How do you factor by grouping $15{{x}^{3}}-6{{x}^{2}}-25x+10$?

Hint: In the given polynomial $15{{x}^{3}}-6{{x}^{2}}-25x+10$, we have to make the two pairs or the two groups of the terms by combining the first two terms and the last two terms. On combining this way, we will get the two groups as $\left( 15{{x}^{3}}-6{{x}^{2}} \right)+\left( -25x+10 \right)$. The factors of $3{{x}^{2}}$ and $-5x$ are common to the respective two groups. So we can take them outside of the respective groups. Then, lastly we will be left with a factor common to both of the groups which can be finally taken outside to completely factor the given polynomial.

Complete step by step solution:
Let us consider the given polynomial as
$\Rightarrow f\left( x \right)=15{{x}^{3}}-6{{x}^{2}}-25x+10$
To factor the above polynomial by grouping, we combine the first two terms and the last to terms to form two pairs or groups of terms as
$\Rightarrow f\left( x \right)=\left( 15{{x}^{3}}-6{{x}^{2}} \right)+\left( -25x+10 \right)$
As we can see that the factor $3{{x}^{2}}$ is common to the first group. Therefore we can take it outside of the first group to get
$\Rightarrow f\left( x \right)=3{{x}^{2}}\left( 5x-2 \right)+\left( -25x+10 \right)$
Similarly, we can take the factor $-5x$ outside of the second group to get
$\Rightarrow f\left( x \right)=3{{x}^{2}}\left( 5x-2 \right)-5\left( 5x-2 \right)$
Now we can take the factor $\left( 5x-2 \right)$ common from both of the groups to get
\[\Rightarrow f\left( x \right)=\left( 5x-2 \right)\left( 3{{x}^{2}}-5 \right)\]
The quadratic factor can be factored further by taking $3$ common to have
\[\begin{align}
  & \Rightarrow f\left( x \right)=\left( 5x-2 \right)3\left( {{x}^{2}}-\dfrac{5}{3} \right) \\
 & \Rightarrow f\left( x \right)=3\left( 5x-2 \right)\left( {{x}^{2}}-\dfrac{5}{3} \right) \\
\end{align}\]
Putting \[\dfrac{5}{3}={{\left( \sqrt{\dfrac{5}{3}} \right)}^{2}}\], we get
\[\Rightarrow f\left( x \right)=3\left( 5x-2 \right)\left( {{x}^{2}}-{{\left( \sqrt{\dfrac{5}{3}} \right)}^{2}} \right)\]
Applying \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get
\[\Rightarrow f\left( x \right)=3\left( 5x-2 \right)\left( x+\sqrt{\dfrac{5}{3}} \right)\left( x-\sqrt{\dfrac{5}{3}} \right)\]
Hence, the given polynomial is factored completely.

Note:
The quadratic factor obtained in the above solution \[\left( {{x}^{2}}-\dfrac{5}{3} \right)\] can also be factored further by considering its roots. For that we need to equate it to zero to get the equation ${{x}^{2}}-\dfrac{5}{3}=0$. On solving this equation by the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we will obtain the roots as $x=\sqrt{\dfrac{5}{3}}$ and $x=-\sqrt{\dfrac{5}{3}}$ and by the factor theorem, the quadratic factor will be factored as \[\left( x+\sqrt{\dfrac{5}{3}} \right)\left( x-\sqrt{\dfrac{5}{3}} \right)\].