Standard +0.3 This is a straightforward binomial expansion question requiring students to expand using the binomial theorem, multiply by a linear factor, and equate coefficients. While it involves multiple steps and algebraic manipulation, it follows a standard template with no novel insight required—slightly easier than average due to the routine nature of the techniques involved.
3. (a) Find the first 4 terms, in ascending powers of \(x\), of the binomial expansion of
$$\left( 2 - \frac { k x } { 4 } \right) ^ { 8 }$$
where \(k\) is a non-zero constant. Give each term in simplest form.
$$f ( x ) = ( 5 - 3 x ) \left( 2 - \frac { k x } { 4 } \right) ^ { 8 }$$
In the expansion of \(\mathrm { f } ( x )\), the constant term is 3 times the coefficient of \(x\).
(b) Find the value of \(k\).
T
Attempts binomial expansion up to at least third (\(x^2\)) term with acceptable structure; correct binomial coefficient combined with correct power of \(x\); allow if powers of 2 incorrect and brackets missing; M0 for descending powers
\(256 - 256kx\)
B1
For \(256 - 256kx\); may be listed, must be simplified; allow \(256(1 - kx + \ldots)\) if \(2^8\) taken out first
\(112k^2x^2\) or \(-28k^3x^3\) (unsimplified)
A1
Correct third or fourth term; need not be simplified but binomial coefficients must be numerical
\(112k^2x^2\) and \(-28k^3x^3\) (simplified)
A1
Correct simplified third and fourth terms; must have \(k^n x^n\) terms
Sets \(5 \times\) their constant term from (a) \(= 3 \times\) their coefficient of \(x\) from \(f(x)\) and solves for \(k\); equation in \(k\) only; allow recovery if \(x\) initially included but later crossed out
\(k = -\dfrac{14}{15}\)
A1
Correct value; must be exact; allow \(-0.9\dot{3}\) but not a terminating decimal
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(2 - \frac{kx}{4}\right)^8 = 2^8 + \binom{8}{1}2^7\left(-\frac{kx}{4}\right) + \binom{8}{2}2^6\left(-\frac{kx}{4}\right)^2 + \binom{8}{3}2^5\left(-\frac{kx}{4}\right)^3 + \ldots$ | M1 | Attempts binomial expansion up to at least third ($x^2$) term with acceptable structure; correct binomial coefficient combined with correct power of $x$; allow if powers of 2 incorrect and brackets missing; M0 for descending powers |
| $256 - 256kx$ | B1 | For $256 - 256kx$; may be listed, must be simplified; allow $256(1 - kx + \ldots)$ if $2^8$ taken out first |
| $112k^2x^2$ **or** $-28k^3x^3$ (unsimplified) | A1 | Correct third or fourth term; need not be simplified but binomial coefficients must be numerical |
| $112k^2x^2$ **and** $-28k^3x^3$ (simplified) | A1 | Correct simplified third and fourth terms; must have $k^n x^n$ terms |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (5-3x)\left(2-\frac{kx}{4}\right)^8 = (5-3x)(256 - 256kx + 112k^2x^2 - 28k^3x^3 + \ldots)$; coefficient of $x$ is $5 \times {-256k} - 3 \times 256$ | M1 | Correct strategy for coefficient of $x$: e.g. $5 \times \text{their } {-256k} - 3 \times \text{their } 256$; terms must have been combined |
| $5 \times 256 = 3(-1280k - 768) \Rightarrow k = \ldots$ | M1 | Sets $5 \times$ their constant term from (a) $= 3 \times$ their coefficient of $x$ from $f(x)$ and solves for $k$; equation in $k$ only; allow recovery if $x$ initially included but later crossed out |
| $k = -\dfrac{14}{15}$ | A1 | Correct value; must be exact; allow $-0.9\dot{3}$ but not a terminating decimal |
3. (a) Find the first 4 terms, in ascending powers of $x$, of the binomial expansion of
$$\left( 2 - \frac { k x } { 4 } \right) ^ { 8 }$$
where $k$ is a non-zero constant. Give each term in simplest form.
$$f ( x ) = ( 5 - 3 x ) \left( 2 - \frac { k x } { 4 } \right) ^ { 8 }$$
In the expansion of $\mathrm { f } ( x )$, the constant term is 3 times the coefficient of $x$.\\
(b) Find the value of $k$.\\
T\\
\hfill \mbox{\textit{Edexcel P2 2022 Q3 [7]}}