Factorisation of polynomials

Given a cubic like $x^3 - 3x^2 - 4x + 12$ , how do you find its factors without blind trial division of every possible expression? The Remainder Theorem and Factor Theorem turn factorisation into a substitution problem: plug a candidate number into the polynomial, read off the remainder for free, and know instantly whether you have found a factor. Once one linear factor is caught this way, the cubic collapses into a linear times a quadratic, and the quadratic is factorised by the familiar Class IX method of splitting the middle term. This article builds that two-stage machine — test, then divide — and applies it to full cubics up to degree 3, the exact scope of the ICSE syllabus, with two completely worked examples showing every step of the coefficient bookkeeping.

The core insight

Substituting $x=a$ into $f(x)$ gives the remainder on division by $(x-a)$ for free, and when that remainder is exactly zero, $(x-a)$ is a factor — this single fact (the Factor Theorem) turns cubic factorisation into "test a value, then divide."

The core idea

Remainder Theorem. If a polynomial $f(x)$ is divided by $(x-a)$ , the remainder is simply $f(a)$ — no long division needed to find just the remainder.

\text{Remainder when } f(x) \div (x-a) = f(a)

where: $f(x)$ is the polynomial being divided, $a$ is any number, and $f(a)$ means substitute $x=a$ into $f(x)$ and evaluate.

Factor Theorem. $(x-a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ .

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

where: the symbol $\iff$ reads "if and only if" — the theorem works in both directions, so a zero remainder guarantees a factor and a factor guarantees a zero remainder. The Factor Theorem is really the Remainder Theorem at the one special value where the remainder happens to vanish. That is why it is used first, as a fast yes/no test: substitute a candidate $a$ — usually a factor of $f(x)$ 's constant term, tried by trial — and check whether $f(a)=0$ . If it is not zero, you have not wasted the work either: $f(a)$ is the remainder you would have got from a full long division by $(x-a)$ , obtained instantly by substitution. Once a linear factor $(x-a)$ is confirmed, the whole strategy for a cubic reduces to three moves. First, divide $f(x)$ by $(x-a)$ — by long division, or faster, by writing $f(x) = (x-a)(x^2+bx+c)$ , expanding the right side, and comparing coefficients to solve for $b$ and $c$ . Second, factorise the resulting quadratic $x^2+bx+c$ by splitting the middle term, exactly as in Class IX, provided it factorises over rationals. Third, write $f(x)$ as the product of all the linear factors found. At Class 10 level $f(x)$ never exceeds degree 3, so this three-move chain — one Factor Theorem hit, one coefficient-comparison division, one middle-term split — is the entire toolkit; synthetic division and quartic-or-higher factorisation are out of scope.

Figure

A flow diagram showing the factor theorem test substitute a into f x then check if the result is zero leading to a confirmed linear factor or a nonzero remainder

Derivation AFactorising a cubic completely using the Factor Theorem

The goal is to take a cubic polynomial with no obvious factors, $f(x) = x^3 - 3x^2 - 4x + 12$ , and show mechanically — using only substitution, coefficient comparison and splitting the middle term, with no trial-and-error beyond one educated guess for $a$ — that it factors completely into three linear pieces, demonstrating the full test-then-divide-then-split chain that this topic is built on.

Step 1. Choose a candidate $a$ from the factors of the constant term $12$ : try $a=2$ . Compute $f(2) = 2^3 - 3(2)^2 - 4(2) + 12 = 8 - 12 - 8 + 12 = 0$ . Since the remainder is zero, the Factor Theorem confirms $(x-2)$ is a factor of $f(x)$ .

Step 2. Divide $f(x)$ by $(x-2)$ using coefficient comparison. Write $f(x) = (x-2)(x^2+bx+c)$ and expand the right side: $(x-2)(x^2+bx+c) = x^3 + (b-2)x^2 + (c-2b)x - 2c$ . Matching this against $x^3 - 3x^2 - 4x + 12$ term by term gives two equations: from the $x^2$ term, $b-2=-3$ , so $b=-1$ ; from the constant term, $-2c=12$ , so $c=-6$ . A quick check on the $x$ term confirms the fit: $c-2b = -6-2(-1) = -4$ , which matches the given coefficient of $x$ .

Step 3. So $f(x) = (x-2)(x^2-x-6)$ . Factorise the quadratic $x^2-x-6$ by splitting the middle term: find two numbers whose product is $-6$ (the constant) and whose sum is $-1$ (the coefficient of $x$ ) — these are $-3$ and $2$ . So $x^2-x-6 = (x-3)(x+2)$ .

Step 4. Combine every factor found: $f(x) = (x-2)(x-3)(x+2)$ . Each factor can be checked independently by substitution — $f(2)=0$ , $f(3)=0$ , $f(-2)=0$ — confirming the complete factorisation is correct. Verified: $f(2)=0$ , $f(3)=3^3-3(3)^2-4(3)+12=27-27-12+12=0$ , $f(-2)=(-2)^3-3(-2)^2-4(-2)+12=-8-12+8+12=0$ ; all three roots check out, so $f(x)=(x-2)(x-3)(x+2)$ is confirmed.

Worked example

Question: Find the value of $k$ if $(x+1)$ is a factor of $f(x) = 2x^3 + kx^2 + x - 4$ . Hence factorise $f(x)$ completely for that value of $k$ .

Step 1. Apply the Factor Theorem. Since $(x+1) = (x-(-1))$ , the theorem requires $f(-1)=0$ . Substitute: $f(-1) = 2(-1)^3 + k(-1)^2 + (-1) - 4 = -2 + k - 1 - 4 = k-7$ .

Step 2. Solve for $k$ . Setting the remainder to zero, $k-7=0$ , so $k=7$ . The polynomial is now fully known: $f(x) = 2x^3 + 7x^2 + x - 4$ .

Step 3. Divide by $(x+1)$ using coefficient comparison. Write $2x^3+7x^2+x-4 = (x+1)(2x^2+bx+c)$ ; expanding the right side gives $2x^3 + (b+2)x^2 + (c+b)x + c$ . Matching constants gives $c=-4$ ; matching the $x^2$ coefficient gives $b+2=7$ , so $b=5$ . A check on the $x$ term confirms it: $c+b = -4+5 = 1$ , matching the given coefficient.

Step 4. So $f(x) = (x+1)(2x^2+5x-4)$ . Check whether $2x^2+5x-4$ factorises further over rationals using the discriminant, $b^2-4ac = 5^2 - 4(2)(-4) = 25+32=57$ . Since $57$ is not a perfect square, this quadratic has no rational factors and the answer stops here.

Final answer: $k=7$ , and $f(x) = (x+1)(2x^2+5x-4)$ , fully factorised over rationals.

Competitive stretch

If instead you were only given that $(x+1)$ is a factor and asked whether the remaining quadratic could ever split for some value of $k$ , you would set the discriminant of the general quotient equal to a perfect square and solve for $k$ — but at Class 10 level this extension is rarely asked; recognising a non-perfect-square discriminant as a valid stopping point is the actual exam skill being tested.

Common traps

Testing a random value of $a$ instead of a factor of the constant term first — always try factors (positive and negative) of the constant term of $f(x)$ for a faster hit.
Sign errors substituting negative values, especially with odd powers such as $(-1)^3=-1$ — track the sign of every term separately before combining.
Assuming every quadratic factor splits nicely — check the discriminant $b^2-4ac$ ; if it is not a perfect square, the quadratic simply does not factorise further over rationals, and reporting it unfactored is a complete, correct answer.
Forgetting the degree cap of this syllabus — $f(x)$ never exceeds degree 3 here, so there is no need for (or expectation of) quartic or higher factorisation, or named synthetic-division shortcuts.

Ask the tutor: Not sure which trial value of $a$ to test first, or want to double-check a coefficient-comparison step on your own polynomial — ask the AI tutor to work through it with you.

How do I know which value of a to try first with the Factor Theorem?

Start with the factors of the constant term of $f(x)$ , trying both positive and negative versions — for $f(x)=x^3-3x^2-4x+12$ , the constant is $12$ , so sensible first tries are $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$ . Small numbers like $1$ , $-1$ , $2$ , $-2$ succeed most often in textbook problems, so test those first before moving to larger factors. If none of the small factors work, the polynomial may not have a rational root at all, which is unusual for this syllabus level.

Why does coefficient comparison work for dividing a polynomial?

Two polynomials are identical only if every power of $x$ has matching coefficients on both sides — this is a basic property of polynomials, not a special trick. When you write $f(x) = (x-a)(x^2+bx+c)$ and expand the right-hand side, you get another cubic in terms of the unknowns $b$ and $c$ . Since this must equal the original $f(x)$ exactly, matching each power of $x$ gives you as many equations as you need to pin down $b$ and $c$ , without ever performing long division by hand.

What if the quadratic left over after dividing does not factorise at all?

That is a completely valid outcome, not a mistake. Once you have the quadratic $x^2+bx+c$ (or $ax^2+bx+c$ ), compute its discriminant $b^2-4ac$ . If it is a perfect square, splitting the middle term will work and you factorise fully into three linear factors. If it is not a perfect square, the quadratic has no rational factors, and the correct final answer is the linear factor times that unfactored quadratic — as happens in the worked example above with discriminant $57$ .

Factorisation of polynomials

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads