Polynomial, Rational and Irrational Functions

Lessons

Senior 4 · Mathematics

Polynomial, Rational and Irrational Functions

5 lessons

Lesson 1 of 5

Polynomials: remainder, factor theorem and factorisation

25 min

A box from a sheet of card

A packaging workshop cuts a square of side $x$ cm from each corner of a 30 cm by 20 cm card and folds up the sides. The open box has volume

V(x) = x(30 - 2x)(20 - 2x) = 4x^3 - 100x^2 + 600x \quad (0 \lt x \lt 10).

A customer wants a box of $1008\,\mathrm{cm^3}$ . That means solving the cubic equation $4 x^{3} - 100 x^{2} + 600 x = 1008$ . There is no simple formula for cubics at this level; instead, you find one root, divide it out, and solve the quadratic that is left. This lesson gives you the tools: the remainder theorem, the factor theorem and synthetic division.

Cutting a square of side $x$ from each corner and folding gives an open box of volume $V (x) = x (30 - 2 x) (20 - 2 x)$ , a polynomial of degree 3.

Definition

Polynomial in $x$ : an expression $P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ , where $n$ is a whole number and the coefficients $a_{i}$ are real numbers. If $a_n \ne 0$ , $n$ is the degree and $a_{n}$ the leading coefficient; $a_{0}$ is the constant term.

Root (or zero) of $P$ : a number $a$ with $P (a) = 0$ .

Example: $P (x) = 4 x^{3} - 100 x^{2} + 600 x$ has degree 3, leading coefficient 4 and constant term 0; $x = 0$ is a root. $\sqrt{x} + 1$ and $\frac{1}{x}$ are not polynomials.

Division, remainders and factors

Dividing $P (x)$ by a non-zero polynomial $D (x)$ gives a quotient $Q (x)$ and a remainder $R (x)$ with

P(x) = D(x)\,Q(x) + R(x), \qquad \deg R \lt \deg D .

When $D (x) = x - a$ has degree 1, the remainder is a constant $r$ : $P (x) = (x - a) Q (x) + r$ . Put $x = a$ : the first term vanishes, so $P (a) = r$ .

Important

Remainder theorem. The remainder when $P (x)$ is divided by $x - a$ is $P (a)$ . (Dividing by $a x - b$ , the remainder is $P\left(\frac{b}{a}\right)$ .)

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

Rational roots. If $P$ has integer coefficients and $\frac{p}{q}$ (in lowest terms) is a root, then $p$ divides the constant term $a_{0}$ and $q$ divides the leading coefficient $a_{n}$ .

Worked example 1: remainders without dividing

Find the remainder when $P (x) = 2 x^{3} - 3 x^{2} + 4 x - 5$ is divided by (a) $x - 2$ (b) $2 x + 1$ .

Solution

$P (2) = 16 - 12 + 8 - 5 = 7$ . The remainder is 7.
$2 x + 1 = 0$ when $x = -\frac{1}{2}$ : $P\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 4\left(-\frac{1}{2}\right) - 5 = -\frac{1}{4} - \frac{3}{4} - 2 - 5 = -8$ .

Synthetic division

To divide by $x - a$ , write the coefficients in a row (with 0 for any missing power), bring down the first, then repeatedly multiply by $a$ and add. The last number is the remainder; the others are the coefficients of the quotient, one degree lower. For $P (x) = x^{3} - 2 x^{2} - 5 x + 6$ divided by $x - 1$ :

$a = 1$	$x^{3}$	$x^{2}$	$x$	constant
coefficients	1	−2	−5	6
multiply by 1, add		1	−1	−6
result	1	−1	−6	0 (remainder)

So $x^{3} - 2 x^{2} - 5 x + 6 = (x - 1) (x^{2} - x - 6)$ .

Paolo Ruffini (1765–1822), the Italian mathematician and doctor who described this shortcut for dividing by x − a in 1804; it is still called Ruffini's rule. Portrait, public domain, via Wikimedia Commons.

Worked example 2: factorise a cubic completely

Factorise $P (x) = x^{3} - 2 x^{2} - 5 x + 6$ and solve $P (x) = 0$ .

Solution Integer roots must divide 6: try $\pm1, \pm2, \pm3, \pm6$ . $P (1) = 1 - 2 - 5 + 6 = 0$ , so $x - 1$ is a factor. The synthetic division above gives the quotient $x^{2} - x - 6 = (x - 3) (x + 2)$ . Hence

P(x) = (x - 1)(x - 3)(x + 2), \qquad P(x) = 0 \iff x \in \{-2, 1, 3\} .

Each factor $x - a$ of $P (x)$ gives an $x$ -intercept $a$ of its graph; the constant term 6 is the $y$ -intercept.

Worked example 3: a fractional root

Solve $2 x^{3} + 3 x^{2} - 8 x + 3 = 0$ .

Solution Possible rational roots: $\pm1, \pm3, \pm\frac{1}{2}, \pm\frac{3}{2}$ . $P (1) = 2 + 3 - 8 + 3 = 0$ . Dividing by $x - 1$ (coefficients $2, 3, - 8, 3$ give $2, 5, - 3$ , remainder 0):

2 x^{3} + 3 x^{2} - 8 x + 3 = (x - 1) (2 x^{2} + 5 x - 3) = (x - 1) (2 x - 1) (x + 3) .

Roots: $1, \frac{1}{2}, -3$ .

Worked example 4: find unknown coefficients

$x - 1$ and $x + 2$ are factors of $P (x) = x^{3} + a x^{2} + b x - 6$ . Find $a$ and $b$ , then factorise $P$ .

Solution $P(1) = 1 + a + b - 6 = 0 \Rightarrow a + b = 5$ . $P(-2) = -8 + 4a - 2b - 6 = 0 \Rightarrow 2a - b = 7$ . Adding: $3 a = 12$ , so $a = 4$ , $b = 1$ .

Then $P (x) = x^{3} + 4 x^{2} + x - 6$ . Dividing by $(x - 1) (x + 2) = x^{2} + x - 2$ leaves $x + 3$ (compare the leading terms and the constant terms: $x^3 \div x^2 = x$ , $-6 \div -2 = 3$ ). So $P (x) = (x - 1) (x + 2) (x + 3)$ .

Worked example 5: the box

Find every $x$ with $0 \lt x \lt 10$ for which the box holds $1008\,\mathrm{cm^3}$ .

Solution $4x^3 - 100x^2 + 600x - 1008 = 0 \iff x^3 - 25x^2 + 150x - 252 = 0$ . Try divisors of 252: $x = 3$ gives $27 - 225 + 450 - 252 = 0$ . Synthetic division by $x - 3$ : coefficients $1, - 25, 150, - 252$ give $1, - 22, 84$ , remainder 0. So

(x - 3)(x^2 - 22x + 84) = 0, \qquad x = 3 \ \text{or}\ x = 11 \pm \sqrt{37} .

$11 + \sqrt{37} \approx 17.08$ is too big ( $x \lt 10$ ), but $11 - \sqrt{37} \approx 4.92$ works. Two boxes are possible: $x = 3\,\mathrm{cm}$ (check: $3 \times 24 \times 14 = 1008$ ✓) or $x \approx 4.92\,\mathrm{cm}$ .

Tip

Before testing possible roots, add the coefficients: if the sum is 0, then $P (1) = 0$ and $x - 1$ is a factor. If the coefficients of the even powers add up to those of the odd powers, then $P (- 1) = 0$ . Also learn $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ and $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ .

Warning

In synthetic division, a missing power needs a 0. For $x^{3} - 7 x + 6$ , the coefficients are $1, 0, - 7, 6$ , not $1, - 7, 6$ . Leaving out the 0 gives a quotient of the wrong degree.

Eddie Woo: Remainder and factor theorem, part 1 (about 9 min). Watch how he finds a remainder by substituting one value instead of doing a long division, and why that works.

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Practice

Questions 7 and 8 are harder.

Find the remainder when $x^{3} + 2 x^{2} - x + 4$ is divided by $x + 3$ .
Show that $x - 2$ is a factor of $x^{3} - 7 x + 6$ , and factorise it completely.
Factorise $x^{3} + 2 x^{2} - 5 x - 6$ .
Solve $2 x^{3} - x^{2} - 13 x - 6 = 0$ .
Factorise $x^{3} - 27$ and $8 x^{3} + 1$ .
When $x^{3} + k x^{2} - 4 x + 1$ is divided by $x - 2$ , the remainder is 5. Find $k$ .
Factorise $x^{4} - 5 x^{3} + 5 x^{2} + 5 x - 6$ .
$P (x)$ leaves remainder 3 when divided by $x - 1$ and remainder $- 1$ when divided by $x + 1$ . Find the remainder when $P (x)$ is divided by $x^{2} - 1$ .

Show answersHide answers

$P (- 3) = - 27 + 18 + 3 + 4 = - 2$ .
$8 - 14 + 6 = 0$ . Coefficients $1, 0, - 7, 6$ divided by $x - 2$ give $1, 2, - 3$ : $(x - 2) (x^{2} + 2 x - 3) = (x - 2) (x - 1) (x + 3)$ .
$P (- 1) = - 1 + 2 + 5 - 6 = 0$ : $(x + 1) (x^{2} + x - 6) = (x + 1) (x - 2) (x + 3)$ .
$P (- 2) = - 16 - 4 + 26 - 6 = 0$ : $(x + 2) (2 x^{2} - 5 x - 3) = (x + 2) (2 x + 1) (x - 3)$ . Roots $-2, -\frac{1}{2}, 3$ .
$(x - 3) (x^{2} + 3 x + 9)$ and $(2 x + 1) (4 x^{2} - 2 x + 1)$ .
$8 + 4k - 8 + 1 = 5 \Rightarrow k = 1$ .
$P (1) = P (- 1) = P (2) = P (3) = 0$ : $(x - 1) (x + 1) (x - 2) (x - 3)$ .
Write $P (x) = (x^{2} - 1) Q (x) + a x + b$ . Then $a + b = 3$ and $- a + b = - 1$ , so $a = 2$ , $b = 1$ : the remainder is $2 x + 1$ .

Quick check

Key points

Remainder theorem: dividing $P (x)$ by $x - a$ leaves $P (a)$ . Factor theorem: $x - a$ is a factor $\iff P(a) = 0$ .
Rational roots $\frac{p}{q}$ of an integer polynomial have $p \mid a_0$ and $q \mid a_n$ .
Synthetic division: multiply by $a$ and add; write 0 for missing powers.
To solve a cubic: find one root, divide it out, solve the quadratic, and keep only roots that make sense.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Image: Paolo Ruffini, portrait, Wikimedia Commons, public domain.
Video: Remainder & Factor Theorem (1 of 2: How do we calculate remainders without dividing?), Eddie Woo, embedded from YouTube.

1A box from a sheet of card

2Division, remainders and factors

3Synthetic division

A box from a sheet of card

Division, remainders and factors

Synthetic division