Polynomial Theorems · Self-Study Quiz

Chapter 1

Remainder Theorem

When f(x) is divided by (x − a), the remainder equals f(a). No actual division needed.

⚡ Memory Point

"PLUG & FIND"

Step 1 — PLUG: Set divisor = 0. If divisor is (x − a), plug in x = a.
Step 2 — FIND: Calculate f(a). That number IS your remainder.
Key check: Remainder = 0 → divisor is a factor. Remainder ≠ 0 → it's not.

✦ Worked Example

Find the remainder when f(x) = x³ − 3x + 2 is divided by (x − 1).

①Divisor: (x − 1) → plug in x = 1

②f(1) = (1)³ − 3(1) + 2 = 1 − 3 + 2 = 0

③Remainder = 0. Also means (x−1) IS a factor!

Q01 Easy

Given f(x) = 2x³ − 3x² + x − 1, what is the remainder when divided by (x − 2)?

Solution

Plug in x = 2:
f(2) = 2(8) − 3(4) + 2 − 1 = 16 − 12 + 2 − 1 = 5
By the Remainder Theorem, remainder = f(2) = 5.

Q02 Easy

What is the remainder when p(x) = x³ + 2x² − x + 3 is divided by (x + 1)?
⚠ Careful: (x + 1) means plug in x = −1

Solution

(x + 1) → set equal to 0 → x = −1
p(−1) = (−1)³ + 2(−1)² − (−1) + 3 = −1 + 2 + 1 + 3 = 5
Common trap: students plug in x = +1 instead of −1. Sign matters!

Q03 Medium

Find the remainder when g(x) = 3x³ − 2x² + 5x − 4 is divided by (x − 1).

Solution

g(1) = 3(1) − 2(1) + 5(1) − 4 = 3 − 2 + 5 − 4 = 2
Remainder = 2. Since ≠ 0, (x − 1) is NOT a factor.

Q04 Hard

h(x) = x³ − kx² + 2x + 1. When divided by (x − 2), the remainder is 5. Find k.

Solution

Set h(2) = 5:
(2)³ − k(2)² + 2(2) + 1 = 5
8 − 4k + 4 + 1 = 5
13 − 4k = 5 → 4k = 8 → k = 2

Chapter 2

Factor Theorem

(x − a) is a factor of f(x) if and only if f(a) = 0. It's the Remainder Theorem with remainder = 0.

⚡ Memory Point

"ZERO = FACTOR"

Rule: (x − a) is a factor ←→ f(a) = 0
BOTH directions: If f(a)=0 → it's a factor. If (x−a) is a factor → f(a)=0.
Strategy: To find factors, TEST ±1, ±2, ±3... until you get 0.

✦ Worked Example

Is (x − 2) a factor of f(x) = x³ − 3x² − x + 3?

①Plug in x = 2: f(2) = 8 − 12 − 2 + 3 = −3

②f(2) = −3 ≠ 0 → (x − 2) is NOT a factor.

③Try x = 1: f(1) = 1 − 3 − 1 + 3 = 0 ✓ → (x − 1) IS a factor!

Q05 Easy

Is (x − 3) a factor of f(x) = x³ − 6x² + 11x − 6?

Solution

f(3) = 27 − 54 + 33 − 6 = 0
Since f(3) = 0, by the Factor Theorem, (x − 3) IS a factor. ✓

Q06 Easy

Is (x + 2) a factor of g(x) = x³ + x² − 4x − 4?
⚠ Remember: (x + 2) → test x = −2

Solution

g(−2) = (−8) + (4) − (−8) − 4 = −8 + 4 + 8 − 4 = 0
Yes, (x + 2) is a factor. Option D is a trap — you must plug in x = −2, not +2!

Q07 Medium

f(x) = x³ + ax² − x − 6. If (x − 2) is a factor, find a.

Solution

Since (x−2) is a factor → f(2) = 0
f(2) = 8 + 4a − 2 − 6 = 0
4a + 0 = 0 → a = 0
So f(x) = x³ − x − 6 = (x−2)(x² + 2x + 3)

Q08 Hard

p(x) = 2x³ + kx² − x + 3. If (x + 1) is a factor, find k.
⚠ (x + 1) is a factor → p(−1) = 0

Solution

p(−1) = 0:
2(−1)³ + k(−1)² − (−1) + 3 = 0
−2 + k + 1 + 3 = 0
k + 2 = 0 → k = −2

Chapter 3

Polynomial Long Division

Divide polynomials step-by-step. Like number long division, but with terms.

⚡ Memory Point

"DMSB"

D — Divide the leading term
M — Multiply divisor × result
S — Subtract (change signs!)
B — Bring down the next term
Repeat until done. The last number = remainder.

✦ Worked Example

Divide (x³ + 2x² − x − 2) by (x − 1)

Dx³ ÷ x = x². Write x² on top.

Mx²(x−1) = x³ − x²

S(x³+2x²) − (x³−x²) = 3x². Bring down: 3x² − x

D3x² ÷ x = 3x. Continue...

✓Result: x² + 3x + 2, remainder 0

Q09 Medium

Divide (x³ − 2x² − 5x + 6) by (x − 3). What is the quotient?

Solution

Using DMSB or synthetic with root 3:
1 | −2 | −5 | 6 (root = 3)
→ 1 | 1 | −2 | 0
Quotient: x² + x − 2, Remainder: 0
Check: (x−3)(x²+x−2) = (x−3)(x+2)(x−1) ✓

Q10 Medium

Divide (2x³ + 3x² − 11x − 6) by (x + 3). What is the quotient?

Solution

(x + 3) → root = −3
Coefficients: 2, 3, −11, −6
Synthetic: 2 | 3+(2×−3)=−3 | −11+(−3×−3)=−2 | −6+(−2×−3)=0
Quotient: 2x² − 3x − 2, Remainder: 0

Q11 Tricky

Divide (x³ − 1) by (x − 1).
⚠ Missing terms! Write as: x³ + 0x² + 0x − 1

Solution

Write: 1·x³ + 0·x² + 0·x − 1, root = 1
Synthetic: 1 | 1 | 1 | 0
Quotient: x² + x + 1, Remainder: 0
This is the famous factoring: x³ − 1 = (x − 1)(x² + x + 1) ✓

Q12 Hard

Find the remainder when 3x³ − 5x² + 2x − 1 is divided by (x − 1).
💡 Shortcut: use Remainder Theorem (no long division needed!)

Solution

By Remainder Theorem: f(1) = 3 − 5 + 2 − 1 = −1
No long division needed! Always check if Remainder Theorem is faster.

Chapter 4

Synthetic Division

A shortcut for long division when the divisor is linear (x − a). Uses only coefficients.

⚡ Memory Point

"COEFFICIENT LADDER"

Setup: Write only coefficients (include 0 for missing terms!)
Root: If divisor = (x − a), use +a. If (x + a), use −a.
Process: Bring down → multiply → add → repeat
Last number = remainder. Rest = quotient coefficients.

✦ Synthetic Division Setup

Divide (2x³ − 3x² + x − 5) by (x − 2), root = 2

2 | 2 -3 1 -5 | 4 2 6 —————————————————— 2 1 3 | 1 \leftarrow remainder

Quotient: 2x² + x + 3, Remainder: 1

Q13 Easy

Use synthetic division to divide (x³ − 6x² + 11x − 6) by (x − 1). What is the remainder?

Solution

Root = 1, Coefficients: 1, −6, 11, −6
1 | 1 → 1+(1×1)=−5 → −5+(1×−5)=6 → 6+(1×−5) wait—
1 | 1 −6 11 −6
1 −5 6
1 −5 6 | 0
Remainder = 0. Quotient: x² − 5x + 6 = (x−2)(x−3)

Q14 Medium

Divide (2x³ + x² − 5x + 2) by (x − 1) using synthetic division. What is the quotient?

Solution

Root = 1 | 2 1 −5 2
2 3 −2
2 3 −2 | 0
Quotient: 2x² + 3x − 2, Remainder: 0
Check: (x−1)(2x²+3x−2) = (x−1)(2x−1)(x+2) ✓

Q15 Hard

Divide (x³ + 3x² − x + 2) by (x + 2). Find the quotient and remainder.
⚠ (x + 2) → use root = −2

Solution

Root = −2 | 1 3 −1 2
−2 −2 6
1 1 −3 | 8
Quotient: x² + x − 3, Remainder: 8

Q16 Tricky

Divide (4x³ − 3x² + 2x − 1) by (x − 1). Express the result in the form:

f(x) = (x - 1) \cdot Q(x) + R

Solution

Root = 1 | 4 −3 2 −1
4 1 3
4 1 3 | 2
So: f(x) = (x − 1)(4x² + x + 3) + 2

Chapter 5

Mixed Challenges

Exam-style questions combining multiple concepts. These are the most commonly missed!

⚡ Top 5 Exam Traps

WATCH OUT!

TRAP 1 Sign error: (x+a) → plug in −a, not +a
TRAP 2 Missing terms: always use 0 as placeholder coefficient
TRAP 3 "Is a factor" ≠ "remainder is 1" — it means remainder = 0
TRAP 4 Fully factoring: after first factor, factor the quotient too!
TRAP 5 Two conditions → two equations → solve simultaneously

Q17 Medium

Fully factor: f(x) = x³ − x² − 4x + 4

Solution

Test x = 1: f(1) = 1 − 1 − 4 + 4 = 0 ✓ → (x − 1) is a factor
Synthetic ÷ (x−1): quotient = x² + 0x − 4 = x² − 4
Factor x² − 4 = (x − 2)(x + 2) (difference of squares!)
Full factoring: (x − 1)(x − 2)(x + 2)

Q18 Hard

p(x) = x³ + ax² + bx − 8. Both (x − 1) and (x + 2) are factors. Find a + b.

Solution

Two factors → two equations:
p(1) = 0: 1 + a + b − 8 = 0 → a + b = 7 ①
p(−2) = 0: −8 + 4a − 2b − 8 = 0 → 2a − b = 8 ②
From ①: a + b = 7. Answer = 7 (no need to solve further!)
(For reference: a = 5, b = 2)

Q19 Easy

Which statement best describes the Factor Theorem?

Solution

The Factor Theorem: (x − a) is a factor ↔ f(a) = 0
Both directions! If you know f(a)=0, you know (x−a) is a factor, and vice versa.
Option D is a common misconception — f(a) must equal zero, not one.

Q20 Tricky

Given f(x) = x⁴ + x³ − 7x² − x + 6 and that (x − 1) and (x + 1) are both factors, fully factor f(x).

Solution

(x−1)(x+1) = x² − 1. Divide f(x) by (x² − 1):
x⁴+x³−7x²−x+6 ÷ (x²−1) = x² + x − 6
Factor x²+x−6: need two numbers that multiply to −6, add to 1 → 3 and −2
x²+x−6 = (x+3)(x−2)
Full: (x−1)(x+1)(x+3)(x−2)