Question

Use long division to find the quotient.
(6x^4-11x^3-7x^2+8x+3)/(3x+2)






A
2x3−5x2+x+12x^3-5x^2+x+12x
3
−5x
2
+x+1

B
2x3−5x2+x+2−13x+22x^3-5x^2+x+2-\frac{1}{3x+2}2x
3
−5x
2
+x+2−
3x+2
1
​


C
2x3−5x2+x+22x^3-5x^2+x+22x
3
−5x
2
+x+2

D
2x3−5x2+x+2+13x+22x^3-5x^2+x+2+\frac{1}{3x+2}2x
3
−5x
2
+x+2+
3x+2
1
​

Answer 1

To perform polynomial long division for the expression \((6x^4 - 11x^3 - 7x^2 + 8x + 3) \div (3x + 2)\), follow these steps:

1. **Divide** the leading term of the dividend \(6x^4\) by the leading term of the divisor \(3x\):
\[
\frac{6x^4}{3x} = 2x^3.
\]

2. **Multiply** \(2x^3\) by the entire divisor \(3x + 2\):
\[
2x^3 \cdot (3x + 2) = 6x^4 + 4x^3.
\]

3. **Subtract** this result from the original polynomial:
\[
(6x^4 - 11x^3 - 7x^2 + 8x + 3) - (6x^4 + 4x^3) = -15x^3 - 7x^2 + 8x + 3.
\]

4. Bring down the next term and **repeat** the process. Divide \(-15x^3\) by \(3x\):
\[
\frac{-15x^3}{3x} = -5x^2.
\]

5. **Multiply** \(-5x^2\) by \(3x + 2\):
\[
-5x^2 \cdot (3x + 2) = -15x^3 - 10x^2.
\]

6. **Subtract** this from the current polynomial:
\[
(-15x^3 - 7x^2 + 8x + 3) - (-15x^3 - 10x^2) = 3x^2 + 8x + 3.
\]

7. Now divide \(3x^2\) by \(3x\):
\[
\frac{3x^2}{3x} = x.
\]

8. **Multiply** \(x\) by \(3x + 2\):
\[
x \cdot (3x + 2) = 3x^2 + 2x.
\]

9. **Subtract**:
\[
(3x^2 + 8x + 3) - (3x^2 + 2x) = 6x + 3.
\]

10. Now divide \(6x\) by \(3x\):
\[
\frac{6x}{3x} = 2.
\]

11. **Multiply**:
\[
2 \cdot (3x + 2) = 6x + 4.
\]

12. **Subtract**:
\[
(6x + 3) - (6x + 4) = -1.
\]

So the final result of the division is:
\[
2x^3 - 5x^2 + x + 2 - \frac{1}{3x + 2}.
\]

In this context, the correct answer using the choices provided would be:
\[
\text{Option B: } 2x^3 - 5x^2 + x + 2 - \frac{1}{3x + 2}.
\]