Question

How do you factor \[2{{x}^{3}}-{{x}^{2}}-162x+81\]?

Answer 1

Hint: In this problem, we have to find the factors of the given cubic equation \[2{{x}^{3}}-{{x}^{2}}-162x+81\]by simple factorization method. We have two methods to solve, one is normal factorization method and another is synthetic division method, here we are going to use normal factorization method. We can check for the possibility of a perfect cube, if it is not a perfect cube, we can split the cubic equation, take out the common terms to find the factor of the given equation.

Complete step by step answer:
We know that the given cubic equation is,
\[2{{x}^{3}}-{{x}^{2}}-162x+81\]……… (1)
Here we can take first two terms from the equation (1), and take common terms out, we get
\[\Rightarrow 2{{x}^{3}}-{{x}^{2}}={{x}^{2}}\left( 2x-1 \right)\]……… (2)
Now we can take next two terms from equation (1), and take common terms out, we get
\[\Rightarrow -162x+81=-81\left( 2x-1 \right)\]……… (3)
Now we can substitute (2) and (3) in the equation (1), then equation (1) becomes
\[\Rightarrow {{x}^{2}}\left( 2x-1 \right)-81\left( 2x-1 \right)\]
We can now take the common terms from the above equation, we get
\[\begin{align}
  & \Rightarrow \left( {{x}^{2}}-81 \right)\left( 2x-1 \right) \\
 & \Rightarrow \left( {{x}^{2}}-{{9}^{2}} \right)\left( 2x-1 \right)......(4) \\
\end{align}\]
We know that the difference of the square formula is
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
We know that \[{{x}^{2}}-{{9}^{2}}=\left( x+9 \right)\left( x-9 \right)\]from the above formula. Substituting in the equation (4), we get
\[\Rightarrow \left( x+9 \right)\left( x-9 \right)\left( 2x-1 \right)\]

Therefore, the factors of \[2{{x}^{3}}-{{x}^{2}}-162x+81\] are \[\left( x+9 \right)\left( x-9 \right)\left( 2x-1 \right)\].

Note: Students may get confused while taking common terms outside the equation. Students should know the basic algebraic formulae to solve these types of problems. Students make mistakes while substituting the correct formulae, in the correct equation. There is also another method to factorize these types of equations, that is the synthetic division method, but it is simpler to factorise using the normal factorization method.