Use long division to find the quotient.

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
​

GPT-4o mini · Answer

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

Divide the leading term of the dividend $6 x^{4}$ by the leading term of the divisor $3 x$ : $\frac{6 x^{4}}{3 x} = 2 x^{3} .$
Multiply $2 x^{3}$ by the entire divisor $3 x + 2$ : $2 x^{3} \cdot (3 x + 2) = 6 x^{4} + 4 x^{3} .$
Subtract this result from the original polynomial: $(6 x^{4} - 11 x^{3} - 7 x^{2} + 8 x + 3) - (6 x^{4} + 4 x^{3}) = - 15 x^{3} - 7 x^{2} + 8 x + 3.$
Bring down the next term and repeat the process. Divide $- 15 x^{3}$ by $3 x$ : $\frac{- 15 x^{3}}{3 x} = - 5 x^{2} .$
Multiply $- 5 x^{2}$ by $3 x + 2$ : $- 5 x^{2} \cdot (3 x + 2) = - 15 x^{3} - 10 x^{2} .$
Subtract this from the current polynomial: $(- 15 x^{3} - 7 x^{2} + 8 x + 3) - (- 15 x^{3} - 10 x^{2}) = 3 x^{2} + 8 x + 3.$
Now divide $3 x^{2}$ by $3 x$ : $\frac{3 x^{2}}{3 x} = x .$
Multiply $x$ by $3 x + 2$ : $x \cdot (3 x + 2) = 3 x^{2} + 2 x .$
Subtract: $(3 x^{2} + 8 x + 3) - (3 x^{2} + 2 x) = 6 x + 3.$
Now divide $6 x$ by $3 x$ : $\frac{6 x}{3 x} = 2.$
Multiply: $2 \cdot (3 x + 2) = 6 x + 4.$
Subtract: $(6 x + 3) - (6 x + 4) = - 1.$

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

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