A) -5, 5
B) 7, 18
C) 7,-18
D) -5,-18
Correct Answer: C
Solution :
Let, \[f(x)={{x}^{3}}+10{{x}^{2}}+mx+n\] Since, \[(x+2)\]and\[(x-1)\]are factors of \[f(x).\] Therefore, by factor theorem. \[f(-2)=0\]and\[f(1)=0\] \[\Rightarrow \]\[{{(-2)}^{3}}+10{{(-2)}^{2}}+m(-2)+n=0\] and \[{{(1)}^{3}}+10{{(1)}^{2}}+m(1)+n=0\] \[\Rightarrow \]\[-8+40-2m+n=0\] and \[1+10+m+n=0\] \[\Rightarrow \]\[-2m+n=-32\] ?(i) and \[m+n=-11\] ?(ii) subtracting (i) form (ii), we get \[3m=21\Rightarrow m=7\] From (ii), \[7+n=-11\Rightarrow n=-18\]
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