How do you factor $4{{x}^{2}}+17x-15$ ?

Hint: We are given a two degree polynomial equation which has to be solved by the method of factoring the equation. We shall break down the one degree term, the x-term into two parts which must add up to the actual value given in the equation. Then, we will group the common terms and form two linear equations out of the given quadratic equation and equate each of them to zero to get our required solution.

Complete step by step solution:
There are four methods for solving quadratic equations, namely, factoring method, completing the square method, taking the square root method and the last method is solving using the various properties of polynomials.
However, we will use the method of factoring the quadratic equation which makes our calculations simpler.
For any quadratic equation $a{{x}^{2}}+bx+c=0$,
the sum of the roots $=-\dfrac{b}{a}$ and the product of the roots $=\dfrac{c}{a}$.
Thus, for the equation, $4{{x}^{2}}+17x-15$,
$\Rightarrow 4{{x}^{2}}+17x-15=0$,
We will find numbers by hit and trial whose product is equal to $4\times \left( -15 \right)=-60$ and whose sum is equal to 17.
Such two numbers are 20 and $-3$ as $-3+20=17$ and $\left( -3 \right)\times 20=-60$.
Now, factoring the equation:
$\Rightarrow 4{{x}^{2}}+20x-3x-15=0$,

Taking common, we get:
$\begin{align}
  & \Rightarrow 4x\left( x+5 \right)-3\left( x+5 \right)=0 \\
 & \Rightarrow \left( x+5 \right)\left( 4x-3 \right)=0 \\
\end{align}$
Hence, $x+5=0$ or $4x-3=0$
$\Rightarrow x=-5$ or $x=\dfrac{3}{5}$
Therefore, the roots of the equation are $x=-5,\dfrac{3}{5}$.

Note: We might also get much more complex quadratic or bi-quadratic or even higher degree polynomials to be factorized. In those cases, we must be careful enough to find the right common factors of the multiple terms given. We must also be careful enough while grouping the terms in order to avoid mistakes. One possible mistake that could be made while transposing terms that -5 could have been written as 5 which would produce incorrect roots of the equation.