Which of the following is equivalent to $\cfrac{3{x^3} - 4x^2 + 3}{x-{4}}$ ?
A. $3x^2 + {8}x + {32} + \cfrac{131}{x-{4}}$
B. $3x^2 + {8}x + \cfrac{35}{x-{4}}$
C. $3x^2 - {16}x - {64} - \cfrac{259}{x-{4}}$
D. $3x^2 - {16}x - \cfrac{67}{x-{4}}$
E. $3x^2 - {16}x + {64} + \cfrac{67}{x-{4}}$
No Solution StepsA. $3x^2 + {8}x + {32} + \cfrac{131}{x-{4}}$
B. $3x^2 + {8}x + \cfrac{35}{x-{4}}$
C. $3x^2 - {16}x - {64} - \cfrac{259}{x-{4}}$
D. $3x^2 - {16}x - \cfrac{67}{x-{4}}$
E. $3x^2 - {16}x + {64} + \cfrac{67}{x-{4}}$
