Easy -1.2 This is a straightforward application of the Remainder Theorem requiring only substitution of x = -3 and solving a simple linear equation. It's a single-step problem testing basic recall of the theorem with minimal algebraic manipulation, making it easier than average.
Or M1 for long division by \((x+3)\) as far as obtaining \(x^2 - x\) and A1 for obtaining remainder as \(k - 24\) (but see below)
\(k = 30\)
A1
Equating coefficients method: M2 for \((x+3)(x^2 - x + 8)\) [+6] o.e. (from inspection or division) eg M2 for obtaining \(x^2 - x + 8\) as quotient in division
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt at $f(-3)$; $-27 + 18 - 15 + k = 6$ | M1, A1 | Or M1 for long division by $(x+3)$ as far as obtaining $x^2 - x$ and A1 for obtaining remainder as $k - 24$ (but see below) |
| $k = 30$ | A1 | Equating coefficients method: M2 for $(x+3)(x^2 - x + 8)$ [+6] o.e. (from inspection or division) eg M2 for obtaining $x^2 - x + 8$ as quotient in division |
7 When $x ^ { 3 } + 2 x ^ { 2 } + 5 x + k$ is divided by ( $x + 3$ ), the remainder is 6 . Find the value of $k$.
\hfill \mbox{\textit{OCR MEI C1 2010 Q7 [3]}}