Yes, 0 has infinite number of factors.
0
The highest common highest common factor of 3 & 0 is 0 as 0 is has no value.
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
8. 300 ends in 0 or 5, so 5 is a factor 300 is divisible by 4 and 5, so 20 is a factor 300 is even and 3 + 0 + 0 = 3, so 6 is a factor Leaving 8 to be the non factor Testing if 8 is a factor: 4 x 3 + 2 x 0 + 0 = 12; 4 x 0 + 2 x 1 + 2 = 4 which is not 8, so 300 is not divisible by 8.
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.
0
The highest common highest common factor of 3 & 0 is 0 as 0 is has no value.
3x3 -3x2 - 48x - 60 = 0 simply factor out a 3... 3(x3 - x2 - 16x - 20) = 0
8. 300 ends in 0 or 5, so 5 is a factor 300 is divisible by 4 and 5, so 20 is a factor 300 is even and 3 + 0 + 0 = 3, so 6 is a factor Leaving 8 to be the non factor Testing if 8 is a factor: 4 x 3 + 2 x 0 + 0 = 12; 4 x 0 + 2 x 1 + 2 = 4 which is not 8, so 300 is not divisible by 8.
No... because 10 is not a factor of 3. 7+0+0+2= 10
If a polynomial has factors x-6 and x-3 it will equal 0 if either factor equals 0 since the other factor then would be multiplied by 0. ie. 0 * (x-6)=0 and 0 * (x-3)=0. so x=3 or 6
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.
Yes, 3 is a factor of 45 because 45 is divisible by 3. 45 = 3 x 15 + 0.
x squared -2x-3 equals 0 is the same as (x + 1)(x - 3) = 0
2x^3(x - 3)(x + 3)
-a^2- 9a - 18 = 0 -(a^2 + 9a + 18) = 0 -(a + 6)(a + 3) = 0
3m^2 - 9m = 0 factor out 3m 3m(m - 3) = 0 m = 0 -------- or m = 3 --------