The remainder when `4a^(3)-12a^(2)+14a-3` is divided by `2a-1` is

Options

A. `(1)/(2)`
B. `(7)/(2)`
C. `(3)/(2)`
D. None of these

Answer

To find the remainder when \(4a^3 - 12a^2 + 14a - 3\) is divided by \(2a - 1\), we can use polynomial long division. Here’s a step-by-step solution:

### Step 1: Set up the division
We want to divide the polynomial \(4a^3 - 12a^2 + 14a - 3\) by \(2a - 1\).

### Step 2: Divide the leading term
Divide the leading term of the dividend \(4a^3\) by the leading term of the divisor \(2a\):
\[ \frac{4a^3}{2a} = 2a^2 \]
This means we will multiply the entire divisor \(2a - 1\) by \(2a^2\).

### Step 3: Multiply and subtract
Now, multiply \(2a^2\) by \(2a - 1\):
\[ 2a^2(2a - 1) = 4a^3 - 2a^2 \]
Subtract this from the original polynomial:
\[ (4a^3 - 12a^2 + 14a - 3) - (4a^3 - 2a^2) = -10a^2 + 14a - 3 \]

### Step 4: Repeat the process
Now, repeat the process with the new polynomial \(-10a^2 + 14a - 3\). Divide the leading term \(-10a^2\) by \(2a\):
\[ \frac{-10a^2}{2a} = -5a \]
Multiply the divisor \(2a - 1\) by \(-5a\):
\[ -5a(2a - 1) = -10a^2 + 5a \]
Subtract:
\[ (-10a^2 + 14a - 3) - (-10a^2 + 5a) = 9a - 3 \]

### Step 5: Repeat again
Now, divide \(9a - 3\) by \(2a - 1\). Divide the leading term \(9a\) by \(2a\):
\[ \frac{9a}{2a} = \frac{9}{2} \]
Multiply the divisor \(2a - 1\) by \(\frac{9}{2}\):
\[ \frac{9}{2}(2a - 1) = 9a - \frac{9}{2} \]
Subtract:
\[ (9a - 3) - (9a - \frac{9}{2}) = -3 + \frac{9}{2} \]
Convert \(-3\) to a fraction:
\[ -3 = -\frac{6}{2} \]
So,
\[ -\frac{6}{2} + \frac{9}{2} = \frac{3}{2} \]

### Conclusion
The remainder when \(4a^3 - 12a^2 + 14a - 3\) is divided by \(2a - 1\) is:
\[ \frac{3}{2} \]