Factorisation of Polynomials: Remainder and Factor Theorems notes: ICSE Class 10 Mathematics

1. Dividing a polynomial by a linear expression

Division algorithm: dividend = divisor × quotient + remainder, that is $f(x) = (x - a)q(x) + r$ .
The remainder has a smaller degree than the divisor, so on division by a linear expression it is a constant.
Long division: divide the leading terms, multiply back, subtract; repeat until the remainder is a constant.

Example: $2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7$ .

This chapter uses polynomials of degree up to $3$ only; polynomials of degree higher than $3$ are not in the ICSE 2027 syllabus.

Quotient $2x^2 + x + 6$, remainder $7$: $2x^3 - 3x^2 + 4x - 5 = (x - 2)(2x^2 + x + 6) + 7$.
Step	Divide the leading terms	Multiply back	Subtract to get
1	$\dfrac{2x^3}{x} = 2x^2$	$2x^2(x - 2) = 2x^3 - 4x^2$	$x^2 + 4x - 5$
2	$\dfrac{x^2}{x} = x$	$x(x - 2) = x^2 - 2x$	$6x - 5$
3	$\dfrac{6x}{x} = 6$	$6(x - 2) = 6x - 12$	$7$

Formulas

Division algorithm
$f(x) = (x - a)\,q(x) + r$

Key terms

Quotient: The polynomial $q(x)$ in $f(x) = (x - a)q(x) + r$ .
Remainder: What is left after dividing; for a linear divisor it is a constant.

2. The remainder theorem

When $f(x)$ is divided by $(x - a)$ , the remainder is $f(a)$ . Put $x = a$ in $f(x) = (x - a)q(x) + r$ : the first term becomes $0$ .

Divisor $(x + a)$ : remainder $f(-a)$ .
Divisor $(ax + b)$ : remainder $f\left(-\dfrac{b}{a}\right)$ ; divisor $(ax - b)$ : remainder $f\left(\dfrac{b}{a}\right)$ .
Rule: put the divisor equal to $0$ and substitute that value.

Example: $x^3 + 4x^2 - 3x + 5$ divided by $(x + 3)$ : $f(-3) = -27 + 36 + 9 + 5 = 23$ .

Long division and substitution give the same remainder every time.
Divisor	Remainder by long division	Value of $f(x)$ at the zero of the divisor
$(x - 2)$	$7$	$f(2) = 7$
$(x - 1)$	$-2$	$f(1) = -2$
$(x + 1)$	$-14$	$f(-1) = -14$
$(x + 3)$	$-98$	$f(-3) = -98$

Substitute the value of $x$ that makes the divisor zero.
Divisor	Put it equal to $0$	Remainder
$(x - 4)$	$x = 4$	$f(4)$
$(x + 3)$	$x = -3$	$f(-3)$
$(2x - 1)$	$x = \dfrac{1}{2}$	$f\left(\dfrac{1}{2}\right)$
$(3x + 2)$	$x = -\dfrac{2}{3}$	$f\left(-\dfrac{2}{3}\right)$

$y = x^3 + 4x^2 - 3x + 5$

At $x = -3$ the graph is at height $23$: the remainder.

Formulas

Divisor x − a
$r = f(a)$
Divisor ax + b
$r = f\left(-\dfrac{b}{a}\right)$

Key terms

Remainder theorem: The remainder when $f(x)$ is divided by $(x - a)$ is $f(a)$ .
Zero of a divisor: The value of $x$ that makes the divisor $0$ ; for $2x - 1$ it is $\dfrac{1}{2}$ .

3. Unknown coefficients from given remainders

Substitute the zero of each divisor, keeping the unknowns as letters.
Set each result equal to the given remainder: one linear equation per condition.
Solve (simultaneously for two unknowns) and substitute back to check.

Examples: $x^3 + kx^2 - 7x + 4$ leaves $6$ on division by $(x - 2)$ : $4k - 2 = 6$ , so $k = 2$ . Equal remainders: set the two remainders equal to each other.

One condition for each unknown.

Two conditions give two linear equations in $p$ and $q$.
Divisor	Substitute	Remainder	Equation
$(x - 2)$	$x = 2$	$8$	$2p + q = 3$
$(x + 1)$	$x = -1$	$-4$	$p - q = 3$

$y = x^3 + 2x^2 - 7x + 4$

With $k = 2$ the graph passes through $(2, 6)$: the remainder is $6$.

Formulas

Remainder R on division by x − a
$f(a) = R$

Key terms

Condition: A given remainder or factor; each gives one equation in the unknowns.

4. The factor theorem

$(x - a)$ is a factor of $f(x)$ exactly when $f(a) = 0$ (the remainder theorem with remainder $0$ ).

$f(-a) = 0 \iff (x + a)$ is a factor; $f\left(\dfrac{b}{c}\right) = 0 \iff (cx - b)$ is a factor.
A zero of $f$ is where the graph of $y = f(x)$ meets the $x$ -axis: each zero gives a factor.
If $f(a) \ne 0$ , then $(x - a)$ is not a factor and $f(a)$ is the remainder.

$y = 2x^3 - x^2 - 7x + 6$

Each zero gives a factor: $-2 \to (x + 2)$, $1 \to (x - 1)$, $\dfrac{3}{2} \to (2x - 3)$.

Value $0$ means a factor; any other value is the remainder.
Expression	Substitute	Value of $f(x)$	Factor?
$(x - 3)$	$x = 3$	$0$	yes
$(x + 1)$	$x = -1$	$-4$	no: remainder $-4$

The factor theorem is the remainder theorem with remainder $0$.
If $f(a)$ is…	then dividing by $(x - a)$ …	and the graph of $y = f(x)$ at $x = a$ …
not $0$	leaves the remainder $f(a)$	is at height $f(a)$ , off the axis
$0$	leaves no remainder: $(x - a)$ is a factor	meets the $x$ -axis: $a$ is a zero

Formulas

Factor theorem
$f(a) = 0 \iff (x - a) \text{ is a factor}$

Key terms

Factor theorem: $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$ .
Zero of a polynomial: A value $a$ with $f(a) = 0$ .

5. Unknown coefficients from given factors

One factor $(x - a)$ and one unknown: solve $f(a) = 0$ .
Two unknowns: use two conditions (two factors, or a factor and a remainder) and solve the two linear equations together.
A divisor such as $x^2 - x - 2$ is the product $(x - 2)(x + 1)$ : use $f(2) = 0$ and $f(-1) = 0$ .

Example: $(x - 2)$ and $(x + 3)$ are factors of $x^3 + ax^2 + bx - 12$ : $2a + b = 2$ and $3a - b = 13$ , so $a = 3$ , $b = -4$ .

Two factors, two equations, two unknowns.
Factor	Condition	Equation
$(x - 2)$	$f(2) = 0$	$2a + b = 2$
$(x + 3)$	$f(-3) = 0$	$3a - b = 13$
Solve	add the equations	$a = 3$ , $b = -4$

$y = 2x^3 - x^2 - 13x - 6$

With $k = -1$: zeros $-2$, $-\dfrac{1}{2}$ and $3$.

Formulas

Given factor x − a
$f(a) = 0$

Key terms

Simultaneous equations: Two linear equations in two unknowns, solved together.

6. Factorising a cubic completely

Trial: try divisors of the constant term ( $\pm 1, \pm 2, \ldots$ ); if the leading coefficient is not $1$ , also try fractions $\dfrac{p}{q}$ with $q$ dividing it.
When $f(a) = 0$ , $(x - a)$ is a factor. Divide (long division or by inspection): $f(x) = (x - a)q(x)$ .
Factorise the quadratic $q(x)$ by splitting the middle term.
If $q(x)$ has a negative discriminant, it cannot be factorised further.

Example: $x^3 - 3x^2 - 10x + 24$ : $f(2) = 0$ , quotient $x^2 - x - 12$ , so $(x - 2)(x - 4)(x + 3)$ .

Trial, divide, factorise, check.

Try $\pm 1$, $\pm 2$, … (divisors of $24$) until the value is $0$.
$x$	$f(x)$	Factor?
$1$	$12$	no
$-1$	$30$	no
$2$	$0$	yes: $(x - 2)$

Quotient $2x^2 + x - 15$, remainder $0$.
Step	Divide the leading terms	Multiply back	Subtract to get
1	$\dfrac{2x^3}{x} = 2x^2$	$2x^2(x - 2) = 2x^3 - 4x^2$	$x^2 - 17x + 30$
2	$\dfrac{x^2}{x} = x$	$x(x - 2) = x^2 - 2x$	$-15x + 30$
3	$\dfrac{-15x}{x} = -15$	$-15(x - 2) = -15x + 30$	$0$

$y = x^3 + x^2 - 4x + 6$

Only one zero: the quadratic factor $x^2 - 2x + 2$ cannot be factorised further.

Formulas

After finding a zero a
$f(x) = (x - a)\,q(x)$

Key terms

Complete factorisation: Writing the polynomial as a product of factors none of which can be factorised further.
Trial value: A value tried in $f(x)$ to find a zero, usually a divisor of the constant term.

7. Quick tests and checking

Sum of coefficients $= 0$ : $f(1) = 0$ , so $(x - 1)$ is a factor.
Sum of coefficients of even powers $=$ sum of coefficients of odd powers: $f(-1) = 0$ , so $(x + 1)$ is a factor.
Check a factorisation by expanding, or by comparing values at $x = 0, 1, -1$ .
Give the answer as factors, e.g. $(2x - 5)$ , not as roots such as $x = \dfrac{5}{2}$ .

Quick tests save time in the exam.
Clue	Then	Example
Coefficients add up to $0$	$f(1) = 0$ : $(x - 1)$ is a factor	$x^3 - 4x^2 + x + 2$ : $1 - 4 + 1 + 2 = 0$
Coefficients of even powers and of odd powers have equal sums	$f(-1) = 0$ : $(x + 1)$ is a factor	$x^3 + 4x^2 + 2x - 1$ : $4 - 1 = 1 + 2$
Leading coefficient $1$	try the divisors of the constant term	$x^3 - 3x^2 - 10x + 24$ : try $\pm 1, \pm 2, \pm 3, \ldots$
Leading coefficient not $1$	also try fractions $\dfrac{p}{q}$ , $q$ dividing the leading coefficient	$2x^3 + 5x^2 - x - 6$ : $x = -\dfrac{3}{2}$ works

A quick value check catches sign slips.
$x$	$f(x) = 2x^3 - 3x^2 - 17x + 30$	$(x - 2)(2x - 5)(x + 3)$	Equal?
$0$	$30$	$(-2) \times (-5) \times 3 = 30$	yes
$1$	$12$	$(-1) \times (-3) \times 4 = 12$	yes
$-1$	$42$	$(-3) \times (-7) \times 2 = 42$	yes

Key terms

Checking by values: Substituting easy values into the polynomial and into the product of factors; they must agree.

Exam tips

How marks are awarded

Remainder (2 marks): value substituted 1; remainder 1.
Unknown coefficient (3 marks): remainder or factor written in terms of $k$ 1; equation 1; value 1.
Two unknowns (4 marks): each equation 1 + 1; solving 1; values 1.
Factorise a cubic (3–4 marks): trial value giving $0$ 1; quotient 1; complete factorisation 1 (+ 1 for showing the division).
10-mark structured question: parts (a), (b), (c), usually 3 + 3 + 4; show all working.

Common mistakes

Substituting $a$ instead of $-a$ for the divisor $(x + a)$ , or $1$ instead of $\dfrac{1}{2}$ for $(2x - 1)$ .
Sign slips with negative values: $(-3)^2 = 9$ but $(-3)^3 = -27$ .
Calling an expression a factor when the value is not $0$ .
Stopping with a quadratic factor that still factorises.
Errors in the quotient: check by expanding.
Writing roots instead of factors, or $\left(x - \dfrac{5}{2}\right)$ instead of $(2x - 5)$ .

1. Dividing a polynomial by a linear expression

Formulas

Key terms

2. The remainder theorem

Formulas

Key terms

3. Unknown coefficients from given remainders

Formulas

Key terms

4. The factor theorem

Formulas

Key terms

5. Unknown coefficients from given factors

Formulas

Key terms

6. Factorising a cubic completely

Formulas

Key terms

7. Quick tests and checking

Key terms

Exam tips

Cookies