CAIE P3 2023 March — Question 3 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFind constants with divisibility condition
DifficultyStandard +0.3 This is a straightforward application of the remainder theorem for polynomial division. Students need to use the fact that p(x) = q(x)(x²-x+1) + 3x+2, then either substitute roots of x²-x+1 or compare coefficients. It's slightly above routine because it involves a quadratic divisor rather than linear, but the method is standard and requires only algebraic manipulation with no novel insight.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division

3 The polynomial \(2 x ^ { 4 } + a x ^ { 3 } + b x - 1\), where \(a\) and \(b\) are constants, is denoted by \(\mathrm { p } ( x )\). When \(\mathrm { p } ( x )\) is divided by \(x ^ { 2 } - x + 1\) the remainder is \(3 x + 2\). Find the values of \(a\) and \(b\).

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Commence division and reach partial quotient \(2x^2 + (a \pm 2)x\)M1 Need \(2x^2 + (a \pm 2)x\); working backwards from remainder: \(2x^2 + (\ldots)x \pm 3\) M1 \(2x^2 - x - 3\) A1
Obtain correct quotient \(2x^2 + (a+2)x + a\)A1 Allow sign error e.g. in \(b-2\)
Set *their* linear remainder equal to part of "\(3x + 2\)" and solve for \(a\) or \(b\)M1 Remainder \(= 3x + 2 = (b-2)x - 1 - a\). Allow for just equating \(x\) term or constant term.
Obtain answer \(a = -3\)A1
Obtain answer \(b = 5\)A1
Alternative Method:
AnswerMarks Guidance
AnswerMarks Guidance
State \(2x^4 + ax^3 + 0x^2 + bx - 1 = (x^2 - x + 1)(2x^2 + Ax + B) + 3x + 2\) and form/solve equations for \(A\) or \(B\)M1 e.g. \(0 = B - A + 2\) and \(-1 = B + 2\)
Obtain \(A = -1\), \(B = -3\)A1
Form and solve equations for \(a\) or \(b\)M1 e.g. \(a = A - 2\) or \(b = -B + A + 3\)
Obtain \(a = -3\)A1
Obtain \(b = 5\)A1
Alternative Method (Remainder Theorem):
AnswerMarks Guidance
AnswerMarks Guidance
Use remainder theorem with \(x = \frac{1 \pm \sqrt{-3}}{2}\) or \(x = \frac{1 \pm i\sqrt{3}}{2}\)M1 Allow correct use with reasonable attempt at either root in exact or decimal form; \(x^2 = \frac{-1+\sqrt{-3}}{2}\), \(x^3 = -1\), \(x^4 = \frac{-1-\sqrt{-3}}{2}\)
Obtain \(-a + \frac{b}{2} \pm \frac{b\sqrt{-3}}{2} \mp \sqrt{-3} - 2 = \frac{7}{2} \pm \frac{3\sqrt{-3}}{2}\) or \(-a + \frac{b}{2} \pm \frac{bi\sqrt{3}}{2} \mp i\sqrt{3} - 2 = \frac{7}{2} \pm \frac{3i\sqrt{3}}{2}\)A1 Expand brackets and obtain exact equation for either root
Solve simultaneous equations, or single equation, for \(a\) or \(b\)M1
Obtain \(a = -3\) from exact workingA1
Obtain \(b = 5\) from exact workingA1
Total5
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Commence division and reach partial quotient $2x^2 + (a \pm 2)x$ | **M1** | Need $2x^2 + (a \pm 2)x$; working backwards from remainder: $2x^2 + (\ldots)x \pm 3$ **M1** $2x^2 - x - 3$ **A1** |
| Obtain correct quotient $2x^2 + (a+2)x + a$ | **A1** | Allow sign error e.g. in $b-2$ |
| Set *their* linear remainder equal to part of "$3x + 2$" and solve for $a$ or $b$ | **M1** | Remainder $= 3x + 2 = (b-2)x - 1 - a$. Allow for just equating $x$ term or constant term. |
| Obtain answer $a = -3$ | **A1** | |
| Obtain answer $b = 5$ | **A1** | |

**Alternative Method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| State $2x^4 + ax^3 + 0x^2 + bx - 1 = (x^2 - x + 1)(2x^2 + Ax + B) + 3x + 2$ and form/solve equations for $A$ or $B$ | **M1** | e.g. $0 = B - A + 2$ and $-1 = B + 2$ |
| Obtain $A = -1$, $B = -3$ | **A1** | |
| Form and solve equations for $a$ or $b$ | **M1** | e.g. $a = A - 2$ or $b = -B + A + 3$ |
| Obtain $a = -3$ | **A1** | |
| Obtain $b = 5$ | **A1** | |

**Alternative Method (Remainder Theorem):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Use remainder theorem with $x = \frac{1 \pm \sqrt{-3}}{2}$ or $x = \frac{1 \pm i\sqrt{3}}{2}$ | **M1** | Allow correct use with reasonable attempt at either root in exact or decimal form; $x^2 = \frac{-1+\sqrt{-3}}{2}$, $x^3 = -1$, $x^4 = \frac{-1-\sqrt{-3}}{2}$ |
| Obtain $-a + \frac{b}{2} \pm \frac{b\sqrt{-3}}{2} \mp \sqrt{-3} - 2 = \frac{7}{2} \pm \frac{3\sqrt{-3}}{2}$ or $-a + \frac{b}{2} \pm \frac{bi\sqrt{3}}{2} \mp i\sqrt{3} - 2 = \frac{7}{2} \pm \frac{3i\sqrt{3}}{2}$ | **A1** | Expand brackets and obtain exact equation for either root |
| Solve simultaneous equations, or single equation, for $a$ or $b$ | **M1** | |
| Obtain $a = -3$ from exact working | **A1** | |
| Obtain $b = 5$ from exact working | **A1** | |
| **Total** | **5** | |

---
3 The polynomial $2 x ^ { 4 } + a x ^ { 3 } + b x - 1$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. When $\mathrm { p } ( x )$ is divided by $x ^ { 2 } - x + 1$ the remainder is $3 x + 2$.

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

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