Pre-U Pre-U 9794/2 2014 June — Question 10 11 marks

Exam BoardPre-U
ModulePre-U 9794/2 (Pre-U Mathematics Paper 2)
Year2014
SessionJune
Marks11
TopicPolynomial Division & Manipulation
TypeSketching Polynomial Curves
DifficultyStandard +0.3 This is a structured multi-part question with clear scaffolding: (i) is trivial substitution, (ii) is routine polynomial division, and (iii) requires factorization and sketching. While it involves multiple steps, each part uses standard A-level techniques without requiring novel insight or complex problem-solving. The scaffolding makes it slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials
10 Let \(\mathrm { f } ( x ) = x ^ { 4 } - 4 x ^ { 3 } - 10 x ^ { 2 } + 28 x - 15\).
  1. Show that \(x = 1\) is a root of the equation \(\mathrm { f } ( x ) = 0\).
  2. Find the quotient and remainder when \(\mathrm { f } ( x )\) is divided by \(x - 5\).
  3. Factorise \(\mathrm { f } ( x )\) fully and hence sketch the graph of \(y = \mathrm { f } ( x )\).
(i) \(f(1) = 1 - 4 - 10 + 28 - 15\)
\(= 0\) hence \(x = 1\) is a root
M1 Substitute \(x = 1\) into the function, or equivalent process to find remainder
A1 [2] Show clearly that it equals zero and conclude, with correct terminology
(ii) \(f(x) = (x-5)(x^3 + x^2 - 5x + 3) + 0\)
M1 Attempt complete division or factorisation
A1 Quotient correct at least as far as \(x^2\) term
A1 Quotient \(x^3 + x^2 - 5x + 3\) soi
B1 [4] Remainder 0 (allow 'no remainder') soi
(iii) \(f(x) = (x-5)(x-1)(x^2 + 2x - 3)\)
M1 Attempt to write \(f(x)\) as product of two linear factors and one quadratic. Could go via \((x-1)(x^3 - 3x^2 - 13x + 15)\)
A1 Obtain correct linear and quadratic factors soi
\(= (x-5)(x-1)^2(x+3)\)
A1 Obtain fully correct factorisation
Sketch showing: a positive quartic intersecting with the \(x\)-axis at \(-3\) and \(5\) and maximum on the \(x\)-axis at \(1\)
M1 Positive quartic, with 3 turning points. Allow \(y \leq 0\) only
A1 [5] \(x\) coords indicated, or implied by scale. No need to see \(-15\) on \(y\)-axis. Allow minimum at \((0, -15)\). Need \(y > 0\) as well, possibly with one arm truncated
(i) $f(1) = 1 - 4 - 10 + 28 - 15$
$= 0$ hence $x = 1$ is a root

M1 Substitute $x = 1$ into the function, or equivalent process to find remainder
A1 [2] Show clearly that it equals zero and conclude, with correct terminology

(ii) $f(x) = (x-5)(x^3 + x^2 - 5x + 3) + 0$

M1 Attempt complete division or factorisation
A1 Quotient correct at least as far as $x^2$ term
A1 Quotient $x^3 + x^2 - 5x + 3$ soi
B1 [4] Remainder 0 (allow 'no remainder') soi

(iii) $f(x) = (x-5)(x-1)(x^2 + 2x - 3)$

M1 Attempt to write $f(x)$ as product of two linear factors and one quadratic. Could go via $(x-1)(x^3 - 3x^2 - 13x + 15)$

A1 Obtain correct linear and quadratic factors soi

$= (x-5)(x-1)^2(x+3)$

A1 Obtain fully correct factorisation

Sketch showing: a positive quartic intersecting with the $x$-axis at $-3$ and $5$ and maximum on the $x$-axis at $1$

M1 Positive quartic, with 3 turning points. Allow $y \leq 0$ only
A1 [5] $x$ coords indicated, or implied by scale. No need to see $-15$ on $y$-axis. Allow minimum at $(0, -15)$. Need $y > 0$ as well, possibly with one arm truncated
10 Let $\mathrm { f } ( x ) = x ^ { 4 } - 4 x ^ { 3 } - 10 x ^ { 2 } + 28 x - 15$.\\
(i) Show that $x = 1$ is a root of the equation $\mathrm { f } ( x ) = 0$.\\
(ii) Find the quotient and remainder when $\mathrm { f } ( x )$ is divided by $x - 5$.\\
(iii) Factorise $\mathrm { f } ( x )$ fully and hence sketch the graph of $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2014 Q10 [11]}}
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