Question

Factorize \[\dfrac{1}{3}{{x}^{2}}-2x-9\] .

Answer 1

Hint: In order to factorize any quadratic equation, it is necessary that the term with degree two has the coefficient one. So, to factorize this equation, we take the L.C.M of this equation and then begin to solve the equation by ‘splitting the middle term method’.

Formula used:
Firstly to simplify the equation, the L.C.M of the terms in the equation is taken in order to make it easy for factorization.
Splitting the middle term method:
In order to factorize $x^2$ + bx + c we have to find numbers p and q such that p + q = b and pq = c and after finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

Complete step by step answer:
Firstly, taking the L.C.M. of the equation.
\[\begin{align}
  & \dfrac{1}{3}{{x}^{2}}-2x-9 \\
 & \Rightarrow \dfrac{{{x}^{2}}-6x-27}{3} \\
 & \Rightarrow \dfrac{1}{3}\left( {{x}^{2}}-6x-27 \right) \\
\end{align}\]
Using the method of ‘splitting the middle term’ for the quadratic equation \[\left( {{x}^{2}}-6x-27 \right)\] i.e. by breaking \[6x\] in two parts such that their sum equals \[-6x\] and product equals \[-27\] .
\[\begin{align}
  & \Rightarrow \dfrac{1}{3}\left( {{x}^{2}}-9x+3x-27 \right) \\
 & \Rightarrow \dfrac{1}{3}\left( x\left( x-9 \right)+3\left( x-9 \right) \right) \\
 & \Rightarrow \dfrac{1}{3}\left( x-3 \right)\left( x-9 \right) \\
\end{align}\]

Thus, the form of equation we got, after factorization is \[ \dfrac{1}{3}\left( x-3 \right)\left( x-9 \right)\]

Note: Alternatively, if you want to factorize directly without taking an LCM
 \[\dfrac{1}{3}{{x}^{2}}-2x-9\]
Then split the middle term in two terms m and n such that their sum equals \[-2x\] and their product equals
\[\begin{align}
  & -9\times \dfrac{1}{3}{{x}^{2}} \\
 & \Rightarrow -3{{x}^{2}} \\
\end{align}\]
Such that the solution becomes
\[\begin{align}
  & \dfrac{1}{3}{{x}^{2}}-3x+x-9 \\
 & \Rightarrow \dfrac{x}{3}\left( x-9 \right)+\left( x-9 \right) \\
 & \Rightarrow \left( \dfrac{x}{3}+1 \right)\left( x-9 \right) \\
 & \Rightarrow \left( \dfrac{x+3}{3} \right)\left( x-9 \right) \\
 & \Rightarrow \dfrac{1}{3}\left( x+3 \right)\left( x-9 \right) \\
\end{align}\]
In this way, we have factorized the equation in a slightly different way but obtained the same factors.
Be careful with the positive and negative signs while splitting the middle term into two terms.