What is the greatest common factor 10 and 20 have in common?

GCF=10

Step-by-step explanation:

1. To solve this problem you must apply the proccedure shown below:

- First, you must find the prime factorization of 10 and 20,as following:

10=2*5

20=2*2*5=2²*5

- Finally, you must write the prime factors common to 20 and 10 and mutpliy them.

- Therefore, you have that the greatest common factor (GCF) of 10 and 20 have in common is:

GCF=2*5=10

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."