A) \[x\left( x+4 \right){{\left( x-2 \right)}^{2}}\]
B) \[x\left( x+4 \right)\left( x-2 \right)\]
C) \[x\left( x+4 \right){{\left( x+2 \right)}^{2}}\]
D) \[x{{\left( x+4 \right)}^{2}}\left( x-2 \right)\]
Correct Answer: A
Solution :
\[{{x}^{2}}+2x-8={{x}^{2}}+4x-2x-8\] = \[x\left( x+4 \right)-2\left( x+4 \right)=\left( x-2 \right)\left( x+4 \right)\] = \[{{x}^{3}}-4{{x}^{2}}+4x={{x}^{3}}-2{{x}^{2}}-2{{x}^{2}}+4x\] = \[{{x}^{2}}(x-2)-2x(x-2)\] = \[({{x}^{2}}-2x)(x-2)\Rightarrow x(x-2)(x-2)\] = \[{{x}^{2}}+4x=x\left( x+4 \right)\] Now, LCM of\[({{x}^{2}}+2x-8)\], \[({{x}^{2}}-4{{x}^{2}}+4x)\] and\[\left( {{x}^{2}}+4x \right)=x\left( x-2 \right)\left( x+4 \right)\left( x-2 \right)\] = \[\mathbf{x}\left( \mathbf{x+4} \right){{\left( \mathbf{x-2} \right)}^{\mathbf{2}}}\]
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