#### Answer

$(3x+1)(5x-4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
15x^2-7x=4
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
15x^2-7x-4=0
\\\\
(3x+1)(5x-4)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
3x+1=0
\text{ OR }
5x-4=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
3x+1=0
\\\\
3x=-1
\\\\
x=-\dfrac{1}{3}
\\\\\text{ OR }\\\\
5x-4=0
\\\\
5x=4
\\\\
x=\dfrac{4}{5}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{1}{3},\dfrac{4}{5} \right\}
.$
$\bf{\text{Supplementary Solution:}}$
To factor the expression, $
15x^2-7x-4
,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
In the expression above the value of $ac$ is $
15(-4)=-60
$ and the value of $b$ is $
-7
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-60\}, \{2,-30\}, \{3,-30\}, \{4,-15\}, \{5,-12\}, \{6,-10\},
\{-1,60\}, \{-2,30\}, \{-3,30\}, \{-4,15\}, \{-5,12\}, \{-6,10\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
5,-12
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
15x^2+5x-12x-4
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(15x^2+5x)-(12x+4)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5x(3x+1)-4(3x+1)
.\end{array}
Factoring the $GCF=
(3x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3x+1)(5x-4)
.\end{array}