Given the rational expression below:
[tex]\frac{12n^2-29n-8}{28n^2-5n-3}[/tex]
The simplification is as shown below:
[tex]\begin{gathered} \frac{12n^2-29n-8}{28n^2-5n-3}=\frac{12n^2-32n+3n-8}{28n^2-12n+7n-3} \\ \frac{12n^2-32n+3n-8}{28n^2-12n+7n-3}=\frac{4n(3n-8)+1(3n-8)}{4n(7n-3)+1(7n-3)} \\ \frac{4n(3n-8)+1(3n-8)}{4n(7n-3)+1(7n-3)}=\frac{(3n-8)(4n+1)}{(7n-3)(4n+1)} \end{gathered}[/tex]
It can be observed from the simplification that (4n+1) is common to both the numerator and denominator. To simplify further we will cross (4n+1) out in the numerator and denominator as shown below:
[tex]\begin{gathered} \frac{(3n-8)(4n+1)}{(7n-3)(4n+1)}=\frac{3n-8}{7n-3} \\ Hence,\frac{12n^2-29n-8}{28n^2-5n-3}=\frac{3n-8}{7n-3} \end{gathered}[/tex]
Hence, after the simplification,
the numerator is 3n - 8, and
The denominator is 7n - 3
