| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.8 This is a straightforward application of the Factor and Remainder Theorems requiring substitution of x-values into the polynomial to create two simultaneous equations. Part (ii) involves routine factorisation once constants are known. The question is easier than average as it follows a standard template with clear conditions and requires only algebraic manipulation without novel insight. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x = -1\), equate to zero and simplify at least as far as \(-8 + a - b - 1 = 0\) | B1 | |
| Substitute \(x = -\frac{1}{2}\) and equate the result to 1 | M1 | |
| Obtain a correct equation in any form, e.g. \(-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1\) | A1 | |
| Solve for \(a\) or for \(b\) | M1 | |
| Obtain \(a = 6\) and \(b = -3\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Commence division by \((x+1)\) reaching a partial quotient \(8x^2 + kx\) | M1 | The M1 is earned if inspection reaches an unknown factor \(8x^2 + Bx + C\) and an equation in \(B\) and/or \(C\), or an unknown factor \(Ax^2 + Bx - 1\) and an equation in \(A\) and/or \(B\) |
| Obtain quadratic factor \(8x^2 - 2x - 1\) | A1 | If linear factors are found by the factor theorem, give B1B1 for \((2x-1)\) and \((4x+1)\), and B1 for the complete factorisation |
| Obtain factorisation \((x+1)(4x+1)(2x-1)\) | A1 | |
| Total: 3 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = -1$, equate to zero and simplify at least as far as $-8 + a - b - 1 = 0$ | B1 | |
| Substitute $x = -\frac{1}{2}$ and equate the result to 1 | M1 | |
| Obtain a correct equation in any form, e.g. $-1 + \frac{1}{4}a - \frac{1}{2}b - 1 = 1$ | A1 | |
| Solve for $a$ or for $b$ | M1 | |
| Obtain $a = 6$ and $b = -3$ | A1 | |
| **Total: 5** | | |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Commence division by $(x+1)$ reaching a partial quotient $8x^2 + kx$ | M1 | The M1 is earned if inspection reaches an unknown factor $8x^2 + Bx + C$ and an equation in $B$ and/or $C$, or an unknown factor $Ax^2 + Bx - 1$ and an equation in $A$ and/or $B$ |
| Obtain quadratic factor $8x^2 - 2x - 1$ | A1 | If linear factors are found by the factor theorem, give B1B1 for $(2x-1)$ and $(4x+1)$, and B1 for the complete factorisation |
| Obtain factorisation $(x+1)(4x+1)(2x-1)$ | A1 | |
| **Total: 3** | | |6 The polynomial $8 x ^ { 3 } + a x ^ { 2 } + b x - 1$, where $a$ and $b$ are constants, is denoted by $\mathrm { p } ( x )$. It is given that $( x + 1 )$ is a factor of $\mathrm { p } ( x )$ and that when $\mathrm { p } ( x )$ is divided by ( $2 x + 1$ ) the remainder is 1 .\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, factorise $\mathrm { p } ( x )$ completely.\\
\hfill \mbox{\textit{CAIE P3 Q6 [8]}}