Question

Use long division to divide the given polynomial.

(8x^3+6x^2-13+3)/(4x-3)






A
2x2−3x+12x^2-3x+12x
2
−3x+1

B
2x2−3x−12x^2-3x-12x
2
−3x−1

C
2x2+3x+12x^2+3x+12x
2
+3x+1

D
2x2+3x −12x^2+3x\ -12x
2
+3x −1

Answer 1

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 \(-\frac{13}{4}\).