0
n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
6
a^17 x b^14 where ^ means "raised to the power of.."
It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form. So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers). Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1. Example: n = 72 = 2*2*2*3*3 = (2^3)*(3^2) so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2. When (a, b) = (0, 0) the factor is 1. (a, b) = (1, 0) the factor is 2. (a, b) = (2, 0) the factor is 4. (a, b) = (3, 0) the factor is 8. (a, b) = (0, 1) the factor is 3. (a, b) = (1, 1) the factor is 6. (a, b) = (2, 1) the factor is 12. (a, b) = (3, 1) the factor is 24. (a, b) = (0, 2) the factor is 9. (a, b) = (1, 2) the factor is 18. (a, b) = (2, 2) the factor is 36. (a, b) = (3, 2) the factor is 72.
6(b + 2)(5b - 2)
n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)
The GCF is 1.
6
a^17 x b^14 where ^ means "raised to the power of.."
It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form. So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers). Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1. Example: n = 72 = 2*2*2*3*3 = (2^3)*(3^2) so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2. When (a, b) = (0, 0) the factor is 1. (a, b) = (1, 0) the factor is 2. (a, b) = (2, 0) the factor is 4. (a, b) = (3, 0) the factor is 8. (a, b) = (0, 1) the factor is 3. (a, b) = (1, 1) the factor is 6. (a, b) = (2, 1) the factor is 12. (a, b) = (3, 1) the factor is 24. (a, b) = (0, 2) the factor is 9. (a, b) = (1, 2) the factor is 18. (a, b) = (2, 2) the factor is 36. (a, b) = (3, 2) the factor is 72.
That would depend on the value of B. The answer will be 1, 2, 3 or 6.
6(b + 2)(5b - 2)
3 times a times a times b
(a + 3)( b + 2)
(a - 2)(b + 3)
(b + 2)(b - 8)
so let's group and then factor ab+2a+3b+6= (ab+2a)+(3b+6)= a(b+2)+3(b+2)= (a+3)(b+2) and that is our final answer! Doctor Chuck