question_answer

                                     Factorise the following :

(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,\[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}\]] \[=\left( 2x-5 \right)\left( 2x-5 \right)~\]