Practice › Number and Algebra › Polynomials, Remainder and Factor Theorems

Polynomials: HKDSE Paper 1 Practice

Fifteen Paper 1 questions on polynomials. The set opens with factorization and algebraic fractions, moves through identities, and finishes with the remainder and factor theorems in the form the exam now favours: find the unknowns, then test a claim about the roots.

15 questions · about 75 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Jump to

Question 1 — factorisation by grouping after a difference of squares
Question 2 — factorising a quadratic in two variables, then grouping
Question 3 — subtracting algebraic fractions using the L.C.M. of the denominators
Question 4 — comparing coefficients in an identity
Question 5 — identity with three unknowns, then complete factorization
Question 6 — remainder theorem with two equal remainders
Question 7 — factor theorem with two given factors, then solving the cubic
Question 8 — one factor and one remainder, then factorizing
Question 9 — divisibility by a quadratic, handled through its two linear factors
Question 10 — finding a cubic from a factor and two remainders; checking a claim about integer roots
Question 11 — rebuilding a polynomial from quotient and remainder; rational roots
Question 12 — dividing algebraic fractions after factorizing
Question 13 — rewriting a quadratic in powers of (x + 1); link to the remainder
Question 14 — quotient and remainder from a quadratic divisor; solving the resulting cubic
Question 15 — equal remainders plus a factor; checking a claim that turns out true

Question 1 · Modelled on 2025 P1 Q5 — the common factor is hidden until the first part is done

Factorize

(a) $9 m^{2} - 16 n^{2}$ ,

(b) $9 m^{2} - 16 n^{2} + 3 m + 4 n$ .

(3 marks)

Show answer

(a) $(3 m + 4 n) (3 m - 4 n)$ (b) $(3 m + 4 n) (3 m - 4 n + 1)$ . In (b), the last two terms are $3 m + 4 n$ , the same factor as in (a).

Question 2 · Modelled on HKDSE P1 Section A(1) factorization items — two-variable trinomial instead of a difference of squares

Factorize

(a) $2 x^{2} - 7 x y + 3 y^{2}$ ,

(b) $2 x^{2} - 7 x y + 3 y^{2} - 4 x + 2 y$ .

(3 marks)

Show answer

(a) $(2 x - y) (x - 3 y)$ (b) $(2 x - y) (x - 3 y - 2)$ , because $- 4 x + 2 y = - 2 (2 x - y)$ .

Question 3 · Written in HKDSE P1 Section A(1) style

Simplify $\frac{4}{x ^{2} - 4 x} - \frac{1}{x - 4}$ .

(3 marks)

Show answer

$- \frac{1}{x}$ . Over the common denominator $x (x - 4)$ , the numerator is $4 - x = - (x - 4)$ , which cancels.

Question 4 · Written in HKDSE P1 Section A(1) style

It is given that $(x + a) (2 x - 3) + b \equiv 2 x^{2} + 5 x - 1$ , where $a$ and $b$ are constants. Find $a$ and $b$ .

(4 marks)

Show answer

$a = 4, b = 11$ . The $x$ coefficient gives $2 a - 3 = 5$ ; the constant term gives $- 3 a + b = - 1$ .

Question 5 · Written in HKDSE P1 style — the identity itself hands you one factor

It is given that $x^{3} - 2 x^{2} + a x + 6 \equiv (x - 1) (x^{2} + b x + c)$ , where $a$ , $b$ and $c$ are constants.

(a) Find $a$ , $b$ and $c$ . (3 marks)

(b) Factorize $x^{3} - 2 x^{2} + a x + 6$ completely. (2 marks)

Show answer

(a) $a = - 5, b = - 1, c = - 6$ (b) $(x - 1) (x - 3) (x + 2)$ .

Question 6 · Modelled on HKDSE P2 remainder-theorem items, extended to a two-part long question

Let $f (x) = 2 x^{3} + k x^{2} - 5 x + 3$ , where $k$ is a constant. When $f (x)$ is divided by $x - 2$ and by $x + 1$ , the two remainders are equal.

(a) Find $k$ . (2 marks)

(b) Find the remainder when $f (x)$ is divided by $2 x - 1$ . (2 marks)

Show answer

(a) $k = - 1$ , from $f (2) = f (- 1)$ , i.e. $9 + 4 k = 6 + k$ . (b) $f (\frac{1}{2}) = \frac{1}{2}$ .

Question 7 · Modelled on 2020 P1 Q13 — two factors are given instead of a factor and a remainder

Let $f (x) = x^{3} + a x^{2} - 7 x + b$ , where $a$ and $b$ are constants. It is given that $f (x)$ is divisible by $x - 1$ and by $x + 3$ .

(a) Find $a$ and $b$ . (3 marks)

(b) Solve the equation $f (x) = 0$ . (2 marks)

Show answer

(a) $a = 0, b = 6$ (b) $x = - 3, 1$ or $2$ , since $f (x) = (x - 1) (x + 3) (x - 2)$ .

Question 8 · Modelled on 2020 P1 Q13 — both unknowns sit in the coefficients

Let $f (x) = x^{3} + p x^{2} + q x - 12$ , where $p$ and $q$ are constants. It is given that $f (x)$ is divisible by $x - 2$ , and that when $f (x)$ is divided by $x + 1$ , the remainder is $- 6$ .

(a) Find $p$ and $q$ . (3 marks)

(b) Factorize $f (x)$ completely. (2 marks)

Show answer

(a) $p = 3, q = - 4$ (b) $(x - 2) (x + 2) (x + 3)$ .

Question 9 · Modelled on 2017 P1 Q14 — the divisor is a quadratic that itself factorizes

Let $f (x) = 3 x^{3} + a x^{2} - 10 x + 8$ , where $a$ is a constant. It is given that $f (x)$ is divisible by $x^{2} + x - 2$ .

(a) Find $a$ . (2 marks)

(b) Factorize $f (x)$ completely. (2 marks)

Show answer

(a) $a = - 1$ , using $f (1) = 0$ and $f (- 2) = 0$ , since $x^{2} + x - 2 = (x - 1) (x + 2)$ . (b) $(x - 1) (x + 2) (3 x - 4)$ .

Question 10 · Modelled on 2020 P1 Q13 — the claim turns out to be false

Let $f (x) = 2 x^{3} + a x^{2} + b x - 6$ , where $a$ and $b$ are constants. It is given that $f (x)$ is divisible by $x + 2$ , and that the remainder when $f (x)$ is divided by $x - 1$ is equal to the remainder when $f (x)$ is divided by $x$ .

(a) Find $a$ and $b$ . (3 marks)

(b) Someone claims that all the roots of the equation $f (x) = 0$ are integers. Do you agree? Explain your answer. (3 marks)

Show answer

(a) $a = 3, b = - 5$ (b) Disagree. $f (x) = (x + 2) (x + 1) (2 x - 3)$ , so one root is $\frac{3}{2}$ , which is not an integer.

Question 11 · Modelled on 2021 P1 Q12 — the claim is about rational roots rather than real roots

When the polynomial $p (x)$ is divided by $x^{2} + 2 x - 1$ , the quotient and the remainder are $2 x - 3$ and $c x - 1$ respectively, where $c$ is a constant. It is given that $p (x)$ is divisible by $x - 2$ .

(a) Find $c$ . (3 marks)

(b) Someone claims that all the roots of the equation $p (x) = 0$ are rational numbers. Is the claim correct? Explain your answer. (3 marks)

Show answer

(a) $c = - 3$ (b) Not correct. $p (x) = 2 x^{3} + x^{2} - 11 x + 2 = (x - 2) (2 x^{2} + 5 x - 1)$ , and $2 x^{2} + 5 x - 1 = 0$ has discriminant $33$ , which is not a perfect square, so its two roots are irrational.

Question 12 · Written in HKDSE P1 Section A(1) style

Simplify $\frac{x ^{2} - 9}{x ^{2} + x - 6} \div \frac{x - 3}{2 x - 4}$ .

(3 marks)

Show answer

$2$ . Factorize: $\frac{( x + 3 ) ( x - 3 )}{( x + 3 ) ( x - 2 )} \times \frac{2 ( x - 2 )}{x - 3}$ , and every bracket cancels.

Question 13 · Written in HKDSE P1 style — connects an identity to the remainder theorem

It is given that $2 x^{2} + 3 x - 5 \equiv A (x + 1)^{2} + B (x + 1) + C$ , where $A$ , $B$ and $C$ are constants.

(a) Find $A$ , $B$ and $C$ . (3 marks)

(b) Using the result of (a), write down the remainder when $2 x^{2} + 3 x - 5$ is divided by $x + 1$ . (1 mark)

Show answer

(a) $A = 2, B = - 1, C = - 6$ (b) $- 6$ . Every other term contains the factor $x + 1$ .

Question 14 · Modelled on 2017 P1 Q14 — asks for the roots instead of a claim

Let $f (x) = 4 x^{3} + a x^{2} - 3 x + b$ , where $a$ and $b$ are constants. When $f (x)$ is divided by $2 x^{2} + x - 1$ , the quotient and the remainder are $2 x + c$ and $- 4 x + 2$ respectively, where $c$ is a constant.

(a) Find $a$ , $b$ and $c$ . (3 marks)

(b) Solve the equation $f (x) + 4 x - 2 = 0$ . (3 marks)

Show answer

(a) $a = 8, b = - 1, c = 3$ (b) $x = - \frac{3}{2}, - 1$ or $\frac{1}{2}$ , since $f (x) + 4 x - 2 = (2 x - 1) (x + 1) (2 x + 3)$ .

Question 15 · Modelled on 2020 P1 Q13 — the same kind of claim, but this time it holds

Let $f (x) = x^{3} + h x^{2} + k x + 6$ , where $h$ and $k$ are constants. The remainder when $f (x)$ is divided by $x - 1$ is equal to the remainder when $f (x)$ is divided by $x + 2$ . It is also given that $f (x)$ is divisible by $x - 3$ .

(a) Find $h$ and $k$ . (4 marks)

(b) Someone claims that all the roots of the equation $f (x) = 0$ are integers. Do you agree? Explain your answer. (3 marks)

Show answer

(a) $h = - 2, k = - 5$ (b) Agree. $f (x) = (x - 3) (x - 1) (x + 2)$ , so the roots are $3, 1$ and $- 2$ .

Continue practising this topic: Try the Paper 1 Challenge for harder variants, or move to Paper 2 multiple-choice — also available as Paper 2 Challenge MC.

Where this sits in the syllabus

Polynomials sit early in the Number and Algebra strand. The factor theorem returns later whenever a cubic has to be solved, and the fraction work here is what makes later algebra in variations and sequences manageable.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

Start with a free discovery session

No cost, no obligation. We look at where your child actually is, and show you what a term of real progress would look like.

Book a free discovery session See the programmes