(a) \[{{a}^{3}}-4{{a}^{2}}+12-3a\] |
(b) \[4{{x}^{2}}-20x+25\] |
Answer:
(a) \[{{a}^{3}}-4{{a}^{2}}+12-3a={{a}^{2}}(a-4)-3a+12\] \[={{a}^{2}}(a-4)-3(a-4)\] \[=(a-4)({{a}^{2}}-3)\] (b) \[4{{x}^{2}}-20x+25={{(2x)}^{2}}-2\times 2x\times 5+{{(5)}^{2}}\] \[={{(2x-5)}^{2}}\] [Since,\[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]] \[=\left( 2x-5 \right)\left( 2x-5 \right)~\]
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