Factorisation of polynomials

A quadratic like $x^2 + 5x + 6$ looks like a dead end until you notice it is secretly a product in disguise: $(x+2)(x+3)$ . Multiply those two brackets back out and the $5x$ and the $6$ reappear exactly. Factorisation is the art of running that multiplication in reverse — writing a polynomial as a product of simpler polynomials instead of a sum of terms. The tool that makes this reliable, rather than a guessing game, is the Factor Theorem. It gives a one-line test — plug in a number, check if the polynomial vanishes — that tells you instantly whether $(x-a)$ divides a polynomial exactly. Combined with splitting the middle term for quadratics, it turns factorisation of quadratics and cubics from trial-and-error into a short, checkable procedure. This article builds the theorem from the Remainder Theorem, applies it to a quadratic and a cubic, and flags the sign slips that cost the most marks.

The core insight

$(x-a)$ is a factor of a polynomial $p(x)$ exactly when $p(a) = 0$ , and this single fact lets you factorise quadratics by splitting the middle term and cubics by finding one integer zero first.

The core idea

Zeroes and factors are two names for the same fact. If a number $a$ makes $p(a) = 0$ , we call $a$ a zero of $p(x)$ ; the Factor Theorem says this is exactly the condition for $(x-a)$ to divide $p(x)$ with no remainder. So "find the zeroes" and "find the linear factors" are the same search.

Statement. Let $p(x)$ be a polynomial of degree $n \geq 1$ and let $a$ be a real number. Then

(x - a)\ \text{is a factor of}\ p(x) \iff p(a) = 0.

where: $p(x)$ is the polynomial being factorised, $a$ is the candidate zero being tested, and $(x-a)$ is the linear expression checked as a factor.

Why it follows from the Remainder Theorem. Dividing $p(x)$ by $(x-a)$ always produces a quotient $q(x)$ and a constant remainder, and that remainder is always $p(a)$ :

p(x) = (x-a)\,q(x) + p(a).

where: $q(x)$ is the quotient polynomial and $p(a)$ is the remainder, which is a fixed number, not an expression in $x$ , because dividing by a linear polynomial always leaves a degree-0 remainder. If $p(a) = 0$ the equation collapses to $p(x) = (x-a)\,q(x)$ , so $(x-a)$ divides $p(x)$ exactly — no leftover. Running the argument the other way, if $(x-a)$ is a factor then $p(x) = (x-a)\,q(x)$ for some $q(x)$ , and substituting $x = a$ gives $p(a) = (a-a)\,q(a) = 0$ . Each direction is a one-line consequence of the division identity.

Splitting the middle term (quadratics). For $ax^2 + bx + c$ with $a \neq 0$ , find two numbers $p$ and $q$ satisfying

p + q = b \quad \text{and} \quad p \times q = ac.

where: $b$ is the coefficient of $x$ and $ac$ is the product of the outer coefficients. Once $p$ and $q$ are found, rewrite $bx$ as $px + qx$ and factor the four resulting terms by grouping.

Cubics via the Factor Theorem. For a cubic $p(x)$ , test small integers — typically the integer factors of the constant term, since these are the only candidates that can give an integer zero. The first $a$ with $p(a) = 0$ hands you a linear factor $(x-a)$ ; the remaining quadratic factor is found by matching coefficients and then split as above.

Figure

Diagram showing a quadratic splitting into two brackets alongside the factor theorem test that a zero of the polynomial gives a linear factor

Derivation AWhy the Factor Theorem holds, and why it forces a testable zero

The goal is to show, starting only from the fact that dividing any polynomial $p(x)$ by a linear polynomial $(x-a)$ always leaves a constant remainder equal to $p(a)$ , that " $(x-a)$ divides $p(x)$ exactly" and " $p(a) = 0$ " are logically the same statement — so testing a single number is enough to certify a factor, with no long division required for the test itself.

Step 1. Write the division identity. For any polynomial $p(x)$ divided by $(x-a)$ , there is always a quotient $q(x)$ and a remainder $r$ , with $p(x) = (x-a)\,q(x) + r$ . Because the divisor $(x-a)$ has degree $1$ , the remainder $r$ must have degree less than $1$ — that is, $r$ is just a constant number, not an expression involving $x$ .

Step 2. Identify the constant $r$ by substitution. Since the identity $p(x) = (x-a)\,q(x) + r$ holds for every value of $x$ , it holds in particular at $x = a$ : $p(a) = (a-a)\,q(a) + r = 0 \cdot q(a) + r = r$ . So the mystery remainder is nothing other than $p(a)$ itself — this is the Remainder Theorem, and it falls out of the division identity with a single substitution.

Step 3. Turn the identity into the forward direction of the Factor Theorem. Substitute $r = p(a)$ back: $p(x) = (x-a)\,q(x) + p(a)$ . If $p(a) = 0$ , this becomes exactly $p(x) = (x-a)\,q(x)$ , which says $(x-a)$ divides $p(x)$ with nothing left over — precisely the definition of $(x-a)$ being a factor.

Step 4. Prove the converse and close the loop. Suppose instead $(x-a)$ is known to be a factor, so $p(x) = (x-a)\,q(x)$ for some polynomial $q(x)$ . Substituting $x = a$ gives $p(a) = (a-a)\,q(a) = 0 \times q(a) = 0$ . Both directions now hold: $(x-a)$ is a factor if and only if $p(a) = 0$ . This is why a single arithmetic check — evaluate $p(a)$ — is enough to certify or rule out a candidate factor, without ever performing the division.

Worked example

Question: Factorise the cubic $p(x) = x^3 - 6x^2 + 11x - 6$ using the Factor Theorem, then verify by splitting the resulting quadratic and checking every root.

Step 1. List candidates. The constant term is $-6$ , whose integer factors are $\pm 1, \pm 2, \pm 3, \pm 6$ — these are the only numbers that can possibly give an integer zero, since any other integer plugged into a polynomial with integer coefficients cannot make the constant term's contribution cancel exactly.

Step 2. Test the smallest candidate first. $p(1) = 1 - 6 + 11 - 6 = 0$ . Since $p(1) = 0$ , the Factor Theorem guarantees $(x-1)$ is a factor of $p(x)$ .

Step 3. Find the remaining quadratic factor by comparing coefficients. Write $p(x) = (x-1)(x^2 + bx + c)$ and expand the right side: $x^3 + (b-1)x^2 + (c-b)x - c$ . Matching the constant terms gives $-c = -6$ , so $c = 6$ ; matching the $x^2$ coefficients gives $b - 1 = -6$ , so $b = -5$ . A quick check on the $x$ coefficient, $c - b = 6-(-5) = 11$ , matches the original — confirming $p(x) = (x-1)(x^2 - 5x + 6)$ .

Step 4. Split the middle term of the quadratic. For $x^2 - 5x + 6$ , seek two numbers with sum $-5$ and product $6$ : these are $-2$ and $-3$ . So $x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = x(x-2) - 3(x-2) = (x-2)(x-3)$ .

Step 5. Combine and verify every root. $p(x) = (x-1)(x-2)(x-3)$ . Checking: $p(2) = 8 - 24 + 22 - 6 = 0$ and $p(3) = 27 - 54 + 33 - 6 = 0$ , both confirming the factorisation. Final answer: $x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$ .

Competitive stretch

If instead the cubic were $p(x) = x^3 - 2x^2 - 5x + 6$ , the same routine applies: test $x = 1$ first, since $1$ is always the cheapest candidate to check. $p(1) = 1 - 2 - 5 + 6 = 0$ , so $(x-1)$ is a factor; dividing out (by the same coefficient-matching method) gives $p(x) = (x-1)(x^2 - x - 6)$ , and splitting $x^2 - x - 6$ (numbers with sum $-1$ , product $-6$ : these are $-3$ and $2$ ) gives $(x-3)(x+2)$ . So $p(x) = (x-1)(x-3)(x+2)$ — the same three-step chain, just different numbers.

Common traps

Testing random or large integers first when hunting for a zero of a cubic — always start with factors of the constant term (both positive and negative), since only these can possibly give $p(a) = 0$ for a polynomial with integer coefficients.
Sign errors while splitting the middle term: for $x^2 - 5x + 6$ , choosing $2$ and $3$ instead of $-2$ and $-3$ . The pair must multiply to the positive $c$ but sum to the negative $b$ — check both conditions, not just the product.
Stopping after finding one linear factor of a cubic and forgetting to factorise the remaining quadratic completely — a cubic with three real integer zeroes needs all three brackets written out.
Skipping verification. After factorising, either expand the brackets back out or re-check $p(a) = 0$ for every claimed root; a single arithmetic slip earlier propagates silently otherwise.

Ask the tutor: Want to try the Factor Theorem on a cubic of your own choosing and get every step checked?

Why do we only try factors of the constant term when hunting for a zero of a cubic?

Because if $p(x) = x^3 + \cdots + k$ has an integer zero $a$ , substituting $x=a$ must make every term combine to exactly cancel the constant $k$ . Working through the algebra, this forces $a$ to divide $k$ exactly — so any integer that is not a factor of the constant term cannot possibly be a zero, and testing it would waste time. Starting with the smallest factors first (like $\pm1$ ) usually finds the zero fastest.

Do I always have to use the Factor Theorem, or can I sometimes spot factors directly?

For simple quadratics, splitting the middle term directly (as in Worked Example, Step 4 above) is usually faster and does not need the Factor Theorem at all. The Factor Theorem becomes essential once a linear factor is not obvious by inspection — especially for cubics, where you genuinely need a systematic way to find the first zero before any splitting can begin.

What if none of the factors of the constant term give p(a) = 0?

Then the cubic does not have a rational (in fact, integer) zero among the numbers you tried, and it cannot be factorised into three linear factors using this trial method. At the Class 9 level, questions are set up so that a small integer zero always exists; if your careful trial of every factor of the constant term (positive and negative) still gives no zero, re-check your arithmetic before concluding the polynomial resists this technique.

Factorisation of polynomials

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads