how many factors are there in these 2 problems?

Question

1) (7^4)a^3b^2c

2) 5(x+y)(x-y)

Answer 1

2) 3
1) it depends on how the problem is written with the exponents. just count each number that is not in superscript as one, and the stuff in the parenthesis as one

Answer 2

You must take care to use parentheses and symbols of operation with precision. I am going to assume your problem is (7^4)*(a^3)*(b^2)*c.

All factors will be of the form (7^w)*(a^x)*(b^y)*(c^z) where w may be any of 0, 1, 2, 3, 4; x may be 0, 1, 2, 3; y may be 0, 1, 2; and z may be 0 or 1.

There are 5 ways to choose w, 4 ways to choose x, 3 ways to choose y, and 2 ways to choose z. In total there arre 5*4*3*2 = 120 ways to choose a factor.