Factorise fully 3a^3b + 12a^2b^2 + 9a^5b^3
The first step here is to find a common factor in each term. If we first look at the coefficients of each term we can see the first common factor is 3, so we can write that first outside of our brackets. 3.....(........). The next common factor in each term is the a, although each term has more than one a so we can pull out a higher power of a. The smallest power of a is in the second term, a2, so we can also take this outside the brackets with the 3 to give 3a2....(.......). The next common factor in each term is now the b left over. The smallest power for b is 1 in the first term so the most we can bring outside is a b by itself. Giving 3a2b(.......) So now to find what goes inside the bracket. The first term we need to get to is 3a3b, so we must look at what to multiply by to get here from the term outside the bracket, we already have a 3 and a b so we do not need to include this but we need to get from an a2 to an a3, so we must multiply by another a. Giving our first term inside the bracket - 3a2b(a ..... ). The next term we need to get to is 12a2b2. The first thing to do is times 3 by 4 to get 12, this will be the coefficient of our second term inside the bracket. Next we must get from b to b2, so we multiply by b, and a is the same so we can leave that. Giving our second term inside the bracket as 4b. Now we have 3a2b(a + 4b ....). So now to do the last term, we must get from 3 to 9, so multiply by 3, from a2 to a5 so multiply by a3 and from b to b3,, so multiply by b2. This give our final answer as: 3a2b(a + 4b + 3a3b2)
