Question

What is the completely factored form of $2{{d}^{4}}+6{{d}^{3}}-18{{d}^{2}}-54d$?

Answer 1

Hint: For solving this question you should know about the factorisation of any expression. For solving this question, we will do the factors by grouping and try to take common power in the bracket variables. And then again, we take the common and by solving them we will get the final answer.

Complete step by step solution:
According to the question, it is asked of us to determine the factored form of $2{{d}^{4}}+6{{d}^{3}}-18{{d}^{2}}-54d$. So, for solving this question we will use the factorisation method to compress our expression. And then we will take the common terms which will help to reduce the expression and then we will solve this. If we do factor by grouping then we get,
$\begin{align}
  & 2{{d}^{4}}+6{{d}^{3}}-18{{d}^{2}}-54d \\
 & =2{{d}^{2}}\left( {{d}^{2}}-9 \right)+6d\left( {{d}^{2}}-9 \right) \\
\end{align}$
Here we can take $\left( {{d}^{2}}-9 \right)$ as a common term from the expression because it is possible to take the common term and it will make our step easy. So, by taking $\left( {{d}^{2}}-9 \right)$ common, we get,
$\left( {{d}^{2}}-9 \right)\left( 2{{d}^{2}}+6d \right)$
Now taking 2 common, we get,
$2\left( {{d}^{2}}-9 \right)\left( {{d}^{2}}+3d \right)$
Now as we know that this is in a form of ${{a}^{2}}-{{b}^{2}}$ and ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, so, we can write it as,
$2\left( d+3 \right)\left( d-3 \right)\left( {{d}^{2}}+3d \right)$
Now taking $d$ common from the expression, we get,
$2d\left( d+3 \right)\left( d-3 \right)\left( d+3 \right)$
This can be written as,
\[\begin{align}
  & 2d\left( d+3 \right)\left( d+3 \right)\left( d-3 \right) \\
 & =2d{{\left( d+3 \right)}^{2}}\left( d-3 \right) \\
\end{align}\]
So, the completely factored form of $2{{d}^{4}}+6{{d}^{3}}-18{{d}^{2}}-54d$ is equal to \[2d{{\left( d+3 \right)}^{2}}\left( d-3 \right)\].

Note: While calculating the factored form of any expression, always be careful about the power of the variables and always take the common terms, whatever is possible because this will reduce the probability of making mistakes and you should make all calculations very carefully.