A) 1
B) \[\left( x-3 \right)\]
C) \[\left( 2x-5 \right)\]
D) \[x-4\]
Correct Answer: A
Solution :
(a) \[\left( {{x}^{2}}-5x+6 \right)\left( {{x}^{2}}-7x+12 \right)and\text{ }{{x}^{2}}+9x+20\] \[{{x}^{2}}-5x+6=\left( x-2 \right)\left( x-3 \right)\] \[{{x}^{2}}-7x+12=\left( x-3 \right)\left( x-4 \right)\] \[{{x}^{2}}+9x+20=\left( x+4 \right)\left( x+5 \right)\] Hence HCF = 1 as no term is common.
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