From a test-taking point of view, you solve it like this.
- you recognize that the fourth term of (a+b)^5 will be some constant times a^2b^3
- you recognize that b is -2, so b^3 will be negative (it is an odd power of a negative number)
- you recognize that the term must contain x^2, because a^2 = (3x)^2 will contain x^2
The only term among the answers that has -x^2 as a factor is the first one, answer (1).
If you actually want to do the math, you recognize that you need to find the multiplier for the term a^2b^3 (as above). That multiplier will be the number of ways 5 objects can be chosen 2 at a time. That is 5!/(2!*(5-2)!) = 5*4/(2*1) = 10
The term is 10*(3x)^2*(-2)^3 = 10*9*(-8)*x^2 = -720x^2
- you recognize that the fourth term of (a+b)^5 will be some constant times a^2b^3
- you recognize that b is -2, so b^3 will be negative (it is an odd power of a negative number)
- you recognize that the term must contain x^2, because a^2 = (3x)^2 will contain x^2
The only term among the answers that has -x^2 as a factor is the first one, answer (1).
If you actually want to do the math, you recognize that you need to find the multiplier for the term a^2b^3 (as above). That multiplier will be the number of ways 5 objects can be chosen 2 at a time. That is 5!/(2!*(5-2)!) = 5*4/(2*1) = 10
The term is 10*(3x)^2*(-2)^3 = 10*9*(-8)*x^2 = -720x^2