Question

How do you factor $3{{x}^{2}}-18x-81=0$?

Answer 1

Hint: To find the factors of $3{{x}^{2}}-18x-81$, first take common ‘3’ from each term. Then find two numbers whose sum or difference will be the coefficient of ‘x’ after taking common i.e. $-6$ and multiplication will be the result of taking common ‘3’ from the constant term i.e. $-27$. Then break the equation and again take common factors out and group them together to obtain the required solution.

Complete step by step solution:
Factorization: Factorization is defined as the breaking or decomposition of an expression into a product of another factors, which when multiplied together give the original expression itself.
The equation we have, $3{{x}^{2}}-18x-81=0$
Taking common ‘3’ from each term, we get
$\Rightarrow 3\left( {{x}^{2}}-6x-27 \right)=0$
Now, we have to find two numbers whose sum is $-6$ and multiplication is $-27$.
Thus, the numbers are 3 and $-9$.
Hence, our equation can be written as
$\begin{align}
  & \Rightarrow 3\left( {{x}^{2}}-6x-27 \right)=0 \\
 & \Rightarrow 3\left( {{x}^{2}}+3x-9x-27 \right)=0 \\
\end{align}$
Taking common ‘x’ from first two terms and ‘$-9$’ from last two terms, we get
$\Rightarrow 3\left( x\left( x+3 \right)-9\left( x+3 \right) \right)=0$
Again taking common $\left( x+3 \right)$ from both the terms, we get
$\Rightarrow 3\left( x+3 \right)\left( x-9 \right)=0$
This is the required factorization of the given question.

Note: As the given equation is equal to zero, so we can solve the equation to find the values of ‘x’ as follows
$\begin{align}
  & \Rightarrow 3\left( x+3 \right)\left( x-9 \right)=0 \\
 & \Rightarrow \left( x+3 \right)\left( x-9 \right)=0 \\
\end{align}$
Either
$\begin{align}
  & x+3=0 \\
 & \Rightarrow x=0-3 \\
 & \Rightarrow x=-3 \\
\end{align}$
Or,
$\begin{align}
  & x-9=0 \\
 & \Rightarrow x=0+9 \\
 & \Rightarrow x=9 \\
\end{align}$
Since it is a second degree equation, hence we are getting two different values of ‘x’ as the solution.