How do you factor $4{{x}^{3}}-2{{x}^{2}}+6x-3$?

Hint: To factor the given polynomial $4{{x}^{3}}-2{{x}^{2}}+6x-3$, we have to use the factoring by grouping method. We can see that the given polynomial consists of four algebraic terms. So we can group two-two terms together to form two pairs of terms. On taking the common factor outside from each of the two pairs, we will get a factor common to both the pairs. Finally, on taking the common factor outside, we will obtain the factored form of the given polynomial.

Complete step by step solution:
The polynomial given in the above question is
$\Rightarrow p\left( x \right)=4{{x}^{3}}-2{{x}^{2}}+6x-3$
Since the above polynomial contains four terms, we can group two-two terms so that we will form two pairs. So, after grouping, we can write the above polynomial as
$\Rightarrow p\left( x \right)=\left( 4{{x}^{3}}-2{{x}^{2}} \right)+\left( 6x-3 \right)$
Now, as we can see that the factor $2{{x}^{2}}$ is common to the first pair, and $3$ is common to the second pair. Taking these outside from the respective pairs, we get
$\Rightarrow p\left( x \right)=2{{x}^{2}}\left( 2x-1 \right)+2\left( 2x-1 \right)$
Now, we have $\left( 2x-1 \right)$ common. Taking it outside, we get
$\Rightarrow p\left( x \right)=\left( 2x-1 \right)\left( 2{{x}^{2}}+2 \right)$
Taking $2$ common from the second factor, we finally obtain
$\begin{align}
  & \Rightarrow p\left( x \right)=\left( 2x-1 \right)2\left( {{x}^{2}}+1 \right) \\
 & \Rightarrow p\left( x \right)=2\left( 2x-1 \right)\left( {{x}^{2}}+1 \right) \\
\end{align}$
Hence, the given polynomial is factored as $2\left( 2x-1 \right)\left( {{x}^{2}}+1 \right)$.

Note:
The quadratic factor $\left( {{x}^{2}}+1 \right)$ cannot be factored further. We must note that factoring does not necessarily mean that we need to express it as the product of one degree terms. We only have to factor to the extent to which the given polynomial can be factored. We can also factor the given polynomial using the hit and trial and then the middle term splitting technique. But the factor by grouping method, if applicable, is the easiest.