Question

How do you factor completely \[6{{a}^{3}}+3{{a}^{2}}-18a\] ?

Answer 1

Hint: To factorize the given equation we have to consider the equation as equation (1). Let us denote the given equation as ‘S’ .Then we have to take common which can be directly. By using any factorization method we have to factorize the equation.

Complete step by step answer:
To solve the given question we are given to factor the equation \[6{{a}^{3}}+3{{a}^{2}}-18a\]. For factoring the equation we have to assume the above equation as ‘S’ and consider it as equation (1).
Let us assume the given equation as ‘S’.
\[\Rightarrow S=6{{a}^{3}}+3{{a}^{2}}-18a\]
Now let us consider the above equation as equation (1), we get
\[\Rightarrow S=6{{a}^{3}}+3{{a}^{2}}-18a................\left( 1 \right)\]
Now by observing the equation (1) we can see that ‘3a’ is common in all the three terms, so let us pull the ‘a’ from the equation (1), we get
\[\Rightarrow S=\left( 3a \right)\left( 2{{a}^{2}}+a-6 \right)\]
Let us consider the above equation as equation (2), we get
\[\Rightarrow S=\left( 3a \right)\left( 2{{a}^{2}}+a-6 \right)........\left( 2 \right)\]
Now we can see that the second part of equation (2) contains a quadratic equation. So we have to factorize the second part for that we have to start by finding two numbers that multiply to \[\left( 2 \right)\left( 6 \right)\] and add to \[\left( 1 \right)\].Use these numbers to split up the \[x\] term. Use grouping to factor the quadratic equation.
Let us do the above steps to factorize equation (2), we get
\[\Rightarrow S=\left( 3a \right)\left( 2{{a}^{2}}+4a-3a-6 \right)\]
\[\Rightarrow S=\left( 3a \right)\left( 2{{a}^{2}}+4a-3a-6 \right)\]
Let us consider the above equation as equation (3), we get
\[\Rightarrow S=\left( 3a \right)\left( 2{{a}^{2}}+4a-3a-6 \right)..........\left( 3 \right)\]
Let us take ‘2a’ as common from first two terms and -3 from next two terms, we get
\[\Rightarrow S=\left( 3a \right)\left( 2a\left( a+2 \right)-3\left( a+2 \right) \right)\]
Now simplifying the above equation a bit, we get
\[\Rightarrow S=\left( 3a \right)\left( 2a-3 \right)\left( a+2 \right)\]
Therefore let us consider the above equation as equation (4), we get
\[\Rightarrow S=\left( 3a \right)\left( 2a-3 \right)\left( a+2 \right)............\left( 4 \right)\]
Hence, equation (4) is the completely factored equation.

Note:
Students should be aware of quadratic equations concepts. Questionnaires may ask this type of problems in many types i.e. sum of the roots or multiplication of roots or in many ways. So students should be aware of all these types of concepts.