Find a and b (given a binomial of x, a and b and a coefficient of a term)

Question

 In the expansion of (x+a)^3(x-b)^6, the coefficient of x^7 is -9 and there is no x^8 term. Find a and b.

Answer 1

Using the binomial theorem: 

(x + a)^3  = (1)x³ + (3)ax² + (3)a²x + (1)a³ 

= x³ + 3ax² + 3a²x + a³

 

(x - b)^6 = (1)x⁶  + (6)(-b)x⁵ + (15)(-b)² x⁴+ (20)(-b)³ x³ + (15)(-b)⁴ x² + (6)(-b)⁵ x + (1)(-b)⁶

= x⁶ - 6bx⁵ + 15b²x⁴ - 20b³x³ + 15b⁴x² - 6b⁵x + b⁶

 

Now rather than multiplying these two together, let's just find what the x^7 term and the x^8 term would be.

The x^8 term would come from multiplying:  

x³(-6bx⁵) + (3ax²)x⁶ 

= -6bx⁸ + 3ax⁸

The coefficient of this 0 (since there is no 8th term)

-6b + 3a = 0

3a = 6b 

a = 2b

 

Now the 7th term:

x³⁽15b²x⁴) +3ax²(-6b)x⁵ + 3a²x(x⁶) 

=15b²x⁷ -18abx⁵ + 6ax⁷

 

The coefficient of this -9

15b² -18ab + 6a = -9 

 

Now let's plug in a = 2b

15b² -18(2b)(b) + 3(2b)² = -9

15b² -36b² + 6b² = -9

-9b² = -9

b² = 1

b = 1

 

Since:  a = 2b

a = 2(1) = 2

 

So finally, we have: (x + 2)³(x - 1)⁶