If 3x^3+ax^2+3x+2 leaves a remainder of 4 on division by (3x-1), find a.

To find the value of "a" in the given polynomial, we can use polynomial division and the remainder theorem.

Step 1: Set up the polynomial division.
Set up the division by writing the dividend, divisor, and quotient:

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3x - 1 | 3x^3 + ax^2 + 3x + 2

Step 2: Perform the polynomial division.
Divide the first term of the dividend (3x^3) by the first term of the divisor (3x). This gives us x^2, which is placed above the line:

x^2
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3x - 1 | 3x^3 + ax^2 + 3x + 2

Multiply the divisor (3x - 1) by x^2, which gives us 3x^3 - x^2. Subtract this from the original dividend:

x^2
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3x - 1 | 3x^3 + ax^2 + 3x + 2
- (3x^3 - x^2)

Simplifying, we obtain:

x^2
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3x - 1 | ax^2 + 4x + 2

Step 3: Apply the Remainder Theorem.
Since the polynomial 3x^3 + ax^2 + 3x + 2 leaves a remainder of 4, the remainder in the polynomial division is also 4. According to the Remainder Theorem, the remainder is equal to the value of the polynomial when the divisor is substituted into it.

In this case, the divisor is (3x - 1). Therefore, we have:

4 = ax^2 + 4x + 2

Step 4: Solve the equation for "a".
To find the value of "a", we need to solve the equation 4 = ax^2 + 4x + 2 for x = 1/3 (based on the divisor, 3x-1):

4 = a(1/3)^2 + 4(1/3) + 2
4 = a/9 + 4/3 + 2

Multiply through by 9 to eliminate fractions:
36 = a + 12 + 18

Simplify:
a = 36 - 12 - 18
a = 6

Therefore, the value of "a" in the given polynomial is 6.