Edexcel AS Paper 1 2020 June — Question 10 10 marks

Exam BoardEdexcel
ModuleAS Paper 1 (AS Paper 1)
Year2020
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeIntegration or area using factorised polynomial
DifficultyStandard +0.3 This is a straightforward multi-part question combining routine factor theorem application, polynomial factorization, and integration. Part (a) is simple substitution, part (b) requires algebraic division and factoring a quadratic (standard AS technique), and part (c) involves integrating a polynomial between roots—all standard textbook exercises with no novel problem-solving required. Slightly above average difficulty only due to the multi-step nature and need to identify correct integration bounds.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals

10. $$g ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 41 x - 70$$
  1. Use the factor theorem to show that \(\mathrm { g } ( x )\) is divisible by \(( x - 5 )\).
  2. Hence, showing all your working, write \(\mathrm { g } ( x )\) as a product of three linear factors. The finite region \(R\) is bounded by the curve with equation \(y = \mathrm { g } ( x )\) and the \(x\)-axis, and lies below the \(x\)-axis.
  3. Find, using algebraic integration, the exact value of the area of \(R\).

Question 10:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(g(5) = 2\times5^3 + 5^2 - 41\times5 - 70 = \ldots\)M1 Attempts to calculate \(g(5)\); attempted division by \((x-5)\) is M0
\(g(5) = 0 \Rightarrow (x-5)\) is a factor, hence \(g(x)\) is divisible by \((x-5)\)A1 Correct calculation, reason and conclusion; must follow M1
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2x^3 + x^2 - 41x - 70 = (x-5)(2x^2 \ldots x \pm 14)\)M1 Attempts quadratic factor by inspection or division
\(= (x-5)(2x^2 + 11x + 14)\)A1 May not see the \((x-5)\) which can be condoned
Attempts to factorise quadratic factordM1 Correct attempt to factorise their \((2x^2+11x+14)\)
\((g(x)) = (x-5)(2x+7)(x+2)\)A1 Must be product of factors, not just statement of three factors
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\int 2x^3 + x^2 - 41x - 70\, dx = \frac{1}{2}x^4 + \frac{1}{3}x^3 - \frac{41}{2}x^2 - 70x\)M1, A1 \(x^n \to x^{n+1}\) for any term; ignore any reference to \(+C\)
Deduces need to use \(\int_{-2}^{5} g(x)\,dx\)M1 Awarded from limits on integral or embedded values
\(-\frac{1525}{3} - \frac{190}{3}\)
Area \(= 571\frac{2}{3}\)A1 Clear algebraic steps leading to answer; oe
# Question 10:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $g(5) = 2\times5^3 + 5^2 - 41\times5 - 70 = \ldots$ | M1 | Attempts to calculate $g(5)$; attempted division by $(x-5)$ is M0 |
| $g(5) = 0 \Rightarrow (x-5)$ is a factor, hence $g(x)$ is divisible by $(x-5)$ | A1 | Correct calculation, reason and conclusion; must follow M1 |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^3 + x^2 - 41x - 70 = (x-5)(2x^2 \ldots x \pm 14)$ | M1 | Attempts quadratic factor by inspection or division |
| $= (x-5)(2x^2 + 11x + 14)$ | A1 | May not see the $(x-5)$ which can be condoned |
| Attempts to factorise quadratic factor | dM1 | Correct attempt to factorise their $(2x^2+11x+14)$ |
| $(g(x)) = (x-5)(2x+7)(x+2)$ | A1 | Must be product of factors, not just statement of three factors |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 2x^3 + x^2 - 41x - 70\, dx = \frac{1}{2}x^4 + \frac{1}{3}x^3 - \frac{41}{2}x^2 - 70x$ | M1, A1 | $x^n \to x^{n+1}$ for any term; ignore any reference to $+C$ |
| Deduces need to use $\int_{-2}^{5} g(x)\,dx$ | M1 | Awarded from limits on integral or embedded values |
| $-\frac{1525}{3} - \frac{190}{3}$ | | |
| Area $= 571\frac{2}{3}$ | A1 | Clear algebraic steps leading to answer; oe |

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10.

$$g ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 41 x - 70$$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $\mathrm { g } ( x )$ is divisible by $( x - 5 )$.
\item Hence, showing all your working, write $\mathrm { g } ( x )$ as a product of three linear factors.

The finite region $R$ is bounded by the curve with equation $y = \mathrm { g } ( x )$ and the $x$-axis, and lies below the $x$-axis.
\item Find, using algebraic integration, the exact value of the area of $R$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q10 [10]}}