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Polynomials: HKDSE Paper 2 MC Challenge

Challenge tier. In each of these fifteen questions one distractor is what you get from a plausible but wrong route: the product instead of the L.C.M., the wrong root of the divisor, a sign dropped from a constant. Work the standard Paper 2 set first.

15 questions · about 25 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 — remainder on division by x² − 1
Question 2 — reading a remainder through a factorized divisor
Question 3 — divisibility by x² − 1
Question 4 — statements about a family of cubics
Question 5 — L.C.M. with repeated factors
Question 6 — recovering a polynomial from H.C.F. and L.C.M.
Question 7 — cubic from three roots and one value
Question 8 — common factor of two quadratics
Question 9 — remainder of a composite
Question 10 — sum of three fractions
Question 11 — dividing fractions in two letters
Question 12 — remainder of a high power on a quadratic divisor
Question 13 — identity: quickest substitution
Question 14 — equal remainders on three divisors
Question 15 — logic of a factor of a product

Question 1 · New technique: remainder on a quadratic divisor from two linear remainders

When $f (x)$ is divided by $x - 1$ and $x + 1$ , the remainders are $3$ and $- 1$ respectively. Find the remainder when $f (x)$ is divided by $x^{2} - 1$ .

$x + 2$
$2 x + 1$
$2 x - 1$
$3 x - 1$

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B. Write the remainder as $a x + b$ : $a + b = 3$ and $- a + b = - 1$ . The distractor $3 x - 1$ just glues the two numbers together.

Question 2 · Challenge: link the linear divisor to the quadratic one

When $f (x)$ is divided by $x^{2} - x - 2$ , the remainder is $3 x - 1$ . Find the remainder when $f (x)$ is divided by $x + 1$ .

$- 4$
$- 1$
$2$
$5$

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A. $x^{2} - x - 2 = (x + 1) (x - 2)$ , so $f (- 1) = 3 (- 1) - 1 = - 4$ . The value $5$ is the remainder on division by $x - 2$ .

Question 3 · Written in HKDSE Paper 2 style

If $x^{4} + a x + b$ is divisible by $x^{2} - 1$ , then $a + b =$

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

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D. $f (1) = 0$ and $f (- 1) = 0$ give $a = 0$ and $b = - 1$ .

Question 4 · New technique: statements that must hold for every k

Let $f (x) = x^{3} + k x^{2} + k x + 1$ , where $k$ is a constant. Which of the following are true?
I. $x + 1$ is a factor of $f (x)$ for every value of $k$ .
II. The remainder when $f (x)$ is divided by $x - 1$ is $2 k + 2$ .
III. When $k = 3$ , $f (x) = (x + 1)^{3}$ .

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

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D. $f (- 1) = 0$ for all $k$ ; $f (1) = 2 k + 2$ ; and $x^{3} + 3 x^{2} + 3 x + 1 = (x + 1)^{3}$ .

Question 5 · Challenge: repeated factors in both expressions

The L.C.M. of $(x + 1)^{2} (x - 2)$ and $(x + 1) (x - 2)^{2} (x + 3)$ is

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

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C. Highest power of each factor. The option with cubes is the product, which double-counts.

Question 6 · Written in HKDSE Paper 2 style

The H.C.F. and the L.C.M. of $x^{2} - 4$ and a quadratic polynomial $B$ with leading coefficient $1$ are $x - 2$ and $(x - 2) (x + 2) (x - 5)$ respectively. $B =$

$x^{2} - 7 x + 10$
$x^{2} - 3 x - 10$
$x^{2} - 4 x + 4$
$x^{2} + 7 x + 10$

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A. $B$ contains $x - 2$ and $x - 5$ but not $x + 2$ : $(x - 2) (x - 5)$ .

Question 7 · Written in HKDSE Paper 2 style

$p (x)$ is a cubic polynomial. If $p (x)$ is divisible by $x - 1$ , $x - 2$ and $x - 3$ , and $p (0) = 12$ , then $p (4) =$

$6$
$- 12$
$12$
$24$

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B. $p (x) = k (x - 1) (x - 2) (x - 3)$ and $p (0) = - 6 k = 12$ , so $k = - 2$ and $p (4) = - 2 \times 6 = - 12$ . The value $12$ forgets the sign of $k$ .

Question 8 · Challenge: check every case, including the degenerate one

If $x^{2} + k x + 4$ and $x^{2} + 4 x + k$ have a common factor of the form $x - a$ , where $a$ is a real constant, then $k =$

$- 5$ only
$4$ only
$5$ or $- 4$
$- 5$ or $4$

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D. Subtracting gives $(k - 4) (x - 1) = 0$ . Either $x = 1$ (so $k = - 5$ ) or $k = 4$ , when both are $(x + 2)^{2}$ .

Question 9 · Written in HKDSE Paper 2 style

When $f (x)$ is divided by $x - 3$ , the remainder is $7$ . Find the remainder when $f (2 x - 1)$ is divided by $x - 2$ .

$3$
$13$
$7$
It cannot be determined.

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C. The remainder is $f (2 \cdot 2 - 1) = f (3) = 7$ .

Question 10 · Written in HKDSE Paper 2 style

$\frac{1}{( x - 1 ) ( x - 2 )} + \frac{1}{( x - 2 ) ( x - 3 )} + \frac{1}{( x - 3 ) ( x - 1 )} =$

$\frac{3}{( x - 1 ) ( x - 3 )}$
$\frac{3}{( x - 1 ) ( x - 2 ) ( x - 3 )}$
$\frac{3 x}{( x - 1 ) ( x - 2 ) ( x - 3 )}$
$\frac{1}{( x - 1 ) ( x - 3 )}$

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A. The numerator is $(x - 3) + (x - 1) + (x - 2) = 3 (x - 2)$ , which cancels one factor.

Question 11 · Written in HKDSE Paper 2 style

$\frac{a - b}{ab} \div \frac{a ^{2} - b ^{2}}{a ^{2} b} =$

$\frac{b}{a + b}$
$\frac{a}{a + b}$
$\frac{1}{a + b}$
$\frac{a}{( a - b ) ( a + b )}$

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B. $\frac{a - b}{ab} \times \frac{a ^{2} b}{( a - b ) ( a + b )} = \frac{a}{a + b}$ .

Question 12 · Challenge: a very high power on a quadratic divisor

Find the remainder when $x^{100}$ is divided by $x^{2} - 1$ .

$x$
$0$
$x + 1$
$1$

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D. With remainder $a x + b$ : $a + b = 1$ and $- a + b = 1$ , so $a = 0, b = 1$ . The answer $x$ belongs to an odd power.

Question 13 · Written in HKDSE Paper 2 style

If $(x - 1) (a x + b) + c \equiv 2 x^{2} - x + 4$ , then $a + b + c =$

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B. $a = 2$ ; $x = 1$ gives $c = 5$ ; the $x$ coefficient gives $b - a = - 1$ , so $b = 1$ .

Question 14 · Written in HKDSE Paper 2 style

$f (x)$ is a cubic polynomial with leading coefficient $1$ . When $f (x)$ is divided by $x - 1$ , $x - 2$ and $x - 3$ , the remainder is $5$ each time. Find $f (4)$ .

$11$
$5$
$6$
$29$

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A. $f (x) - 5$ has roots $1, 2, 3$ , so $f (x) = (x - 1) (x - 2) (x - 3) + 5$ and $f (4) = 6 + 5$ .

Question 15 · New technique: what a factor of a product does and does not force

Let $f (x)$ and $g (x)$ be polynomials and $a$ be a real constant. If $x - a$ is a factor of $f (x) g (x)$ , which of the following must be true?
I. $x - a$ is a factor of $f (x)$ or of $g (x)$ .
II. $f (a) + g (a) = 0$ .
III. $f (a) g (a) = 0$ .

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

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C. $f (a) g (a) = 0$ forces $f (a) = 0$ or $g (a) = 0$ . II fails, e.g. $f (x) = x - a, g (x) = 1$ .

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

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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