| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants with divisibility condition |
| Difficulty | Standard +0.8 This question requires polynomial division or equating coefficients to find the constant 'a', then complete factorization including finding complex roots. While the factor theorem is standard A-level content, working with a quartic polynomial where the given factor is quadratic (not linear) adds complexity, and part (ii) requires understanding of complex roots and discriminant analysis. This is moderately above average difficulty for A-level. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.02g Conjugate pairs: real coefficient polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Either: Attempt division by \(x^2 - x + 3\) reaching a partial quotient \(x^2 + x\) | B1 | |
| Complete division and equate constant remainder to zero | M1 | |
| Obtain answer \(a = -6\) | A1 | |
| Or: Commence inspection and reach unknown factor of \(x^2 + x + c\) | B1 | |
| Obtain \(3c = a\) and an equation in \(c\) | M1 | |
| Obtain answer \(a = -6\) | A1 | |
| Then: State or obtain factor \(x^2 + x - 2\) | B1 | |
| State or obtain factors \(x + 2\) and \(x - 1\) | B1 + B1 | 6 |
| (ii) State that \(x^2 + x - 2 = 0\) has two (real) roots | B1 | |
| Show that \(x^2 - x + 3 = 0\) has no (real) roots | B1 | 2 |
**(i)** **Either:** Attempt division by $x^2 - x + 3$ reaching a partial quotient $x^2 + x$ | B1 |
Complete division and equate constant remainder to zero | M1 |
Obtain answer $a = -6$ | A1 |
**Or:** Commence inspection and reach unknown factor of $x^2 + x + c$ | B1 |
Obtain $3c = a$ and an equation in $c$ | M1 |
Obtain answer $a = -6$ | A1 |
**Then:** State or obtain factor $x^2 + x - 2$ | B1 |
State or obtain factors $x + 2$ and $x - 1$ | B1 + B1 | 6
**(ii)** State that $x^2 + x - 2 = 0$ has two (real) roots | B1 |
Show that $x^2 - x + 3 = 0$ has no (real) roots | B1 | 25 The polynomial $x ^ { 4 } + 5 x + a$ is denoted by $\mathrm { p } ( x )$. It is given that $x ^ { 2 } - x + 3$ is a factor of $\mathrm { p } ( x )$.\\
(i) Find the value of $a$ and factorise $\mathrm { p } ( x )$ completely.\\
(ii) Hence state the number of real roots of the equation $\mathrm { p } ( x ) = 0$, justifying your answer.
\hfill \mbox{\textit{CAIE P3 2005 Q5 [8]}}