A) 9
B) -3
C) 6
D) 3
Correct Answer: D
Solution :
Given polynomial is\[\therefore \] Let one zero be\[H.C.F.=\frac{\text{Product of the numbers}}{L.C.M.}\]. Then, according to the problem, the other zero is\[\text{5474}=\text{2}\times \text{7}\times \text{17}\times \text{23}\]. \[\text{9775}={{\text{5}}^{\text{2}}}\times \text{17}\times \text{23}\]Product of zeros\[\text{1173}0=\text{2}\times \text{3}\times \text{5}\times \text{17}\times \text{23}\] But product of zeros\[\therefore \] \[=\text{17}\times \text{23}=\text{391}\]\[=\text{2}\times \text{3}\times {{\text{5}}^{\text{2}}}\times \text{7}\times \text{l7}\times \text{23}=\text{41}0\text{55}0\]\[\pi \]\[\frac{22}{7}\]\[\frac{22}{7}\]\[\pi \] Hence, the value of \[\text{4}0={{\text{2}}^{\text{3}}}\times \text{5}\]
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