replace x with a for this exercise
we use this formula to factor
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a=5, b=-3 and c=-14
[tex]x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4(5)(-14)}}{2(5)}[/tex][tex]\begin{gathered} x=\frac{3\pm\sqrt[]{9+280}}{10} \\ \\ x=\frac{3\pm\sqrt[]{289}}{10} \\ \\ x=\frac{3\pm17}{10} \end{gathered}[/tex]we have two roots
[tex]\begin{gathered} x=\frac{3+17}{10} \\ x=2 \end{gathered}[/tex]and
[tex]\begin{gathered} x=\frac{3-17}{10} \\ \\ x=-\frac{7}{5} \end{gathered}[/tex]so the simplified equation is
[tex](x-2)(x+\frac{7}{5})[/tex]now replace x for a
[tex](a-2)(a+\frac{7}{5})[/tex]