A) \[-\,\,3,6\]
B) \[3,-\,\,6\]
C) \[6,-\,\,3\]
D) None of these
Correct Answer: A
Solution :
\[p\,\,(x)=(x+2)(x+1)({{x}^{2}}+2x+a)\] |
\[q\,\,(x)=(x+3)(x+4)({{x}^{2}}+7x+b)\] |
As HCF is \[(x+1)(x+3),\] then both \[(x+1)\] and \[(x+3)\] must be factors of p(x) and g(x) |
For p(x) (x + 1) is already factor, so (x + 3) must be a factor of \[{{x}^{2}}+2x+a\] |
So, \[p\,\,(-\,\,3)={{(-\,\,3)}^{2}}+2\,\,(-\,\,3)+a=0\] |
\[\Rightarrow \] \[9-6+a=0\Rightarrow a=-\,\,3\] |
For q (x) (x+ 3) is already factor. |
\[\therefore \] \[(x+1)\] must be a factor of \[{{x}^{2}}+7x+b\] |
\[\therefore \] \[q\,\,(-\,\,1)={{(-\,\,1)}^{2}}+7\,\,(-\,\,1)+b=0\Rightarrow b=6\] |
So, \[a=-\,\,3\] and \[b=6\] is solution. |
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