Which of the following is equivalent to $\cfrac{3{x^3} - 3x^2 + 1}{x-{4}}$ ?
A. $3x^2 + {9}x + {36} + \cfrac{145}{x-{4}}$
B. $3x^2 + {9}x + \cfrac{37}{x-{4}}$
C. $3x^2 - {15}x - {60} - \cfrac{241}{x-{4}}$
D. $3x^2 - {15}x - \cfrac{61}{x-{4}}$
E. $3x^2 - {15}x + {60} + \cfrac{61}{x-{4}}$
No Solution StepsA. $3x^2 + {9}x + {36} + \cfrac{145}{x-{4}}$
B. $3x^2 + {9}x + \cfrac{37}{x-{4}}$
C. $3x^2 - {15}x - {60} - \cfrac{241}{x-{4}}$
D. $3x^2 - {15}x - \cfrac{61}{x-{4}}$
E. $3x^2 - {15}x + {60} + \cfrac{61}{x-{4}}$
