The Factor Theorem and Remainder Theorem

The factor theorem and remainder theorem are the two pillars of polynomial factorisation beyond degree 2. Together they give you a systematic way to find roots of cubics, quartics and higher-degree polynomials without long division — by evaluation alone. Mastery of these theorems unlocks every later topic that depends on factorising: algebraic fractions, partial fractions, integration of rational functions, sketching polynomial curves, and the binomial theorem in disguise.

---

Spec mapping

Edexcel 9MA0 specification, Pure Mathematics Paper 1 — Section 2.4 / 2.5 covers the structure of mathematical proof, proceeding from given assumptions through a series of logical steps to a conclusion; use the factor theorem and remainder theorem to factorise polynomials and to evaluate values of polynomials (refer to the official specification document for exact wording). This material sits at the heart of Paper 1 and is examined every series in some form — most often a 6–10 mark cubic question that asks you to find an unknown coefficient via the factor theorem, factorise fully, and either solve or sketch.

The factor and remainder theorems also underpin Section 4 — Sequences and series (binomial expansion as a polynomial identity), Section 7 — Algebraic fractions and partial fractions (denominator factorisation is a prerequisite for partial-fraction decomposition), Section 9 — Differentiation (locating turning points of polynomial curves) and Section 8 — Integration (integrating rational functions almost always begins with factorising the denominator). The Edexcel formula booklet does not list either theorem — they must be memorised in their generalised forms.

---

The Remainder Theorem

Statement

If a polynomial $P(x)$P(x) is divided by a linear divisor $(x - a)$(x−a), the remainder is $P(a)$P(a).

That is: there exist a quotient polynomial $Q(x)$Q(x) and a constant remainder $r$r such that

$P(x) = (x - a)\,Q(x) + r, \qquad r = P(a).$P(x)=(x−a)Q(x)+r,r=P(a).

Why it works — proof sketch via the division algorithm

The polynomial division algorithm says: given $P(x)$P(x) and a non-zero divisor $D(x)$D(x), there exist unique polynomials $Q(x)$Q(x) and $R(x)$R(x) with

$P(x) = D(x)\,Q(x) + R(x), \qquad \deg R < \deg D.$P(x)=D(x)Q(x)+R(x),degR<degD.

This is the polynomial analogue of long division of integers. When $D(x) = (x - a)$D(x)=(x−a) has degree 1, the remainder $R(x)$R(x) has degree strictly less than 1, so it must be a constant. Substituting $x = a$x=a into both sides:

$P(a) = (a - a)\,Q(a) + R = 0 + R = R.$P(a)=(a−a)Q(a)+R=0+R=R.

So the constant remainder equals $P(a)$P(a). The argument is a one-line consequence of the division algorithm — you do not need to perform the division to find the remainder.

Generalised form for divisor $(ax - b)$(ax−b)

If $P(x)$P(x) is divided by the linear divisor $(ax - b)$(ax−b) (with $a \neq 0$a=0), the remainder is

$r = P\!\left(\dfrac{b}{a}\right).$r=P(ab​).

The proof is identical: the unique remainder is a constant, and substituting $x = b/a$x=b/a — the unique root of $ax - b = 0$ax−b=0 — into $P(x) = (ax - b)Q(x) + r$P(x)=(ax−b)Q(x)+r gives $P(b/a) = r$P(b/a)=r.

Exam Tip: When a divisor has leading coefficient other than 1, do not substitute $x = b$x=b — substitute $x = b/a$x=b/a. The most frequent slip on (2x − 1) divisors is computing P(1) instead of P(1/2).

Worked Example 1 — finding a remainder

Find the remainder when $P(x) = 2x^3 - 5x^2 + 3x - 7$P(x)=2x3−5x2+3x−7 is divided by $(x - 2)$(x−2).

By the remainder theorem, the remainder equals $P(2)$P(2):

$P(2) = 2(8) - 5(4) + 3(2) - 7 = 16 - 20 + 6 - 7 = -5.$P(2)=2(8)−5(4)+3(2)−7=16−20+6−7=−5.

The remainder is $-5$−5. No long division required.

---

The Factor Theorem

Statement

$(x - a)$(x−a) is a factor of the polynomial $P(x)$P(x) if and only if $P(a) = 0$P(a)=0.

Why it follows from the remainder theorem

By the remainder theorem, dividing $P(x)$P(x) by $(x - a)$(x−a) gives $P(x) = (x - a)Q(x) + P(a)$P(x)=(x−a)Q(x)+P(a). The divisor $(x - a)$(x−a) is a factor precisely when the remainder is zero — that is, when $P(a) = 0$P(a)=0. The "if and only if" is symmetric: a zero of $P$P at $x = a$x=a produces a factor $(x - a)$(x−a), and conversely the factor $(x - a)$(x−a) forces a zero at $x = a$x=a.

Generalised form

$(ax - b)$(ax−b) is a factor of $P(x)$P(x) if and only if $P(b/a) = 0$P(b/a)=0. Same proof, same caveat: substitute the root of the divisor, not the constant term.

Worked Example 2 — factor theorem with non-monic divisor

Show that $(2x - 1)$(2x−1) is a factor of $P(x) = 6x^3 + x^2 - 4x + 1$P(x)=6x3+x2−4x+1.

The divisor $(2x - 1) = 0$(2x−1)=0 at $x = \tfrac{1}{2}$x=21​. Compute $P(\tfrac{1}{2})$P(21​):

$P\!\left(\tfrac{1}{2}\right) = 6 \cdot \tfrac{1}{8} + \tfrac{1}{4} - 4 \cdot \tfrac{1}{2} + 1 = \tfrac{3}{4} + \tfrac{1}{4} - 2 + 1 = 1 - 1 = 0.$P(21​)=6⋅81​+41​−4⋅21​+1=43​+41​−2+1=1−1=0.

Since $P(\tfrac{1}{2}) = 0$P(21​)=0, by the (generalised) factor theorem $(2x - 1)$(2x−1) is a factor of $P(x)$P(x).

---

Systematic factorising of cubics

A general cubic $P(x) = ax^3 + bx^2 + cx + d$P(x)=ax3+bx2+cx+d has at most three real roots. The factor theorem reduces factorisation to a two-step procedure:

1. Find one root by trial substitution. Test small integers ($\pm 1, \pm 2, \pm 3, \dots$±1,±2,±3,…) and small rationals dictated by the rational root theorem (below).
2. Divide the cubic by the linear factor $(x - a)$(x−a) to obtain a quadratic quotient. Then factorise (or apply the quadratic formula to) the quadratic.

The division step can be done by polynomial long division, by inspection (assume the quotient has the form $ax^2 + px + q$ax2+px+q and equate coefficients), or — historically — by Ruffini's rule (1804), the synthetic-division shortcut still taught in many continental syllabuses.

Worked Example 3 — factorising a cubic

Factorise $f(x) = x^3 - 4x^2 + x + 6$f(x)=x3−4x2+x+6 completely, and hence solve $f(x) = 0$f(x)=0.

Step 1 — find a root. Try $x = 1$x=1: $1 - 4 + 1 + 6 = 4 \neq 0$1−4+1+6=4=0. Try $x = -1$x=−1: $-1 - 4 - 1 + 6 = 0$−1−4−1+6=0. So $(x + 1)$(x+1) is a factor.

Step 2 — divide. Assume $f(x) = (x + 1)(x^2 + px + q)$f(x)=(x+1)(x2+px+q). Expanding the right-hand side:

$(x + 1)(x^2 + px + q) = x^3 + px^2 + qx + x^2 + px + q = x^3 + (p + 1)x^2 + (p + q)x + q.$(x+1)(x2+px+q)=x3+px2+qx+x2+px+q=x3+(p+1)x2+(p+q)x+q.

Equating coefficients with $f(x) = x^3 - 4x^2 + x + 6$f(x)=x3−4x2+x+6:

• $x^2$x2: $p + 1 = -4 \implies p = -5$p+1=−4⟹p=−5.
• constant: $q = 6$q=6.
• check the $x$x coefficient: $p + q = -5 + 6 = 1$p+q=−5+6=1. ✓

So $f(x) = (x + 1)(x^2 - 5x + 6)$f(x)=(x+1)(x2−5x+6). The quadratic factor factorises further: $x^2 - 5x + 6 = (x - 2)(x - 3)$x2−5x+6=(x−2)(x−3). Hence

$f(x) = (x + 1)(x - 2)(x - 3),$f(x)=(x+1)(x−2)(x−3),

and $f(x) = 0$f(x)=0 when $x = -1$x=−1, $x = 2$x=2 or $x = 3$x=3.

---

The Rational Root Theorem and bounded search

Statement (rational root theorem)

If a polynomial with integer coefficients

$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$P(x)=an​xn+an−1​xn−1+⋯+a1​x+a0​

has a rational root $x = p/q$x=p/q in lowest terms, then $p$p divides the constant term $a_0$a0​ and $q$q divides the leading coefficient $a_n$an​.

This is a corollary of Gauss's lemma on primitive polynomials (a result from elementary number theory, normally proved in Year 1 ring theory). The practical effect at A-Level is dramatic: rather than guessing roots blindly, you have a finite list of rational candidates to test.

How to apply it

Given a cubic $2x^3 + 3x^2 - 8x + 3$2x3+3x2−8x+3:

• Constant term $a_0 = 3$a0​=3, with divisors $\pm 1, \pm 3$±1,±3.
• Leading coefficient $a_n = 2$an​=2, with divisors $\pm 1, \pm 2$±1,±2.
• Possible rational roots $p/q$p/q: $\pm 1, \pm 3, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$±1,±3,±21​,±23​.

Eight candidates total. Substitute each into $P(x)$P(x) and any that gives zero is a root.

Rational-root search table

Polynomial	Constant term	Leading coefficient	Candidates $p/q$p/q
$x^3 + 2x^2 - 5x - 6$x3+2x2−5x−6	$-6$−6	$1$1	$\pm 1, \pm 2, \pm 3, \pm 6$±1,±2,±3,±6
$2x^3 - 3x^2 - 11x + 6$2x3−3x2−11x+6	$6$6	$2$2	$\pm 1, \pm 2, \pm 3, \pm 6, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$±1,±2,±3,±6,±21​,±23​
$3x^3 + x^2 - x + 2$3x3+x2−x+2	$2$2	$3$3	$\pm 1, \pm 2, \pm \tfrac{1}{3}, \pm \tfrac{2}{3}$±1,±2,±31​,±32​
$x^4 - 5x^2 + 4$x4−5x2+4	$4$4	$1$1	$\pm 1, \pm 2, \pm 4$±1,±2,±4

The theorem bounds the search — but it does not guarantee that a rational root exists. A polynomial like $x^3 - 2$x3−2 has only the irrational root $\sqrt[3]{2}$32​, and the rational candidates $\pm 1, \pm 2$±1,±2 all fail. In that case the polynomial is irreducible over the rationals, and exact factorisation requires moving to a larger number field — a topic for university algebra.

Exam Tip: Edexcel cubics are deliberately constructed to have at least one small integer root (typically $\pm 1$±1 or $\pm 2$±2). Always test these first before reaching for fractions. If no integer root works, suspect a misprint — or that the question wants you to use the rational root theorem to eliminate possibilities.

---

Using the theorems to find unknown coefficients

A signature Edexcel question pattern: a polynomial contains one or more unknown letters (typically $a$a, $b$b, $k$k), and you are given factor or remainder information. Each piece of information is one equation; with two unknowns, you need two equations. With one unknown, one equation suffices.

Worked Example 4 — two unknowns from two conditions

Given $f(x) = 2x^3 + ax^2 + bx - 6$f(x)=2x3+ax2+bx−6, where $(x - 1)$(x−1) is a factor of $f(x)$f(x) and the remainder when $f(x)$f(x) is divided by $(x + 1)$(x+1) is $-12$−12, find the values of $a$a and $b$b.

Condition 1 (factor): $(x - 1)$(x−1) is a factor, so $f(1) = 0$f(1)=0:

$f(1) = 2 + a + b - 6 = 0 \implies a + b = 4. \quad (1)$f(1)=2+a+b−6=0⟹a+b=4.(1)

Condition 2 (remainder): dividing by $(x + 1)$(x+1) gives remainder $f(-1) = -12$f(−1)=−12:

$f(-1) = -2 + a - b - 6 = -12 \implies a - b = -4. \quad (2)$f(−1)=−2+a−b−6=−12⟹a−b=−4.(2)

Solve simultaneously. Adding (1) and (2): $2a = 0 \implies a = 0$2a=0⟹a=0. Subtracting (1) − (2): $2b = 8 \implies b = 4$2b=8⟹b=4.

Verify. With $a = 0$a=0 and $b = 4$b=4: $f(x) = 2x^3 + 4x - 6$f(x)=2x3+4x−6.

• $f(1) = 2 + 4 - 6 = 0$f(1)=2+4−6=0 ✓
• $f(-1) = -2 - 4 - 6 = -12$f(−1)=−2−4−6=−12 ✓

So $a = 0$a=0 and $b = 4$b=4.

---

Solving polynomial equations after factorisation

Once a polynomial is fully factorised over the reals, solving $P(x) = 0$P(x)=0 is mechanical: set each factor equal to zero. The subtlety lies in how many real roots emerge:

• A linear factor $(x - r)$(x−r) contributes the real root $x = r$x=r.
• An irreducible-over-$\mathbb{R}$R quadratic factor $(x^2 + px + q)$(x2+px+q) with negative discriminant contributes no real roots (its zeros are complex conjugates, studied at Further Maths).
• A repeated linear factor $(x - r)^2$(x−r)2 contributes a single real root $x = r$x=r where the curve is tangent to the x-axis.

A cubic always has at least one real root (the intermediate value theorem guarantees this for any odd-degree polynomial with real coefficients), but may have only one if the quadratic factor is irreducible. A quartic may have 0, 2 or 4 real roots, counting multiplicity.

Sketching after factorising

Once $f(x) = a(x - r_1)(x - r_2)(x - r_3)$f(x)=a(x−r1​)(x−r2​)(x−r3​) is fully factorised:

1. Mark the x-intercepts at $r_1, r_2, r_3$r1​,r2​,r3​.
2. Find the y-intercept by computing $f(0)$f(0).
3. Determine end-behaviour from the leading coefficient: as $x \to \infty$x→∞, $f(x) \to +\infty$f(x)→+∞ if $a > 0$a>0 and $-\infty$−∞ if $a < 0$a<0.
4. Connect the intercepts smoothly, respecting end-behaviour.

For a cubic with three distinct real roots and $a > 0$a>0, the graph crosses the x-axis three times in a characteristic "down-up-down-up" pattern from left to right.

---

Quartics with quadratic factors

Some quartics factorise as a product of two quadratics rather than four linear factors. The factor theorem alone cannot find the linear factors if none exist — but combined with the division algorithm, you can extract a quadratic factor by finding two roots and dividing by their product.

Worked Example 5 — quartic with four real roots

Factorise $g(x) = x^4 + 2x^3 - 7x^2 - 8x + 12$g(x)=x4+2x3−7x2−8x+12 completely, and solve $g(x) = 0$g(x)=0.

Step 1 — find roots. Constant term 12 has divisors $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$±1,±2,±3,±4,±6,±12. Test:

• $g(1) = 1 + 2 - 7 - 8 + 12 = 0$g(1)=1+2−7−8+12=0 ✓ — so $(x - 1)$(x−1) is a factor.
• $g(-2) = 16 - 16 - 28 + 16 + 12 = 0$g(−2)=16−16−28+16+12=0 ✓ — so $(x + 2)$(x+2) is a factor.

Step 2 — divide by the quadratic $(x - 1)(x + 2) = x^2 + x - 2$(x−1)(x+2)=x2+x−2. Assume

$g(x) = (x^2 + x - 2)(x^2 + px + q).$g(x)=(x2+x−2)(x2+px+q).

Expanding the right-hand side:

$(x^2 + x - 2)(x^2 + px + q) = x^4 + px^3 + qx^2 + x^3 + px^2 + qx - 2x^2 - 2px - 2q$(x2+x−2)(x2+px+q)=x4+px3+qx2+x3+px2+qx−2x2−2px−2q $= x^4 + (p + 1)x^3 + (q + p - 2)x^2 + (q - 2p)x - 2q.$=x4+(p+1)x3+(q+p−2)x2+(q−2p)x−2q.

Equate coefficients with $x^4 + 2x^3 - 7x^2 - 8x + 12$x4+2x3−7x2−8x+12:

• $x^3$x3: $p + 1 = 2 \implies p = 1$p+1=2⟹p=1.
• constant: $-2q = 12 \implies q = -6$−2q=12⟹q=−6.
• check $x^2$x2: $q + p - 2 = -6 + 1 - 2 = -7$q+p−2=−6+1−2=−7 ✓
• check $x$x: $q - 2p = -6 - 2 = -8$q−2p=−6−2=−8 ✓

Step 3 — factorise the second quadratic. $x^2 + x - 6 = (x - 2)(x + 3)$x2+x−6=(x−2)(x+3). Hence

$g(x) = (x - 1)(x + 2)(x - 2)(x + 3),$g(x)=(x−1)(x+2)(x−2)(x+3),

and $g(x) = 0$g(x)=0 when $x \in \{-3, -2, 1, 2\}$x∈{−3,−2,1,2}.

When a quadratic factor has negative discriminant, the second factor is irreducible over the reals and contributes no further real roots. The technique still works — you just stop after extracting the irreducible quadratic.

---

Common question types

Q-type	Trigger phrase	Technique
Find remainder	"Find the remainder when … is divided by …"	Apply remainder theorem: substitute the root of the divisor
Show factor	"Show that (x − a) is a factor"	Apply factor theorem: verify P(a) = 0
Factorise cubic	"Factorise f(x) completely"	Find one root by trial, divide by the linear factor, factorise the resulting quadratic
Solve cubic	"Solve f(x) = 0"	Factorise completely, then set each factor to zero
Find unknown coefficient(s)	"Given (x − a) is a factor and remainder is r when divided by (x − b), find …"	Set up simultaneous equations from each condition
Sketch polynomial	"Sketch y = f(x), showing intercepts"	Factorise, mark x-intercepts and y-intercept, use end-behaviour
Quartic factorisation	"Factorise the quartic g(x)"	Find two rational roots, divide by their quadratic product

---