Use long division to divide the given polynomial.

To perform long division on the polynomial \( (8x^3 + 6x^2 - 13 + 3) \) by \( (4x - 3) \), we first rewrite the polynomial:

\[ P(x) = 8x^3 + 6x^2 - 10 \]

Now, we'll divide \( P(x) \) by \( 4x - 3 \).

Step 1: Divide the leading terms

Divide the leading term of the dividend \( 8x^3 \) by the leading term of the divisor \( 4x \):

\[ \frac{8x^3}{4x} = 2x^2 \]

Step 2: Multiply and subtract

Next, multiply \( 2x^2 \) by the entire divisor:

\[ 2x^2(4x - 3) = 8x^3 - 6x^2 \]

Now, subtract this from the polynomial:

\[ (8x^3 + 6x^2 - 10) - (8x^3 - 6x^2) = 6x^2 + 6x^2 - 10 = 12x^2 - 10 \]

Step 3: Repeat the process

Now, take \( 12x^2 \) and divide it by \( 4x \):

\[ \frac{12x^2}{4x} = 3x \]

Multiply \( 3x \) by the divisor:

\[ 3x(4x - 3) = 12x^2 - 9x \]

Now, subtract:

\[ (12x^2 - 10) - (12x^2 - 9x) = -10 + 9x = 9x - 10 \]

Step 4: Repeat again

Now divide \( 9x \) by \( 4x \):

\[ \frac{9x}{4x} = \frac{9}{4} \]

Multiply by the divisor:

\[ \frac{9}{4}(4x - 3) = 9x - \frac{27}{4} \]

Subtract:

\[ (9x - 10) - (9x - \frac{27}{4}) = -10 + \frac{27}{4} = -\frac{40}{4} + \frac{27}{4} = -\frac{13}{4} \]

Final Result

So, we have:

\[ P(x) = (4x - 3)(2x^2 + 3x + \frac{9}{4}) - \frac{13}{4} \]

Thus, the quotient is:

\[ 2x^2 + 3x + \frac{9}{4} \]

and the remainder is \(-\frac{13}{4}\).

Now, looking at the options provided, we can identify that none of the options match our result directly.

Please cross-check the options because it seems that either I misunderstood how the options are given or the options might be incorrect. Nonetheless, from our calculations, the quotient is \( 2x^2 + 3x + \frac{9}{4} \) with a remainder of \(-\frac{13}{4}\).