Factorise the following expression in prime factors

Question

2a⅖-7a⅕/5+3b

Answer 1

2a^⅖-(7a^⅕/5)+3b or (2a^⅖-7a^⅕)/(5+3b)?

The expression needs to be evaluated (reduced to a single quantity, which must be an integer so as to have prime factors) in order to factorise it and we have two unknowns, a and b, which prevent us from doing so. The ambiguity in the presentation of the expression creates a further problem.

If we take the second interpretation as the intended one, then it factorises:

a^⅕(2a^⅕-7)/(5+3b). If 5+3b=1, then b=-4/3 so the expression becomes a^⅕(2a^⅕-7) and the prime factors are a^⅕ and 2a^⅕-7.

If 2a^⅕<7, then 2a^⅕-7 will be negative, so, for this factor to be positive 2a^⅕>7, 32a>16807, a>16807/32, a>525.22. a^⅕ must also be an integer, so a^⅕>7/2, making a^⅕∈ℕ, a^⅕≥4, a≥1024. If a=1024, the factors in this example would be 4 and 1, and the prime factors 2 and 1. But if 1 is not considered a prime, then we move to a=3125, making a^⅕=5. This creates the factors 5 and 3 which are both primes. The expression (2a^⅖-7a^⅕)/(5+3b) evaluates to 15, when a=3125 and b=-4/3.

If we now take a=7776, a^⅕=6 and the factors become 6 and 5, and the prime factors: 2, 3 and 5, and the expression evaluates to 30, while b remains at -4/3.

If b=-1 the denominator becomes 2. If a remains at 7776, then we have 6×5/2=15 with prime factors 3 and 5. If b=-⅔ and a remains at 7776, then the denominator is 3 so the expression evaluates to 6×5/3=10 with prime factors 2 and 5.

If b=-2, the denominator becomes -1. The expression can only be positive when a<525.22, a<7/2, so a can only be 2⁵=32 or 3⁵=243. The expression evaluates to -6/-1=6=2×3 or -3/-1=3 (a prime).

These examples generate different prime factors depending on the values of a and b.

To consider the interpretation 2a^⅖-(7a^⅕/5)+3b, we know that the expression must evaluate to an integer because a fraction can't have prime factors. If b is an integer then a^⅕ must be divisible by 5, making a a multiple of 5. It must also be a perfect 5th power. a=3125 fits this requirement and the expression evaluates to 50-7+3b=43+3b. Examples: b=1, expression=46=2×23; b=2, expression=49=7²; b=3, expression=52=2²×13; b=-1, expression=40=2³×5, b=-3, expression=34=2×17. If b=⅓, expression=44=2²×11, etc. There are many other combinations, for example, when b is a multiple of ⅕ so that -(7a^⅕/5)+3b reduces to an integer for particular values of a^⅕ which are not divisible by 5.

I hope this analysis is useful.