This problem was very hard indeed. However I got the hang of it when I was revising quadratic equations and doing the puzzle: 'power quady.' I developed my confidence in factorising the quadratics and I used that technique to solve the puzzle.
Problem 1 : (n2-5n+5)(n^2-11n +30)=1
In quadratic equations, there must be an equality with 0. In order to do this we must minus 1 from each side. Now we must solve the set of brackets: (n2-5n+4). I shall now factorise the quadratics:
(n-4)(n-1). This means that two solutions for the first problem are: 4 and 1.
The second set of brackets are: (n2-11n+30). I will also factorise this set of brackets. This makes: (n-5)(n-6). This means that another two solutions for the problem are: 5 and 6.
However, we need 6 solutions! I figured that the least two answers could be found if you changed the answer "1" to "-1". So this would make the calculation: (n2-5n+5)(n^2-11n+30)=-1. To make both sides 0 we must add one to each side which makes the first set of brackets : (n2-5n+6). However, the second set of brackets doesn’t change. As usual, I factorised the set of brackets making: (n-2)(n-3). This means that the last pair of solutions are : 2 and 3
The second problem was : (n2-7n+11)(n^2-13n+42)=1. I also used the exact same method for this problem as the first one. I factorised the quadratics as usual and I split up the set of brackets.
So, I first subtract 1 from each side to get 0. So the formula becomes: (n2-7n+10)(n^2-13n+42)=0.
Now , to factorise the first set of brackets, I get : (n-2)(n-5). So a pair of solutions to the problem are : 2 and 5.
Now if we take the second set of brackets, (n2-13n+42), and factorise that we get: (n-6)(n-7). So another set of solutions are: 6 and 7.
As I said earlier on, we need 6 solutions, however we have 4 solutions. The last pair of solutions can come from the calculation: (n2—7n+11)(n^2-13n+42)=-1. So we need to add one to each side so we get: (n2-7n+12)(n^2-13n+42)=0. If we factorise the first set of brackets, we get: (n-2)(n-5). So the last pairs of solutions are:2 and 5.