Question

find the sum of all possible natural numbers x (where x>2) such that (x-2) divide (3x^2-2x+10)​

Answer 1

Answer:

here's your answer.

Step-by-step explanation:

here's your example

STEP1:Equation at the end of step 1 (((5•(x2))+2)-((0-(4•(x2)))+7))+((0-3x2)-5)

STEP 2 :Equation at the end of step2: (((5•(x2))+2)-((0-22x2)+7))+(-3x2-5)

STEP 3 :Equation at the end of step3: ((5x2 + 2) - (7 - 4x2)) + (-3x2 - 5)

STEP4:

STEP5:Pulling out like terms

 5.1     Pull out like factors :

   6x2 - 10  =   2 • (3x2 - 5) 

Trying to factor as a Difference of Squares:

 5.2      Factoring:  3x2 - 

Answer 2

Given : (x-2) divide (3x²-2x+10)​

To Find : sum of all possible natural numbers x (where x>2)

Solution:

let say x = 2 + k   where k > 0  k ∈ Integers

x - 2 = k

3x²-2x+10

= 3(2 + k)² - 2(2 + k) + 10

= 3(k² + 4k + 4) - 4 - 4k  + 10

= 3k² + 8k  + 18

on diving by k

= 3k  + 8  + 18/k

18/k must be an integer

Hence  k can be 1 , 2 , 3 , 6 , 9 and 18

x = 2 + k

x  = 3 ,  4 , 5 , 8 , 11 and 20

Sum = ( 3 + 4 + 5 + 8 + 11 + 20)  = 51

sum of all possible natural numbers x (where x>2)  = 51

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