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