| \[{{({{a}^{2}}+2a)}^{2}}+12\,({{a}^{2}}+2a)-45\] can be expressed as |
A) \[(a-1)(a-3)({{a}^{2}}+2a+15)\]
B) \[(a-1)(a+3)({{a}^{2}}+2a+15)\]
C) \[(a+1)(a+3)({{a}^{2}}+2a+15)\]
D) \[(a+1)(a-3)({{a}^{2}}+2a+15)\]
Correct Answer: B
Solution :
| Given, \[{{({{a}^{2}}+2a)}^{2}}+12\,({{a}^{2}}+2a)-45\] |
| Let \[({{a}^{2}}+2a)=x\] |
| Then, \[{{x}^{2}}+12x-45={{x}^{2}}+15x-3x-45\] |
| \[=x\,(x+15)-3\,(x+15)\] |
| \[=(x+15)(x-3)\] |
| Now, putting the value of x in these factors. |
| \[=({{a}^{2}}+2a+15)({{a}^{2}}+2a-3)\] |
| \[=({{a}^{2}}+3a-a-3)({{a}^{2}}+2a+15)\] |
| \[=[a\,(a+3)-1\,(a+3)({{a}^{2}}+2a+15)\] |
| \[=(a-1)(a+3)({{a}^{2}}+2a+15)\] |
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