CAIE P3 2023 November — Question 3 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeTwo unknowns, direct system
DifficultyModerate -0.8 This is a straightforward application of the Remainder Theorem requiring students to substitute two values into the polynomial and solve simultaneous linear equations. The arithmetic involves fractions but is routine, and the method is a standard textbook exercise with no problem-solving insight required—easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

3 The polynomial \(2 x ^ { 3 } + a x ^ { 2 } + b x + 6\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). When \(\mathrm { p } ( x )\) is divided by \(( x + 2 )\) the remainder is - 38 and when \(\mathrm { p } ( x )\) is divided by \(( 2 x - 1 )\) the remainder is \(\frac { 19 } { 2 }\). Find the values of \(a\) and \(b\).

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(2(-2)^3 + a(-2)^2 + b(-2) + 6 = -38\)M1 Substitute \(x = -2\) and equate the result to \(-38\), or divide by \(x+2\) to obtain quadratic quotient, and equate constant remainder to \(-38\).
Obtain a correct evaluated equation, e.g. \(-16 + 4a - 2b + 6 = -38\) or \(4a - 2b = -28\)A1
\(2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 6 = \frac{19}{2}\)M1 Substitute \(x = \frac{1}{2}\) and equate the result to \(\frac{19}{2}\), or divide by \(2x-1\) to obtain quadratic quotient, and equate constant remainder to \(\frac{19}{2}\).
Obtain a correct evaluated equation, e.g. \(\frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 6 = \frac{19}{2}\) or \(\frac{a}{4} + \frac{b}{2} = \frac{13}{4}\)A1
Obtain \(a = -3\) and \(b = 8\)A1 ISW
5
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2(-2)^3 + a(-2)^2 + b(-2) + 6 = -38$ | M1 | Substitute $x = -2$ and equate the result to $-38$, or divide by $x+2$ to obtain quadratic quotient, and equate constant remainder to $-38$. |
| Obtain a correct evaluated equation, e.g. $-16 + 4a - 2b + 6 = -38$ or $4a - 2b = -28$ | A1 | |
| $2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 6 = \frac{19}{2}$ | M1 | Substitute $x = \frac{1}{2}$ and equate the result to $\frac{19}{2}$, or divide by $2x-1$ to obtain quadratic quotient, and equate constant remainder to $\frac{19}{2}$. |
| Obtain a correct evaluated equation, e.g. $\frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 6 = \frac{19}{2}$ or $\frac{a}{4} + \frac{b}{2} = \frac{13}{4}$ | A1 | |
| Obtain $a = -3$ and $b = 8$ | A1 | ISW |
| | **5** | |

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3 The polynomial $2 x ^ { 3 } + a x ^ { 2 } + b x + 6$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. When $\mathrm { p } ( x )$ is divided by $( x + 2 )$ the remainder is - 38 and when $\mathrm { p } ( x )$ is divided by $( 2 x - 1 )$ the remainder is $\frac { 19 } { 2 }$.

Find the values of $a$ and $b$.\\

\hfill \mbox{\textit{CAIE P3 2023 Q3 [5]}}