SSC Sample Paper Mock Test-7 SSC CGL Tear-II Paper-1

Find the values of a and b so that the polynomials \[p\,\,(x)\]and \[q\,\,(x)\]have \[(x+1)(x+3)\]as their HCF \[p\,\,(x)=({{x}^{2}}+3x+2)({{x}^{2}}+2x+a)\] \[q\,\,(x)=({{x}^{2}}+7x+12)({{x}^{2}}+7x+b)\]

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.