Q.9 The coefficient of x4 in the polynomial (x − 1)3(x − 2)3 is equal to ______.
- 33
- −3
- 30
- 21
The coefficient of x4 in the polynomial (x−1)3(x−2)3 is 33.
Step-by-Step Solution
Expand each binomial using the binomial theorem. For (x−1)3, the terms are x3−3x2+3x−1. For (x−2)3, the terms are x3−6x2+12x−8.
The full polynomial is degree 6, so x4 arises from products of terms whose degrees sum to 4: x3⋅x, x2⋅x2, and x⋅x3.
- From first x3 and second x: coefficient 1⋅12=12
- From first x2 and second x2: (−3)⋅(−6)=18
- From first x and second x3: 3⋅1=3
Total: 12+18+3=33.
Option Analysis
- 33: Correct, as calculated above.
- -3: Incorrect; possibly confuses sign from single term like (−1)⋅3, ignores other contributions.
- 30: Incorrect; close but misses the +3 from x⋅x3, yielding 12+18=30.
- 21: Incorrect; might sum only two terms (18+3) or miscalculate binomial coefficients.
Binomial Expansion Method
Apply binomial theorem: (x+a)n=∑k=0n(nk)xn−kak.
(x−1)3=∑r=03(3r)x3−r(−1)r=x3−3x2+3x−1
(x−2)3=∑s=03(3s)x3−s(−2)s=x3−6x2+12x−8
General term in product: (3r)(−1)r⋅(3s)(−2)sx(3−r)+(3−s). Set (6−r−s)=4, so r+s=2.
Contributing Pairs (r,s)
| r | s | Coefficient Contribution |
|---|---|---|
| 0 | 2 | (30)(−1)0⋅(32)(−2)2=1⋅12=12 |
| 1 | 1 | (31)(−1)1⋅(31)(−2)1=(−3)⋅(−6)=18 |
| 2 | 0 | (32)(−1)2⋅(30)(−2)0=3⋅1=3 |
Sum: 33. This matches exam options and verifies via full multiplication.
Exam Tips
- Practice identifying degree-matching pairs efficiently.
- Recognize [(x−1)(x−2)]3=(x2−3x+2)3, but direct expansion suits low powers.
