Use long division to find the quotient and the remainder. $\frac { 6 x ^ { 3 } - 3 x + 3 } { 2 x + 1 }$
A) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } \frac { 15 } { 4 }$
B) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x + \frac { 1 } { 4 } ; \text { remainder: } \frac { 15 } { 4 }$
C) $\text { quotient: } 15 x ^ { 2 } + \frac { 3 } { 2 } x - \frac { 1 } { 4 } ; \text { remainder: } \frac { 3 } { 4 }$
D) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } - \frac { 15 } { 4 }$
E) $\text { quotient: } 3 x ^ { 2 } - \frac { 3 } { 2 } x - \frac { 3 } { 4 } ; \text { remainder: } \frac { 7 } { 4 }$

Question

Quizplus · Accepted Answer

3x^2 - (3/2)x + (1/4)
     ___________________________
2x + 1 | 6x^3 + 0x^2 - 3x + 3   
           6x^3 + 3x^2  
           _____________
                   -3x^2 - 3x
                  -3x^2 - (3/2)x  
                  _____________
                            -3/2x + 3 
                            -3/2x - (3/4)
                            ___________
                                          15/4

Therefore, the quotient is 3x^2 - (3/2)x + (1/4) and the remainder is 15/4. Choice A matches the result obtained.

Use Long Division to Find the Quotient and the Remainder 6x3−3x+32x+1\frac { 6 x ^ { 3 } - 3 x + 3 } { 2 x + 1 }2x+16x3−3x+3​

Use Long Division to Find the Quotient and the Remainder $\frac { 6 x ^ { 3 } - 3 x + 3 } { 2 x + 1 }$