Answer
$x=\left\{ -\dfrac{2}{3},\dfrac{4}{15} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
5(3x-1)^2+3=-16(3x-1)
,$ use substitution and then the method on factoring trinomials. Then equate each factor to zero (Zero Product Property) and solve for the value of the variable. Finally use back-substitution and then solve for the value of the original variable.
$\bf{\text{Solution Details:}}$
Let $z=3x-1.$ Then the equation above is equivalent to
\begin{array}{l}\require{cancel}
5z^2+3=-16z
\\\\
5z^2+16z+3=0
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
5(3)=15
$ and the value of $b$ is $
16
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,15\}, \{3,5\},
\\
\{-1,-15\}, \{-3,-5\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,15
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5z^2+z+15z+3=0
.\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}
(5z^2+z)+(15z+3)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
z(5z+1)+3(5z+1)=0
.\end{array}
Factoring the $GCF=
(5z+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(5z+1)(z+3)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then the solutions are
\begin{array}{l}\require{cancel}
5z+1=0
\\\\\text{OR}\\\\
z+3=0
.\end{array}
Since $z=3x-1,$ by back-substitution, the solutions are
\begin{array}{l}\require{cancel}
5(3x-1)+1=0
\\\\\text{OR}\\\\
(3x-1)+3=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
5(3x-1)+1=0
\\\\=
15x-5+1=0
\\\\=
15x-4=0
\\\\=
15x=4
\\\\=
x=\dfrac{4}{15}
\\\\\text{OR}\\\\
(3x-1)+3=0
\\\\
3x-1+3=0
\\\\
3x+2=0
\\\\
3x=-2
\\\\
x=-\dfrac{2}{3}
.\end{array}
Hence, the solutions are $
x=\left\{ -\dfrac{2}{3},\dfrac{4}{15} \right\}
.$
