P2.6Polynomials, algebraic division and the factor theorem
Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Polynomials, algebraic division and the factor theorem. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \(\mathrm{f}(x) = x^{3} - 4x^{2} - 15x + 18\)
(a) Use the factor theorem to show that \((x - 1)\) is a factor of \(\mathrm{f}(x)\).[2]
(b) Hence fully factorise \(\mathrm{f}(x)\).[4]
Show the answer and mark scheme
(a)Answer: \(\mathrm{f}(1) = 0\), so \((x - 1)\) is a factor.
M1 for attempting \(\mathrm{f}(1)\)
A1* for \(\mathrm{f}(1) = 0\) with a conclusion, e.g. so \((x - 1)\) is a factor
Worked solution: \(\mathrm{f}(1) = 1^3 - 4(1)^2 - 15(1) + 18 = 1 - 4 - 15 + 18 = 0\). Since \(\mathrm{f}(1) = 0\), \((x - 1)\) is a factor of \(\mathrm{f}(x)\).
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \(\mathrm{f}(x) = 3x^{3} - 13x^{2} + 2x + 8\)
(a) Use the factor theorem to show that \((3x + 2)\) is a factor of \(\mathrm{f}(x)\).[2]
(b) Hence fully factorise \(\mathrm{f}(x)\).[4]
(c) Hence solve \(\mathrm{f}(x) = 0\)[1]
Show the answer and mark scheme
(a)Answer: \(\mathrm{f}\left(-\frac{2}{3}\right) = 0\), so \((3x + 2)\) is a factor.
M1 for attempting \(\mathrm{f}\left(-\frac{2}{3}\right)\)
A1* for \(\mathrm{f}\left(-\frac{2}{3}\right) = 0\) with a conclusion, e.g. so \((3x + 2)\) is a factor
Worked solution: \(\mathrm{f}\left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right)^3 - 13\left(-\frac{2}{3}\right)^2 + 2\left(-\frac{2}{3}\right) + 8 = -\frac{8}{9} - \frac{52}{9} - \frac{4}{3} + 8 = 0\). Since \(\mathrm{f}\left(-\frac{2}{3}\right) = 0\), \((3x + 2)\) is a factor of \(\mathrm{f}(x)\).
Worked solution: From the factorised form, \(x = -\frac{2}{3}, \ 1, \ 4\).
Question 3Hard9 marks
(a) Explain why, when a polynomial is divided by a quadratic, the remainder can be written in the form \(ax + b\).[1]
(b) Find the remainder when \(x^{100}\) is divided by \(x^2 - 3x + 2\). Give your answer in the form \(ax + b\), where \(a\) and \(b\) are expressed in terms of powers of 2.[5]
(c) When the polynomial \(\mathrm{q}(x)\) is divided by \((x - 1)\) the remainder is 3, and when \(\mathrm{q}(x)\) is divided by \((x + 2)\) the remainder is \(-9\). Find the remainder when \(\mathrm{q}(x)\) is divided by \(x^2 + x - 2\).[3]
Show the answer and mark scheme
(a)Answer: The remainder has a lower degree than the divisor, so its degree is at most 1.
B1 for the degree of the remainder is less than the degree of the divisor (2), so it is linear or constant
Worked solution: In polynomial division the remainder always has a smaller degree than the divisor. The divisor has degree 2, so the remainder has degree at most 1 and can be written \(ax + b\) (with \(a\) possibly 0).
M1 for \(\mathrm{q}(x) = (x - 1)(x + 2)\mathrm{S}(x) + cx + d\) and using the remainder theorem: \(\mathrm{q}(1) = 3\), \(\mathrm{q}(-2) = -9\)
M1 for \(c + d = 3\) and \(-2c + d = -9\) solved
A1 for \(4x - 1\)
Worked solution: \(x^2 + x - 2 = (x - 1)(x + 2)\), so \(\mathrm{q}(x) = (x - 1)(x + 2)\mathrm{S}(x) + cx + d\). By the remainder theorem, \(\mathrm{q}(1) = c + d = 3\) and \(\mathrm{q}(-2) = -2c + d = -9\). So \(3c = 12\), \(c = 4\), \(d = -1\): the remainder is \(4x - 1\).