^2 + 7x + 12$.

Look for $p + q = 7, pq = 12$. The pair $(3, 4)$ works. So $x^2 + 7x + 12 = (x+3)(x+4)$.

Example 2. Factorise $4x^2 - 9$.

A difference of squares: $4x^2 - 9 = (2x)^2 - 3^2 = (2x - 3)(2x + 3)$.

Example 3. Factorise $x^3 - 64$.

Difference of cubes: $x^3 - 4^3 = (x - 4)(x^2 + 4x + 16)$.

Example 4. Factorise $2x^3 - x^2 - 4x + 3$.

Try $x = 1$: $2 - 1 - 4 + 3 = 0$. So $(x - 1)$ is a factor. Divide: quotient $2x^2 + x - 3$. Factor: $a \cdot c = -6$, look for $\{p, q\}$ with sum $1$ and product $-6$ , pair $(3, -2)$. So $2x^2 + x - 3 = 2x^2 + 3x - 2x - 3 = x(2x + 3) - 1(2x + 3) = (x - 1)(2x + 3)$. Final: $p(x) = (x - 1)(x - 1)(2x + 3) = (x - 1)^2 (2x + 3)$.

Example 5. Factorise $x^4 - 81$.

Difference of squares applied twice: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$. The factor $x^2 + 9$ does not factorise further over the reals.

Try it yourself

1. Factorise $x^2 - 11x + 30$.
2. Factorise $3x^2 + 8x + 4$.
3. Factorise $x^2 - 169$.
4. Factorise $x^3 - 125$.
5. Factorise $8x^3 + 27y^3$.
6. Factorise $x^3 - 23x^2 + 142x - 120$ (one zero is among $1, 2, 3, 5, 8, 10, 12, 15$).
7. Factorise $x^4 - 1$.
8. Factorise $x^3 - 3x^2 - 9x - 5$ (try small integer zeros).
9. Factorise $4y^2 - 12y + 9$.
10. Factorise $x^3 + x^2 - 4x - 4$.

Pitfalls / Insight

• Pull out the GCF first. Forgetting this turns simple problems into ugly ones.
• Don't stop too early. $x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)$. The middle step is not the final answer.
• Watch sign conventions when splitting. With $a > 0, c > 0, b > 0$, both $p$ and $q$ are positive. With $a > 0, c > 0, b < 0$, both are negative. Mixed signs of $c$ mean one positive, one negative.

Insight. Factorisation is just reading a polynomial backwards through its identities and its zeros. Each lesson of this chapter contributed one piece: standard identities give patterns to spot, the Factor Theorem promises a linear factor for every zero, and polynomial division does the bookkeeping. Once factored, the polynomial is essentially solved , every zero is visible, every behaviour predictable.