You have the following trinomial:
[tex]5x^2+39x-8[/tex]The general form of a trinomial is:
[tex]ax^2+bx+c[/tex]In order to find the factors you use the quadratic formula, which is given by:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]By comparing the given trinomial with the general form you have;
a = 5
b = 39
c = -8
You replace the previous values into the quadratic formula:
[tex]\begin{gathered} x=\frac{-39\pm\sqrt[]{(39)^2-4(5)(-8)}}{2(5)} \\ x=\frac{-39\pm\sqrt[]{1521+160}}{10} \\ x=\frac{-39\pm\sqrt[]{1681}}{10}=\frac{-39\pm41}{10} \end{gathered}[/tex]Then, you have the following two solutions:
[tex]\begin{gathered} x_1=\frac{-39+41}{10}=\frac{2}{10}=\frac{1}{5} \\ x_2=\frac{-39-41}{10}=\frac{-80}{10}=-8 \end{gathered}[/tex]The factor of the trinomial are of the form:
[tex](x-x_1)(x-x_2)[/tex]Finally, by replacing you obtain:
[tex]5x^2+39x-8=(x-\frac{1}{5})(x-(-8))=(x-\frac{1}{5})(x+8)[/tex]Hence, the factors are (x - 1/5)( x + 8)
