A) 0
B) 1
C) 2
D) 4
Correct Answer: A
Solution :
Here, \[p\,\,(x)={{x}^{3}}+{{x}^{2}}-17x+15\] and, \[(x-3)\] will be a factor of p(x), if p(3) = 0 So,\[p\,\,(3)={{(3)}^{3}}+{{(3)}^{2}}-17\times 3+15\] \[=27+9-51+15=51-51=0\] \[\therefore \] \[x-3\] is a factor of p(x).You need to login to perform this action.
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