A) \[15{{x}^{2}}+x-2\]
B) \[15{{x}^{2}}+5x-6\]
C) \[15{{x}^{2}}-5x+6\]
D) \[15{{x}^{2}}-x+2\]
Correct Answer: A
Solution :
Substituting the given zeros in \[\text{12}0=\text{1}0\text{4}\times \text{1}+\text{16}\],we get \[\text{1}0\text{4}=\text{16}\times \text{6}+\text{8}\] \[16=8\times 2+0\] [\[\text{256}=\text{8}\times \text{32}+0\]has the same zeros as\[3465={{3}^{2}}\times 5\times 7\times 11\]\[\text{546}0=\text{22}\times \text{3}\times \text{5}\times \text{7}\times \text{l3}\]]
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