It is given that the expansion of (ax + b)3 is 8x3 − px2 + 18x − 3√3, where a, b and p are real constants.
What is the value of p?
- A−12√3
- B−6√3
- C−4√3
- D−√3
- E√3
- F4√3
- G6√3
- H12√3
Show the answer and worked solution
answer · H
- A−12√3
- B−6√3
- C−4√3
- D−√3
- E√3
- F4√3
- G6√3
- H12√3
Expand: (ax+b)3 = a3x3 + 3a2b x2 + 3ab2 x + b3. The two end terms pin down a and b on their own: a3 = 8 gives a = 2, and b3 = −3√3 gives b = −√3, since (−√3)3 = −3√3. Check against the x term: 3ab2 = 3⋅ 2⋅ 3 = 18, as required. The x2 coefficient is 3a2b = 3⋅ 4⋅(−√3) = −12√3, and the expansion writes that coefficient as −p, so p = 12√3. The sign trap is dropping the minus in −px2 and answering −12√3.