Easy -1.2 This is a straightforward application of the binomial theorem with n=4, requiring only direct substitution into the formula and basic algebraic simplification. It's a routine C1 exercise with no problem-solving element, making it easier than average but not trivial since students must handle negative terms and coefficients carefully.
M3 for 4 terms correct, or all coefficients correct except for sign errors, or correct answer seen then further 'simplified', or all terms correct eg seen in table but not combined; condone eg \(+(-96x)\) or \(+-96x\) instead of \(-96x\)
M2
M2 for 3 terms correct or for correct expansion seen without correct evaluation of coefficients [if brackets missing in elements such as \((3x)^2\) there must be evidence from calculation that \(9x^2\) has been used]; binomial coefficients such as \(^4C_2\) or \(\binom{4}{2}\) are not sufficient – must show understanding of these symbols by at least partial evaluation
M1
M1 for \(1\ 4\ 6\ 4\ 1\) row of Pascal's triangle seen
## Question 7:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $81x^4 - 216x^3 + 216x^2 - 96x + 16$ | M3 | M3 for 4 terms correct, or all coefficients correct except for sign errors, or correct answer seen then further 'simplified', or all terms correct eg seen in table but not combined; condone eg $+(-96x)$ or $+-96x$ instead of $-96x$ |
| | M2 | M2 for 3 terms correct or for correct expansion seen without correct evaluation of coefficients [if brackets missing in elements such as $(3x)^2$ there must be evidence from calculation that $9x^2$ has been used]; binomial coefficients such as $^4C_2$ or $\binom{4}{2}$ are not sufficient – must show understanding of these symbols by at least partial evaluation |
| | M1 | M1 for $1\ 4\ 6\ 4\ 1$ row of Pascal's triangle seen |