How do you factor $3{{x}^{2}}-8x+15$?

Hint: The process of finding out factors of the given polynomial or in this case quadratic equation is called factorization. It can be done in two ways. Either by splitting the middle term or by directly applying the quadratic equation form and find its roots. The quadratic equation formula is :
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $\sqrt{{{b}^{2}}-4ac}$ is called the discriminant, $D$, of the of the quadratic equation.

Complete step by step answer:
Now before finding out the roots of any quadratic equation, it is very important to check the nature of its root. It means we need to check out whether it’s discriminant is greater than or less than or equal to $0$.
The roots of a quadratic equation are said to be :
$1)$ real and different or unique if $\sqrt{{{b}^{2}}-4ac}\ge 0$
$2)$ real and repeated if $\sqrt{{{b}^{2}}-4ac}=0$
$3)$ imaginary if $\sqrt{{{b}^{2}}-4ac}\le 0$
Now let us check the nature of the roots of $3{{x}^{2}}-8x+15$.
Let us consider $f\left( x \right)=3{{x}^{2}}-8x+15=0$.
Let’s make use of a standard form of a quadratic equation to simplify things.
Let $h\left( x \right)=a{{x}^{2}}+bx+c$
Upon comparing $f\left( x \right)$ with $h\left( x \right)$ , we conclude :
$\begin{align}
  & \Rightarrow a=3 \\
 & \Rightarrow b=-8 \\
 & \Rightarrow c=15 \\
\end{align}$
Now, let us plug-in these values in $\sqrt{{{b}^{2}}-4ac}$.
 \[\begin{align}
  & \Rightarrow D=\sqrt{{{b}^{2}}-4ac} \\
 & \Rightarrow D=\sqrt{{{\left( 8 \right)}^{2}}-4\left( 15 \right)\left( 3 \right)} \\
 & \Rightarrow D=\sqrt{64-180} \\
 & \Rightarrow D=\sqrt{64-180}=\sqrt{-116}\le 0 \\
\end{align}\]
As we can see that $D\le 0$, we can conclude that the roots of this quadratic equation are imaginary.
So now, let us directly use the formula to find it’s imaginary root.
$\begin{align}
  & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{{{\left( 8 \right)}^{2}}-4\left( 15 \right)\left( 3 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{64-180}}{6} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{64-180}}{6} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{-116}}{6} \\
 & \Rightarrow x=\dfrac{8\pm 116i}{6} \\
\end{align}$
We know that $\sqrt{-1}=i$.
$\begin{align}
  & \Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{{{\left( 8 \right)}^{2}}-4\left( 15 \right)\left( 3 \right)}}{2\left( 3 \right)} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{64-180}}{6} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{64-180}}{6} \\
 & \Rightarrow x=\dfrac{8\pm \sqrt{-116}}{6} \\
 & \Rightarrow x=\dfrac{8\pm 116i}{6} \\
 & \Rightarrow x=\dfrac{4\pm 58i}{3} \\
\end{align}$
So the imaginary roots are $\dfrac{4+58i}{3}$ ,$\dfrac{4-58i}{3}$ .
$\therefore $ Hence the factors of the quadratic equation $3{{x}^{2}}-8x+15$ are $x-\dfrac{4+58i}{3}$ ,$x-\dfrac{4-58i}{3}$.

Note: It is very important to check the discriminant of quadratic before proceeding to find its roots. It is good practice. For these kinds of questions, it is better to just go with the formula and find out the roots rather than splitting the middle term as it is time taking. So when a quadratic equation has imaginary roots, it’s graph doesn’t cut the $x$-axis. It’s either completely above or below $x$-axis depending upon the sign of $a$.