CAIE P3 2024 November — Question 1 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2024
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypePolynomial with equal remainders
DifficultyStandard +0.3 This is a straightforward application of the factor and remainder theorems requiring students to set up two equations from the given conditions and solve simultaneously. The factor condition gives p(-1/2)=0, and the equal remainders condition gives p(4)=3p(2). While it involves some algebraic manipulation, it's a standard textbook exercise with no novel insight required, making it slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

1 The polynomial \(4 x ^ { 3 } + a x ^ { 2 } + 5 x + b\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). It is given that \(( 2 x + 1 )\) is a factor of \(\mathrm { p } ( x )\). When \(\mathrm { p } ( x )\) is divided by \(( x - 4 )\) the remainder is equal to 3 times the remainder when \(\mathrm { p } ( x )\) is divided by \(( x - 2 )\). Find the values of \(a\) and \(b\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Substitute \(x = -\frac{1}{2}\) and equate the result to zeroM1
Obtain a correct equation, e.g. \(-\frac{4}{8} + \frac{a}{4} - \frac{5}{2} + b = 0\)A1 \(\left(\frac{a}{4} + b = 3\right)\) Any equivalent form.
Substitute \(x = 2\) and \(x = 4\) and use \(p(4) = 3p(2)\)M1 If using long division, M1 is for correct use of two constant remainders. Condone if 3 is on the wrong side.
Obtain a correct equation, e.g. \(3(32 + 4a + 10 + b) = 256 + 16a + 20 + b\)A1 \((-2a + b = 75)\) Any equivalent form.
Obtain \(a = -32\) and \(b = 11\)A1
Total5
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -\frac{1}{2}$ and equate the result to zero | M1 | |
| Obtain a correct equation, e.g. $-\frac{4}{8} + \frac{a}{4} - \frac{5}{2} + b = 0$ | A1 | $\left(\frac{a}{4} + b = 3\right)$ Any equivalent form. |
| Substitute $x = 2$ and $x = 4$ and use $p(4) = 3p(2)$ | M1 | If using long division, M1 is for correct use of two constant remainders. Condone if 3 is on the wrong side. |
| Obtain a correct equation, e.g. $3(32 + 4a + 10 + b) = 256 + 16a + 20 + b$ | A1 | $(-2a + b = 75)$ Any equivalent form. |
| Obtain $a = -32$ and $b = 11$ | A1 | |
| **Total** | **5** | |
1 The polynomial $4 x ^ { 3 } + a x ^ { 2 } + 5 x + b$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( 2 x + 1 )$ is a factor of $\mathrm { p } ( x )$. When $\mathrm { p } ( x )$ is divided by $( x - 4 )$ the remainder is equal to 3 times the remainder when $\mathrm { p } ( x )$ is divided by $( x - 2 )$.

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

\hfill \mbox{\textit{CAIE P3 2024 Q1 [5]}}