Prove that 2x3 + 3x² + x is always divisible by 6 if x is an integer. PLEASE I NEED THIS ANSWERED ASAP.
USE PROOF BY EXHAUSTION.

 see proof below

Step-by-step explanation:

Proof by exhaustion means to input several numbers that satisfy the claim and if they all prove to be true then you have proven the claim.

Claim: (2x³ + 3x² + x)/(6) ∈ Z (integer)

Let's choose x = {1, 2, 3, 4)

Case 1: x = 1

2(1)³ + 3(1)² + (1) = 12

12/6 = 2

2 ∈ Z

Case 2: x = 2

2(2)³ + 3(2)² + (2) = 30

30/6 = 5

5 ∈ Z

Case 3: x = 3

2(3)³ + 3(3)² + (3) = 84

84/6 = 14

14 ∈ Z

Case 4: x = 4

2(4)³ + 3(4)² + (4) = 180

180/6 = 30

30 ∈ Z

Since each case was shown to be true we have proven the claim is true by exhaustion.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.