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Polynomials: HKDSE Paper 2 MC Practice

Twenty-five multiple-choice questions covering the whole topic: factorizing, identities, the remainder and factor theorems, H.C.F. and L.C.M., and simplifying algebraic fractions. Most can be done in under a minute and a half once the right substitution is spotted.

25 questions · about 35 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.

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Question 1 — factorizing by grouping
Question 2 — perfect square minus a square
Question 3 — comparing coefficients
Question 4 — remainder theorem
Question 5 — remainder theorem with a non-monic divisor
Question 6 — factor theorem
Question 7 — remainder from a combination of unknowns
Question 8 — remainder in terms of a letter
Question 9 — two factors, two unknowns
Question 10 — identifying a factor
Question 11 — H.C.F. of monomials
Question 12 — L.C.M. of polynomials
Question 13 — H.C.F. of polynomials
Question 14 — the third expression from an H.C.F. and an L.C.M.
Question 15 — subtracting fractions
Question 16 — cancelling common factors
Question 17 — fractions with a factorizable denominator
Question 18 — identity with two unknowns
Question 19 — identity with a completed square
Question 20 — remainder of a sum
Question 21 — remainder read from a division statement
Question 22 — statements about factors
Question 23 — complex fraction
Question 24 — factor theorem with a high power
Question 25 — H.C.F. given, unknown coefficients

Question 1 · Written in HKDSE Paper 2 style

$a^{2} - b^{2} - a - b =$

$(a - b) (a + b - 1)$
$(a + b) (a - b + 1)$
$(a + b) (a - b - 1)$
$(a - b) (a + b + 1)$

Show answer

C. $a^{2} - b^{2} - a - b = (a + b) (a - b) - (a + b)$ .

Question 2 · Written in HKDSE Paper 2 style

$4 x^{2} - 12 x y + 9 y^{2} - 1 =$

$(2 x - 3 y + 1) (2 x - 3 y - 1)$
$(2 x + 3 y + 1) (2 x + 3 y - 1)$
$(2 x - 3 y + 1)^{2}$
$(4 x - 9 y + 1) (x - y - 1)$

Show answer

A. The first three terms are $(2 x - 3 y)^{2}$ ; then use the difference of squares.

Question 3 · Written in HKDSE Paper 2 style

If $(x + 2) (x + a) \equiv x^{2} + b x - 6$ , then $b =$

$- 5$
$1$
$5$
$- 1$

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D. The constant term gives $2 a = - 6$ , so $a = - 3$ and $b = 2 + a = - 1$ .

Question 4 · Written in HKDSE Paper 2 style

Find the remainder when $x^{3} - 2 x^{2} + 3 x - 4$ is divided by $x + 1$ .

$- 10$
$- 8$
$- 2$
$2$

Show answer

A. Substitute $x = - 1$ : $- 1 - 2 - 3 - 4 = - 10$ .

Question 5 · Written in HKDSE Paper 2 style

Find the remainder when $4 x^{3} - 2 x + 1$ is divided by $2 x - 1$ .

$1$
$\frac{1}{2}$
$\frac{3}{2}$
$4$

Show answer

B. Substitute $x = \frac{1}{2}$ , not $x = 1$ : $\frac{1}{2} - 1 + 1 = \frac{1}{2}$ .

Question 6 · Written in HKDSE Paper 2 style

If $x - 2$ is a factor of $x^{3} + k x^{2} - 4$ , then $k =$

$- 2$
$1$
$- 1$
$3$

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C. $8 + 4 k - 4 = 0$ , so $k = - 1$ .

Question 7 · Modelled on 2026 P2 Q8 — two unknowns, but only their sum is needed

Let $g (x) = h x^{3} + x^{2} + k x - 3$ , where $h$ and $k$ are constants. If $g (x)$ is divisible by $x - 1$ , find the remainder when $g (x)$ is divided by $x + 1$ .

$- 2$
$0$
$2$
$- 4$

Show answer

D. $g (1) = 0$ gives $h + k = 2$ ; then $g (- 1) = - (h + k) - 2 = - 4$ .

Question 8 · Modelled on 2022 P2 Q9 — the constant term is the unknown this time

Let $f (x) = x^{2} - 3 a x + b$ , where $a$ and $b$ are constants. If $f (x)$ is divisible by $x - a$ , find the remainder when $f (x)$ is divided by $x + a$ .

$6 a^{2}$
$0$
$2 a^{2}$
$4 a^{2}$

Show answer

A. $f (a) = 0$ gives $b = 2 a^{2}$ ; then $f (- a) = a^{2} + 3 a^{2} + 2 a^{2} = 6 a^{2}$ .

Question 9 · Written in HKDSE Paper 2 style

Let $f (x) = x^{3} + a x^{2} + b x + 6$ , where $a$ and $b$ are constants. If $f (x)$ is divisible by both $x - 1$ and $x + 2$ , then $a =$

$- 5$
$2$
$5$
$- 2$

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D. $a + b = - 7$ and $4 a - 2 b = 2$ , so $a = - 2$ and $b = - 5$ .

Question 10 · Written in HKDSE Paper 2 style

Which of the following is a factor of $x^{3} - 3 x^{2} - 4 x + 12$ ?

$x + 3$
$x - 3$
$x - 1$
$x + 1$

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B. Group: $x^{2} (x - 3) - 4 (x - 3) = (x - 3) (x - 2) (x + 2)$ .

Question 11 · Written in HKDSE Paper 2 style

The H.C.F. of $12 x^{3} y^{2}$ , $18 x^{2} y^{4}$ and $30 x y^{3} z$ is

$6 x^{2} y^{2}$
$6 x y^{2} z$
$6 x y^{2}$
$180 x^{3} y^{4} z$

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C. Take the lowest power of each letter: $x^{1}$ , $y^{2}$ , and no $z$ because two terms have none.

Question 12 · Written in HKDSE Paper 2 style

The L.C.M. of $x^{2} - 4$ , $x^{2} + 4 x + 4$ and $x - 2$ is

$(x - 2) (x + 2)^{2}$
$(x - 2) (x + 2)$
$(x - 2)^{2} (x + 2)^{2}$
$(x - 2)^{2} (x + 2)^{3}$

Show answer

A. The factors are $(x - 2) (x + 2)$ , $(x + 2)^{2}$ , $(x - 2)$ ; take the highest power of each.

Question 13 · Written in HKDSE Paper 2 style

The H.C.F. of $x^{2} - 9$ , $x^{2} + x - 6$ and $2 x^{2} + 6 x$ is

$x - 3$
$2 (x + 3)$
$x + 3$
$(x + 3) (x - 3)$

Show answer

C. The factorizations are $(x + 3) (x - 3)$ , $(x + 3) (x - 2)$ and $2 x (x + 3)$ ; only $x + 3$ is shared.

Question 14 · Modelled on 2026 P2 Q32 — the powers of x are fixed and y, z vary

The H.C.F. and the L.C.M. of three expressions are $x^{2} y$ and $x^{4} y^{3} z$ respectively. If the first expression and the second expression are $x^{2} y^{3}$ and $x^{4} y z$ respectively, then the third expression could be

$x y^{2}$
$x^{3} y^{4}$
$x^{2} y z^{2}$
$x^{2} y z$

Show answer

D. The third expression needs at least $x^{2}$ and $y$ , and at most $x^{4}$ , $y^{3}$ and $z$ . Only $x^{2} y z$ fits.

Question 15 · Written in HKDSE Paper 2 style

$\frac{1}{x - 1} - \frac{1}{x + 1} =$

$\frac{2 x}{x ^{2} - 1}$
$\frac{2}{x ^{2} - 1}$
$\frac{- 2}{x ^{2} - 1}$
$0$

Show answer

B. The numerator is $(x + 1) - (x - 1) = 2$ .

Question 16 · Written in HKDSE Paper 2 style

$\frac{x ^{2} - x - 6}{x ^{2} - 9} =$

$\frac{x - 2}{x - 3}$
$\frac{x + 2}{x + 3}$
$\frac{x + 2}{x - 3}$
$\frac{x - 6}{- 9}$

Show answer

B. $\frac{( x - 3 ) ( x + 2 )}{( x - 3 ) ( x + 3 )}$ .

Question 17 · Written in HKDSE Paper 2 style

$\frac{2}{x ^{2} - x} - \frac{1}{x - 1} =$

$\frac{2 - x}{x ( x - 1 )}$
$\frac{1}{x ( x - 1 )}$
$\frac{x - 2}{x ( x - 1 )}$
$\frac{1}{x ^{2} - x}$

Show answer

A. The L.C.M. of the denominators is $x (x - 1)$ , giving numerator $2 - x$ .

Question 18 · Written in HKDSE Paper 2 style

If $A (x - 1) + B (x + 2) \equiv 4 x - 1$ , then $A =$

$1$
$- 1$
$4$
$3$

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D. Put $x = - 2$ : $- 3 A = - 9$ , so $A = 3$ .

Question 19 · Written in HKDSE Paper 2 style

If $x^{2} + 6 x + k \equiv (x + m)^{2} - 2$ , then $k =$

$3$
$9$
$7$
$11$

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C. $m = 3$ , so $k = m^{2} - 2 = 7$ .

Question 20 · Written in HKDSE Paper 2 style

When $f (x)$ is divided by $x - 3$ , the remainder is $2$ . Find the remainder when $f (x) + x^{2}$ is divided by $x - 3$ .

$11$
$2$
$9$
$5$

Show answer

A. $f (3) + 3^{2} = 2 + 9 = 11$ .

Question 21 · Written in HKDSE Paper 2 style

It is given that $f (x) = (x - 1) (x + 2) Q (x) + 2 x - 3$ , where $Q (x)$ is a polynomial. Find the remainder when $f (x)$ is divided by $x + 2$ .

$- 3$
$- 7$
$1$
$2 x - 3$

Show answer

B. At $x = - 2$ the first term vanishes, leaving $2 (- 2) - 3 = - 7$ .

Question 22 · Written in HKDSE Paper 2 style

Let $f (x) = x^{3} - 4 x^{2} + x + 6$ . Which of the following are true?
I. $x + 1$ is a factor of $f (x)$ .
II. $x - 2$ is a factor of $f (x)$ .
III. $x + 2$ is a factor of $f (x)$ .

I only
I and II only
II and III only
I, II and III

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B. $f (- 1) = 0$ and $f (2) = 0$ , but $f (- 2) = - 20$ .

Question 23 · Written in HKDSE Paper 2 style

$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{1}{a} + \frac{1}{b}} =$

$a + b$
$\frac{a - b}{ab}$
$a - b$
$\frac{1}{a - b}$

Show answer

C. Multiply top and bottom by $ab$ : $\frac{a ^{2} - b ^{2}}{a + b} = a - b$ .

Question 24 · Written in HKDSE Paper 2 style

If $x + 1$ is a factor of $x^{101} + x + k$ , then $k =$

$- 2$
$0$
$1$
$2$

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D. $(- 1)^{101} = - 1$ , so $- 1 - 1 + k = 0$ .

Question 25 · Written in HKDSE Paper 2 style

The H.C.F. of $3 x^{2} - 5 x + a$ and $x^{2} + b x - 2$ is $x - 2$ , where $a$ and $b$ are constants. Find $a + b$ .

$- 3$
$- 1$
$1$
$3$

Show answer

A. Substitute $x = 2$ into each: $a = - 2$ and $b = - 1$ .

Continue practising this topic: Try the Paper 1 long questions for harder variants, or move to Paper 1 Challenge — 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.

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