A) \[({{x}^{2}}-3x+2),({{x}^{2}}+2x+3)\]
B) \[({{x}^{2}}-3x-2),({{x}^{2}}-2x+3)\]
C) \[({{x}^{2}}-3x+2),({{x}^{2}}+2x-3)\]
D) \[({{x}^{2}}+3x+2),({{x}^{2}}+2x+3)\]
Correct Answer: C
Solution :
Putting \[x-1=0\]i.e. \[x=1\]in \[\left( x-1 \right)\]respectively. Remainder \[={{\left( +1 \right)}^{3}}-7\left( 1 \right)+6=1-7+6=0\] \[\therefore (x-1)\] is a factor of expression \[{{x}^{3}}-7x+6\] Now, \[{{x}^{3}}-7x+6={{x}^{2}}(x-1)+x(x-1)-6(x-1)\] \[=(x-1)({{x}^{2}}+x-6)\] \[=(x-1)[{{x}^{2}}+3x-2x-6]\] \[=(x-1)[x(x+3)-2(x+3)]\] \[=(x-1)(x-2)(x+3)\] LCM \[={{x}^{3}}-7x+6=(x-1)(x-2)(x+3)\]and their HCF \[=(x-1)\] \[\therefore (x-1)\] is common in both. \[\therefore \] First expression \[=(x-1)(x-2)={{x}^{2}}-3x+2\]and second expression \[=(x-1)(x+3)={{x}^{2}}+2x-3\]
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