Factorize the given equation: $3{{a}^{2}}b-12{{a}^{2}}-9b+36$

Question

Vedantu Content Team · Accepted Answer

Hint: Observe the terms of the given expression. Take $3{{a}^{2}}$ as common from two terms and hence, use the algebraic identity of $\left( {{a}^{2}}-{{b}^{2}} \right)$ wherever required. It is given as${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$Complete step-by-step answer:Given expression in the problem is$3{{a}^{2}}b-12{{a}^{2}}-9b+36=0.................\left( i \right)$As we need to factorize the relation given in equation (i) so, we have to observe the relation among the terms of the equation. So, we can observe that if we take $3{{a}^{2}}$ common from the first two terms and – 3 from the last two terms remaining in the brackets, that would be 3b – 4. Hence, let us take $3{{a}^{2}}$ as common from the first two terms and – 3 from the last two terms. So, we can rewrite the expression (i) as$ 3{{a}^{2}}\left( b-4 \right)-9\left( b-4 \right)=0 $$ \Rightarrow \left( 3{{a}^{2}}-9 \right)\left( b-4 \right)=0 $Now, we can take ‘3’ as common in the first bracket of the above equation. So, we get$ 3\left( {{a}^{2}}-3 \right)\left( b-4 \right)=0 $$  \Rightarrow 3\left( {{a}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)\left( b-4 \right)=0....................\left( ii \right) $Now, we know algebraic identity of ${{x}^{2}}-{{y}^{2}}$ can be given as$\left( {{x}^{2}}-{{y}^{2}} \right)=\left( x-y \right)\left( x+y \right).............................\left( iii\right)$So, we can observe the left hand side of the above equation and first bracket $\left({{a}^{2}}-{{\left( \sqrt{3} \right)}^{2}} \right)$ of the equation (ii) hence, we getx = a and y = 1Hence, we can re-write the equation (ii) with the help of equation (iii) as $3\left( a-\sqrt{3} \right)\left( a+\sqrt{3} \right)\left( b-4 \right)=0$So, the factorization of the given expression in the problem is given by the above equation.Hence, $3\left( a-\sqrt{3} \right)\left( a+\sqrt{3} \right)\left( b-4 \right)$ is the answer.Note: Note: Another approach to factorize the given expression would be$ 3{{a}^{2}}b-9b-12{{a}^{2}}+36=0 $$  3b\left( {{a}^{2}}-3 \right)-12\left( {{a}^{2}}-3 \right)=0 $$  \left( 3b-12 \right)\left( {{a}^{2}}-3 \right)=0 $$ 3\left( b-4 \right)\left( a-\sqrt{3} \right)\left( a+\sqrt{3} \right)=0 $So, we can factorize the expression by observing the relation between first and third term and between second and fourth term. Hence it can be another approach. Observing the terms in the given expression is the key point of the problem.