Remainder theorem and factor theorem — proof and applications: Exercises

This page is a self-test. Each problem is graded by difficulty: L1 Recognition → L2 Application → L3 Analysis → L4 Synthesis → L5 Mastery. Read the problem, try it yourself, THEN open the collapsible solution.

Everything here rests on the two tools you built in the parent note:

Before starting, look at this map of what "plug in the root of the divisor" means visually — the remainder is just the height of the polynomial's graph at that special input.

Figure — Remainder theorem and factor theorem — proof and applications

The green vertical line sits at $x = a$ (where $x - a = 0$ ). Where it meets the curve, that height is $p (a)$ — and that height IS the remainder. If the height is zero (curve crosses the axis there), $(x - a)$ is a factor.

Level 1 — Recognition

Can you spot which number to plug in?

L1.1

State the value of $x$ you must substitute to find the remainder when $p (x)$ is divided by each divisor: (a) $(x - 4)$ (b) $(x + 7)$ (c) $(3 x - 2)$ (d) $x$ (i.e. divisor is just $x$ ).

Recall Solution

The rule: set the divisor equal to zero and solve for $x$ . That root is what you plug in.

(a) $x - 4 = 0 \Rightarrow x = 4$ . Substitute $== 4 ==$ .
(b) $x + 7 = 0 \Rightarrow x = - 7$ . Rewrite $(x + 7) = (x - (- 7))$ , so $a = - 7$ . Substitute $== - 7 ==$ .
(c) $3 x - 2 = 0 \Rightarrow x = \frac{2}{3}$ . Substitute $== \frac{2}{3} ==$ .
(d) $x = 0$ is already the divisor's root. Substitute $== 0 ==$ . (This just gives the constant term of $p$ .)

L1.2

$p (x) = x^{3} - 2 x^{2} + 5 x - 1$ . Without dividing, find the remainder when $p (x)$ is divided by $(x - 1)$ .

Recall Solution

Divisor zero at $x = 1$ , so remainder $= p (1)$ . $p (1) = 1^{3} - 2 (1)^{2} + 5 (1) - 1 = 1 - 2 + 5 - 1 = 3.$ Remainder $= 3$ .

Level 2 — Application

Run the machine end-to-end.

L2.1

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

Recall Solution

$(x + 2) = 0 \Rightarrow x = - 2$ . Remainder $= p (- 2)$ . $p (- 2) = (- 2)^{4} - 3 (- 2)^{3} + 2 (- 2) - 5.$ Compute each piece, watching signs:

$(- 2)^{4} = 16$
$- 3 (- 2)^{3} = - 3 (- 8) = + 24$
$2 (- 2) = - 4$
constant $- 5$ $p (- 2) = 16 + 24 - 4 - 5 = 31.$ Remainder $= 31$ .

L2.2

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

Recall Solution

Coefficient on $x$ , so solve the divisor: $2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$ . Remainder $= p (\frac{3}{2})$ . $p (\frac{3}{2}) = 2 (\frac{3}{2})^{3} + 3 (\frac{3}{2})^{2} - 5 (\frac{3}{2}) + 1.$

$(\frac{3}{2})^{3} = \frac{27}{8}$ , times $2 = \frac{27}{4}$ .
$(\frac{3}{2})^{2} = \frac{9}{4}$ , times $3 = \frac{27}{4}$ .
$- 5 \cdot \frac{3}{2} = - \frac{15}{2} = - \frac{30}{4}$ .
$+ 1 = \frac{4}{4}$ . $= \frac{27 + 27 - 30 + 4}{4} = \frac{28}{4} = 7.$ Remainder $= 7$ .

L2.3

Is $(x - 3)$ a factor of $p (x) = x^{3} - 4 x^{2} + x + 6$ ?

Recall Solution

Factor Theorem: $(x - 3)$ is a factor $⟺ p (3) = 0$ . $p (3) = 27 - 4 (9) + 3 + 6 = 27 - 36 + 3 + 6 = 0.$ Since $p (3) = 0$ , yes, $(x - 3)$ is a factor.

Level 3 — Analysis

Reason backwards to find missing pieces.

L3.1

$(x - 2)$ is a factor of $p (x) = x^{3} + k x^{2} - 7 x + 2$ . Find $k$ .

Recall Solution

Factor $\Rightarrow p (2) = 0$ . $p (2) = 8 + k (4) - 14 + 2 = 4 k - 4.$ Set to zero: $4 k - 4 = 0 \Rightarrow k = 1$ . $k = 1$ .

L3.2

When $p (x) = x^{3} + a x^{2} + b x + 6$ is divided by $(x - 1)$ the remainder is $8$ , and $(x + 1)$ is a factor. Find $a$ and $b$ .

Recall Solution

Two conditions give two equations. Condition 1: remainder $8$ on division by $(x - 1)$ means $p (1) = 8$ : $1 + a + b + 6 = 8 ⟹ a + b = 1.$ Condition 2: $(x + 1)$ a factor means $p (- 1) = 0$ : $- 1 + a - b + 6 = 0 ⟹ a - b = - 5.$ Add the two equations: $2 a = - 4 \Rightarrow a = - 2$ . Then $b = 1 - a = 3$ . $a = - 2, b = 3$ .

L3.3

The polynomial $p (x) = x^{3} - 3 x^{2} + c x - 4$ leaves the same remainder when divided by $(x - 1)$ and by $(x + 2)$ . Find $c$ .

Recall Solution

"Same remainder" means $p (1) = p (- 2)$ . $p (1) = 1 - 3 + c - 4 = c - 6.$ $p (- 2) = - 8 - 12 - 2 c - 4 = - 24 - 2 c .$ Set equal: $c - 6 = - 24 - 2 c \Rightarrow 3 c = - 18 \Rightarrow c = - 6$ . $c = - 6$ .

Level 4 — Synthesis

Combine the theorem with factoring and division.

L4.1

Factor completely: $p (x) = x^{3} - 2 x^{2} - 5 x + 6$ .

Recall Solution

Step 1 — find one root. By the Rational Root Theorem, any rational root divides the constant $6$ : try $\pm 1, \pm 2, \pm 3, \pm 6$ . $p (1) = 1 - 2 - 5 + 6 = 0. ✓$ So $(x - 1)$ is a factor. Step 2 — divide out $(x - 1)$ (via Synthetic Division or Polynomial Long Division): $p (x) = (x - 1) (x^{2} - x - 6) .$ Step 3 — factor the quadratic. Need two numbers multiplying to $- 6$ , adding to $- 1$ : those are $- 3$ and $2$ . $x^{2} - x - 6 = (x - 3) (x + 2) .$ Final: $p (x) = (x - 1) (x - 3) (x + 2)$ . Roots $x = 1, 3, - 2$ .

L4.2

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

Recall Solution

Step 1 — rational root candidates. Leading coefficient $2$ , constant $6$ . Candidates are $\pm \frac{factor of 6}{factor of 2}$ : $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$ . Try $x = 3$ : $p (3) = 2 (27) - 3 (9) - 33 + 6 = 54 - 27 - 33 + 6 = 0. ✓$ So $(x - 3)$ is a factor. Step 2 — divide: $p (x) = (x - 3) (2 x^{2} + 3 x - 2)$ . Step 3 — factor the quadratic $2 x^{2} + 3 x - 2$ : split $3 x$ as $4 x - x$ : $2 x^{2} + 4 x - x - 2 = 2 x (x + 2) - 1 (x + 2) = (2 x - 1) (x + 2) .$ Final: $p (x) = (x - 3) (2 x - 1) (x + 2)$ . Roots $x = 3, \frac{1}{2}, - 2$ .

Level 5 — Mastery

Full-power reasoning; multiple ideas at once.

L5.1

A polynomial $p (x)$ leaves remainder $5$ when divided by $(x - 1)$ and remainder $- 1$ when divided by $(x + 2)$ . Find the remainder when $p (x)$ is divided by $(x - 1) (x + 2)$ .

Recall Solution

Why a new form of remainder? Dividing by the quadratic $(x - 1) (x + 2)$ gives a remainder of degree $< 2$ — so a linear remainder $R (x) = m x + n$ (two unknowns). Write $p (x) = (x - 1) (x + 2) q (x) + (m x + n)$ . Kill the quadratic by plugging in each root — that's the same trick as the basic proof.

$x = 1$ : $p (1) = m + n$ . Given $p (1) = 5$ , so $m + n = 5$ .
$x = - 2$ : $p (- 2) = - 2 m + n$ . Given $p (- 2) = - 1$ , so $- 2 m + n = - 1$ . Subtract: $(m + n) - (- 2 m + n) = 5 - (- 1) \Rightarrow 3 m = 6 \Rightarrow m = 2$ , then $n = 3$ . Remainder $= 2 x + 3$ .

L5.2

Find all values of $k$ for which $(x - 1)$ is a factor of $p (x) = x^{3} - k^{2} x^{2} + 3 k x - 3$ .

Recall Solution

Factor $\Rightarrow p (1) = 0$ : $1 - k^{2} + 3 k - 3 = 0 ⟹ - k^{2} + 3 k - 2 = 0 ⟹ k^{2} - 3 k + 2 = 0.$ Factor: $(k - 1) (k - 2) = 0 \Rightarrow k = 1$ or $k = 2$ . $k = 1$ or $k = 2$ . (Both are valid — check: for $k = 1$ , $p (1) = 1 - 1 + 3 - 3 = 0$ ✓; for $k = 2$ , $p (1) = 1 - 4 + 6 - 3 = 0$ ✓.)

L5.3

Show that $(x - a)$ can be a factor of $p (x) = x^{n} - a^{n}$ for every positive integer $n$ , and state the remainder when $p (x) = x^{n} + a^{n}$ is divided by $(x - a)$ .

Recall Solution

Part 1 (factor of $x^{n} - a^{n}$ ): By the Factor Theorem, $(x - a)$ is a factor $⟺ p (a) = 0$ . Here $p (a) = a^{n} - a^{n} = 0 for every n .$ So $(x - a)$ divides $x^{n} - a^{n}$ for all positive integers $n$ . (Its full factorization is $x^{n} - a^{n} = (x - a) (x^{n - 1} + x^{n - 2} a + \dots + a^{n - 1})$ — Factoring Polynomials.) Part 2 (remainder of $x^{n} + a^{n}$ ): the remainder is $p (a)$ where $p (x) = x^{n} + a^{n}$ : $p (a) = a^{n} + a^{n} = 2 a^{n} .$ So the remainder is $2 a^{n}$ — which is $0$ only if $a = 0$ . Hence $(x - a)$ is a factor of $x^{n} + a^{n}$ only in the degenerate case $a = 0$ .

Recall Quick self-grade checklist

Set divisor to zero before substituting ::: Always — never copy the visible digit "Factor" translates to ::: $p (a) = 0$ "Remainder $r$ " translates to ::: $p (a) = r$ Leading coefficient $\neq = 1$ means ::: also test fractional roots $p / q$ Dividing by a degree-2 polynomial leaves a remainder of degree ::: at most $1$ (form $m x + n$ )