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

Polynomials: HKDSE Paper 1 Challenge

Challenge tier. These ten questions state a condition and leave the method to you: a remainder on a quadratic divisor, a squared factor, an H.C.F. that must have a given degree. Work the standard Paper 1 set first if the remainder theorem is not yet automatic.

10 questions · about 70 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 a quadratic, from two linear remainders
Question 2 — a factor that works for every k; repeated roots
Question 3 — divisibility by a squared factor
Question 4 — recovering two polynomials from their H.C.F., then the L.C.M.
Question 5 — building a cubic from three remainders
Question 6 — a claim about the number of distinct roots for every k
Question 7 — working backwards from an H.C.F. and an L.C.M.
Question 8 — remainder on division by x² − 1 using x² = 1
Question 9 — testing a general claim about cubics
Question 10 — choosing a parameter so that the H.C.F. has degree 2

Question 1 · New technique: the remainder on division by a quadratic is written as ax + b

When the polynomial $f (x)$ is divided by $x - 1$ , the remainder is $5$ . When $f (x)$ is divided by $x + 2$ , the remainder is $- 1$ . Find the remainder when $f (x)$ is divided by $x^{2} + x - 2$ .

(4 marks)

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$2 x + 3$ . Write $f (x) = (x - 1) (x + 2) Q (x) + a x + b$ ; then $a + b = 5$ and $- 2 a + b = - 1$ .

Question 2 · New technique: a parameter that cancels out, and two different ways a root can repeat

Let $f (x) = x^{3} + k x^{2} + k x + 1$ , where $k$ is a real constant.

(a) Show that $x + 1$ is a factor of $f (x)$ for every value of $k$ . (2 marks)

(b) Find all values of $k$ such that the equation $f (x) = 0$ has a repeated root. (4 marks)

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(a) $f (- 1) = - 1 + k - k + 1 = 0$ . (b) $k = - 1$ or $k = 3$ . The other factor is $x^{2} + (k - 1) x + 1$ : it has a repeated root when $k = - 1$ or $3$ , and it has $- 1$ as a root when $k = 3$ (giving $(x + 1)^{3}$ ). Both routes must be checked.

Question 3 · New technique: divisibility by a squared factor

The polynomial $x^{4} + a x + b$ is divisible by $(x - 1)^{2}$ , where $a$ and $b$ are constants. Find $a$ and $b$ .

(4 marks)

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$a = - 4, b = 3$ . Long division gives $x^{4} = (x - 1)^{2} (x^{2} + 2 x + 3) + 4 x - 3$ , so the remainder is $(4 + a) x + (b - 3)$ , which must be zero. Using only $f (1) = 0$ is not enough.

Question 4 · New technique: the H.C.F. is given, and the student must turn it into a factor-theorem condition

The H.C.F. of $x^{2} + p x + 6$ and $x^{2} - x + q$ is $x - 3$ , where $p$ and $q$ are constants. Find the L.C.M. of the two polynomials.

(4 marks)

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$(x - 3) (x - 2) (x + 2)$ . Since $x - 3$ divides both, $p = - 5$ and $q = - 6$ , giving $(x - 2) (x - 3)$ and $(x - 3) (x + 2)$ .

Question 5 · New technique: building a polynomial from its remainders

$p (x)$ is a cubic polynomial. When $p (x)$ is divided by $x - 1$ , $x - 2$ and $x - 3$ , the remainders are $2$ , $4$ and $6$ respectively. It is also given that $p (0) = 6$ . Find $p (x)$ .

(5 marks)

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$p (x) = - x^{3} + 6 x^{2} - 9 x + 6$ . The cubic $p (x) - 2 x$ has roots $1, 2, 3$ , so $p (x) = k (x - 1) (x - 2) (x - 3) + 2 x$ , and $p (0) = 6$ gives $k = - 1$ .

Question 6 · New technique: disprove a 'for every value' claim with one counterexample

Let $f (x) = x^{3} - (k + 2) x^{2} + (2 k + 1) x - k$ , where $k$ is a real constant.

(a) Factorize $f (x)$ completely. (3 marks)

(b) Someone claims that for every value of $k$ , the equation $f (x) = 0$ has exactly two distinct real roots. Do you agree? Explain your answer. (2 marks)

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(a) $(x - 1)^{2} (x - k)$ (b) Disagree. When $k = 1$ , $f (x) = (x - 1)^{3}$ and there is only one distinct root.

Question 7 · New technique: the powers of each factor decide what the unknown polynomial may contain

Let $A = (x + 1)^{2} (x - 2)$ . A polynomial $B$ with leading coefficient $1$ is such that the H.C.F. of $A$ and $B$ is $x + 1$ and the L.C.M. of $A$ and $B$ is $(x + 1)^{2} (x - 2) (x + 3)$ .

(a) Someone claims that $B$ could be $(x + 1)^{2} (x + 3)$ . Is the claim correct? Explain your answer. (2 marks)

(b) Find $B$ . (2 marks)

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(a) Not correct: the H.C.F. would then be $(x + 1)^{2}$ , not $x + 1$ . (b) $B = (x + 1) (x + 3)$ . $B$ must contain $x + 3$ , exactly one factor $x + 1$ , and no factor $x - 2$ .

Question 8 · New technique: reduce the polynomial using the divisor, rather than long division

When $2 x^{3} + a x^{2} + b x + 3$ is divided by $x^{2} - 1$ , the remainder is $5 x + 1$ , where $a$ and $b$ are constants. Find $a$ and $b$ .

(4 marks)

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$a = - 2, b = 3$ . Put $x = 1$ and $x = - 1$ : $5 + a + b = 6$ and $1 + a - b = - 4$ .

Question 9 · New technique: a general statement about all integer coefficients, not a specific polynomial

Let $f (x) = x^{3} + p x + q$ , where $p$ and $q$ are integers. Someone claims: “If $x - 1$ is a factor of $f (x)$ , then the equation $f (x) = 0$ has three integer roots.” Is the claim correct? Explain your answer.

(3 marks)

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Not correct. Take $p = 0, q = - 1$ : $x - 1$ is a factor of $x^{3} - 1 = (x - 1) (x^{2} + x + 1)$ , but $x^{2} + x + 1 = 0$ has no real roots.

Question 10 · New technique: a condition on the degree of the H.C.F.

Let $f (x) = x^{3} + 2 x^{2} - x - 2$ and $g (x) = x^{3} + k x^{2} - 4$ , where $k$ is a constant. It is given that the H.C.F. of $f (x)$ and $g (x)$ is a quadratic polynomial.

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

(b) Find the L.C.M. of $f (x)$ and $g (x)$ . (2 marks)

Show answer

(a) $k = 3$ . $f (x) = (x - 1) (x + 1) (x + 2)$ ; $k = 3$ makes both $x - 1$ and $x + 2$ factors of $g (x)$ . The value $k = 5$ also gives a common factor, $x + 1$ , but only one. (b) $(x - 1) (x + 1) (x + 2)^{2}$ .

Continue practising this topic: Try the Paper 1 long questions 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.

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