A) \[{{p}^{3}}-6\]
B) \[{{p}^{3}}-6{{p}^{2}}+6\]
C) \[{{p}^{3}}+6{{p}^{2}}+11p+6\]
D) \[{{p}^{3}}-6{{p}^{3}}-11p-6\]
Correct Answer: C
Solution :
Let\[16=8\times 2+0\]. If (-1), (-2) and (-3) are the zeros of\[\text{256}=\text{8}\times \text{32}+0\], then f (-1), f (-2) and f (-3) are all zero.
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