Edexcel C1 — Question 4 5 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeMultiple unknowns with derivative condition
DifficultyStandard +0.3 This is a standard C1 factor theorem question requiring students to use the given roots to form equations. Since the curve touches at x=3, students must recognize this means (x-3) is a repeated factor. The algebra involves solving simultaneous equations from f(-1)=0, f(3)=0, and f'(3)=0, which is routine for this topic with clear scaffolding from the 'show that' structure.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

\includegraphics{figure_1} Figure 1 shows the curve with equation \(y = x^3 + ax^2 + bx + c\), where \(a\), \(b\) and \(c\) are constants. The curve crosses the \(x\)-axis at the point \((-1, 0)\) and touches the \(x\)-axis at the point \((3, 0)\). Show that \(a = -5\) and find the values of \(b\) and \(c\). [5]

AnswerMarks Guidance
cubic, coeff of \(x^3 = 1\), crosses x-axis at \((-1, 0)\), touches at \((3, 0)\)M1 A1
\(\therefore y = (x + 1)(x - 3)^2\)
\(= (x + 1)(x^2 - 6x + 9)\)
\(= x^3 - 6x^2 + 9x + x^2 - 6x + 9\)M1
\(= x^3 - 5x^2 + 3x + 9\)
\(\therefore a = -5, b = 3, c = 9\)A2 (5)
cubic, coeff of $x^3 = 1$, crosses x-axis at $(-1, 0)$, touches at $(3, 0)$ | M1 A1 | 
$\therefore y = (x + 1)(x - 3)^2$ | 
$= (x + 1)(x^2 - 6x + 9)$ | 
$= x^3 - 6x^2 + 9x + x^2 - 6x + 9$ | M1 |
$= x^3 - 5x^2 + 3x + 9$ | 
$\therefore a = -5, b = 3, c = 9$ | A2 | (5)
\includegraphics{figure_1}

Figure 1 shows the curve with equation $y = x^3 + ax^2 + bx + c$, where $a$, $b$ and $c$ are constants. The curve crosses the $x$-axis at the point $(-1, 0)$ and touches the $x$-axis at the point $(3, 0)$.

Show that $a = -5$ and find the values of $b$ and $c$. [5]

\hfill \mbox{\textit{Edexcel C1  Q4 [5]}}