How do you factor $3{{x}^{2}}-7x-20$ .

Hint: Now we want to factorize the given expression. Now first we will find the roots of the equation by completing the square method. Hence we will first make the coefficient of ${{x}^{2}}$ as 1 and then add and subtract the expression of the form $a{{x}^{2}}+bx+c$ by ${{\left( \dfrac{b}{2a} \right)}^{2}}$ . Now we will simplify the equation by using ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and take square root to eliminate the powers. Hence we have the roots of the expression. Now we will write the factors corresponding to the roots.

Complete step by step solution:
Now the given expression is a quadratic expression in x.
Now to find the factors of the roots of the given expression.
Now to find the roots of the above expression we will use the complete square method.
Now consider the equation $3{{x}^{2}}-7x-20=0$ .
Now let us first divide the whole equation by 3 so that we get the coefficient of ${{x}^{2}}$ as 1.
Hence we get the expression as ${{x}^{2}}-\dfrac{7}{3}x-\dfrac{20}{3}=0$
Now adding ${{\left( \dfrac{-7}{2\left( 3 \right)} \right)}^{2}}=\dfrac{49}{36}$ on both sides we get,
$\Rightarrow {{x}^{2}}-\dfrac{7}{3}x+\dfrac{49}{36}-\dfrac{49}{36}-\dfrac{20}{3}=0$
Now using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ we get,
$\Rightarrow {{\left( x-\dfrac{7}{6} \right)}^{2}}-\dfrac{49}{36}-\dfrac{240}{36}=0$
Now shifting the terms on RHS and simplifying we get,
$\begin{align}
  & \Rightarrow {{\left( x-\dfrac{7}{6} \right)}^{2}}=\dfrac{240+49}{36} \\
 & \Rightarrow {{\left( x-\dfrac{7}{6} \right)}^{2}}=\dfrac{289}{36} \\
\end{align}$
Now taking square root on both sides we get,
$\Rightarrow \left( x-\dfrac{7}{6} \right)=\pm \dfrac{17}{6}$
Now shifting $\dfrac{7}{6}$ on RHS we get,
$\Rightarrow x=\dfrac{7}{6}\pm \dfrac{17}{6}$
Hence the roots of the expression are $\dfrac{24}{6}$ and $\dfrac{-10}{6}$
Hence we have the roots of the expression are $\dfrac{-5}{3}$ and 4.
Now we know that $\alpha $ is the root of the expression, then $x-\alpha $ is the factor of the expression.
Hence the factors of the given expression are $x+\dfrac{5}{3}$ and $x-4$

Note: Now note that here we have found the roots of the quadratic expression by using the completing square method. Now note that we can also directly find the roots of the quadratic by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence substituting the values of a, b and c obtained by comparing the given expression with $a{{x}^{2}}+bx+c$ we get the roots of the equation.