Remainder theorem and factor theorem — proof and applications: Worked examples

This page is the drill ground for the parent topic. We will not re-prove anything — we will instead hunt down every kind of divisor and every kind of trap the Remainder Theorem and Factor Theorem can throw at you, and solve one representative of each.

Before the examples, two tiny reminders so every symbol on this page is earned:

The scenario matrix

Every problem in this whole topic is one of the cells below. The trick is realising a hard-looking question is just a familiar cell wearing a costume.

#	Case class	What makes it tricky	Example
A	Plain divisor $(x - a)$ , $a > 0$	nothing — the warm-up	Ex 1
B	Sign trap: $(x + a)$	root is $- a$ , not $+ a$	Ex 2
C	Leading coefficient: $(a x - b)$	root is $\frac{b}{a}$ , a fraction	Ex 3
D	Zero / degenerate: divide by $x$ itself, or constant polynomial	$a = 0$ ; remainder = constant term	Ex 4
E	Two conditions → two unknowns	solve a system from two roots	Ex 5
F	Full factorisation (all real roots)	chain Factor Theorem + division	Ex 6
G	Real-world word problem	translate a story into $p (a) = 0$	Ex 7
H	Exam twist: remainders as clues	reverse-engineer $p (x)$ from two remainders	Ex 8

Below, each example is stamped with [Cell X] so you can see the matrix filling up.

Worked example Example 3 — [Cell C] leading coefficient, fractional root

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

Forecast: the remainder theorem's root will be a fraction here. Which fraction?

Solve the divisor for zero. $2 x - 1 = 0 \Rightarrow x = \frac{1}{2}$ . Why this step? For $(a x - b)$ the root is $\frac{b}{a} = \frac{1}{2}$ , not the visible $1$ or $- 1$ .
Substitute $x = \frac{1}{2}$ , one power at a time. $p (\frac{1}{2}) = 2 \cdot \frac{1}{8} + 3 \cdot \frac{1}{4} - 5 \cdot \frac{1}{2} + 1 = \frac{1}{4} + \frac{3}{4} - \frac{5}{2} + 1.$ Why this step? $(\frac{1}{2})^{3} = \frac{1}{8}$ and $(\frac{1}{2})^{2} = \frac{1}{4}$ ; keep fractions exact.
Add up. $\frac{1}{4} + \frac{3}{4} = 1$ , then $1 - \frac{5}{2} + 1 = 2 - \frac{5}{2} = - \frac{1}{2}$ .

Answer: remainder $= - \frac{1}{2}$ .

Verify: the remainder is still a constant even though the root is fractional — good, because $(2 x - 1)$ has degree $1$ so the leftover must have degree $0$ . Related tool: Synthetic Division with $\frac{1}{2}$ then rescale.

Worked example Example 5 — [Cell E] two conditions, two unknowns

$p (x) = x^{3} + a x^{2} + b x - 6$ has both $(x - 1)$ and $(x + 2)$ as factors. Find $a$ and $b$ .

Forecast: two factors give two equations. Guess: will $a, b$ come out to whole numbers?

Turn each factor into a zero-condition. By the Factor Theorem, $p (1) = 0$ and $p (- 2) = 0$ . Why this step? A factor $(x - r)$ forces $p (r) = 0$ — that is the whole content of the theorem.
Build equation one from $p (1) = 0$ . $1 + a + b - 6 = 0 \Rightarrow a + b = 5.$
Build equation two from $p (- 2) = 0$ . $(- 2)^{3} + a (- 2)^{2} + b (- 2) - 6 = 0 \Rightarrow - 8 + 4 a - 2 b - 6 = 0 \Rightarrow 4 a - 2 b = 14 \Rightarrow 2 a - b = 7.$ Why this step? Two unknowns need two independent equations; each factor supplies one.
Solve the system. Add $a + b = 5$ and $2 a - b = 7$ : $3 a = 12 \Rightarrow a = 4$ , then $b = 1$ .

Answer: $a = 4, b = 1$ , so $p (x) = x^{3} + 4 x^{2} + x - 6$ .

Verify: $p (1) = 1 + 4 + 1 - 6 = 0$ ✓ and $p (- 2) = - 8 + 16 - 2 - 6 = 0$ ✓. Both factors confirmed.

Worked example Example 6 — [Cell F] complete factorisation

Factor completely: $p (x) = x^{3} - 6 x^{2} + 11 x - 6$ .

Forecast: it will split into three linear factors. Guess the three roots before diving in.

Hunt a first root among divisors of the constant. By the Rational Root Theorem, any integer root divides $6$ : candidates $\pm 1, \pm 2, \pm 3, \pm 6$ . Test $x = 1$ : $p (1) = 1 - 6 + 11 - 6 = 0$ . ✓ Why this step? The Factor Theorem turns " $p (1) = 0$ " straight into " $(x - 1)$ is a factor" — a cheap first crack.
Divide out $(x - 1)$ . Using Synthetic Division / Polynomial Long Division: $p (x) = (x - 1) (x^{2} - 5 x + 6) .$ Why this step? Removing one factor drops the degree so the leftover quadratic is easy.
Factor the quadratic. Numbers multiplying to $6$ and adding to $- 5$ are $- 2, - 3$ : $x^{2} - 5 x + 6 = (x - 2) (x - 3) .$ See Factoring Polynomials.

Answer: $p (x) = (x - 1) (x - 2) (x - 3)$ ; roots $1, 2, 3$ .

Verify: Fundamental Theorem of Algebra promises exactly $3$ roots for a cubic — we found $3$ . Check $p (2) = 8 - 24 + 22 - 6 = 0$ ✓, $p (3) = 27 - 54 + 33 - 6 = 0$ ✓. This underlies Solving Polynomial Equations.

Worked example Example 7 — [Cell G] a real-world story (with a picture)

A stone is thrown so its height in metres after $t$ seconds is $h (t) = - 5 t^{2} + 20 t + 60$ . An engineer claims the stone is at ground level ( $h = 0$ ) at $t = 6$ s. Is $(t - 6)$ a factor of $h (t)$ — i.e., is the claim exactly true?

Forecast: look at the parabola. Does it cross zero at $t = 6$ , or before?

Translate the claim into the Factor Theorem. "Ground level at $t = 6$ " means $h (6) = 0$ , which means $(t - 6)$ is a factor. Why this step? A root of a height function is a time when height $= 0$ ; the Factor Theorem tests exactly that.
Evaluate $h (6)$ . $h (6) = - 5 (36) + 20 (6) + 60 = - 180 + 120 + 60 = 0.$ Why this step? If this is $0$ , the claim (and the factor) is exact.

Answer: $h (6) = 0$ , so $(t - 6)$ is a factor and the engineer is right.

Verify: factor fully — $h (t) = - 5 (t^{2} - 4 t - 12) = - 5 (t - 6) (t + 2)$ . The other root $t = - 2$ is negative (before the throw), so physically only $t = 6$ matters. Units: $t$ in seconds, $h$ in metres — the red dot in the figure sits exactly on the $t$ -axis at $6$ . ✓

Worked example Example 8 — [Cell H] exam twist: rebuild from two remainders

When a polynomial $p (x)$ is divided by $(x - 1)$ the remainder is $5$ , and divided by $(x + 2)$ the remainder is $- 4$ . Find the remainder when $p (x)$ is divided by $(x - 1) (x + 2)$ .

Forecast: dividing by a degree-2 thing means the remainder can be degree $1$ : shape $r (x) = c x + d$ . Guess whether $c$ is positive.

Write the general division statement. $p (x) = (x - 1) (x + 2) q (x) + c x + d$ . Why this step? The remainder's degree must be less than the divisor's degree $2$ , so at most $1$ : a line $c x + d$ .
Kill the big product by substituting the two roots. At $x = 1$ and $x = - 2$ the $(x - 1) (x + 2)$ term is $0$ , leaving only $c x + d$ : $p (1) = c + d, p (- 2) = - 2 c + d .$ Why this step? Same "make the divisor vanish" trick as the basic theorem — now applied at both roots.
Feed in the known remainders. From the given data $p (1) = 5$ and $p (- 2) = - 4$ : $c + d = 5, - 2 c + d = - 4.$
Solve. Subtract: $3 c = 9 \Rightarrow c = 3$ , then $d = 2$ .

Answer: remainder $= 3 x + 2$ .

Verify: at $x = 1$ : $3 (1) + 2 = 5$ ✓; at $x = - 2$ : $3 (- 2) + 2 = - 4$ ✓. Both original single-divisor remainders reproduced.

Recall Self-test (reveal after guessing)

Divisor $(x + 5)$ , which value do you substitute? ::: $x = - 5$ Divisor $(3 x - 2)$ , which value do you substitute? ::: $x = \frac{2}{3}$ Remainder when any $p (x)$ is divided by $x$ ? ::: the constant term $p (0)$ $(x - a)$ is a factor exactly when…? ::: $p (a) = 0$ Dividing by a degree-2 polynomial, the remainder has degree at most…? ::: $1$ (a line $c x + d$ )