A) \[{{a}^{2}}-8a-20=0\]
B) \[{{a}^{2}}+30a-40=0\]
C) \[{{a}^{2}}-80a-30=0\]
D) \[{{a}^{2}}+7a-30=0\]
Correct Answer: A
Solution :
We have, \[\frac{{{a}^{2}}-13a+30}{(a-10)}=\frac{{{a}^{2}}+4a+4}{(a+2)}\] \[(a+2)({{a}^{2}}-13a+30)=(a-10)({{a}^{2}}+4a+4)\] \[\Rightarrow {{a}^{3}}-13{{a}^{2}}+30a+2{{a}^{2}}-26a+60\] \[={{a}^{3}}+4{{a}^{2}}+4a-10{{a}^{2}}-40a-40\] \[\Rightarrow {{a}^{3}}-11{{a}^{2}}+4a+60={{a}^{3}}-6{{a}^{2}}-36a-40\]\[\Rightarrow -11{{a}^{2}}+4a+60+6{{a}^{2}}+36a+40=0\] \[\Rightarrow -5{{a}^{2}}+40a+100=0\] \[\Rightarrow 5{{a}^{2}}-40a-100=0\Rightarrow {{a}^{2}}-8a-20=0\]
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