Use the quadratic formula.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Where a = 3, b = -31, and c = -60.
[tex]x=\frac{-(-31)\pm\sqrt[]{(-31)^2-4(3)(-60)}}{2(3)}[/tex]
Solve to find both solutions.
[tex]x=\frac{31\pm\sqrt[]{961+720}}{6}=\frac{31\pm\sqrt[]{1681}}{6}=\frac{31\pm41}{6}[/tex]
Rewrite the expression as two.
[tex]\begin{gathered} x_1=\frac{31+41}{6}=\frac{72}{6}=12 \\ x_2=\frac{31-41}{6}=\frac{-10}{6}=-\frac{5}{3} \end{gathered}[/tex]
Once we have the solutions, we express them as factors. To do that, we have to move the constant to the right side of each equation.
[tex]\begin{gathered} x=12\to(x-12) \\ x=-\frac{5}{3}\to(3x+5)_{} \end{gathered}[/tex]
As can observe, the factor of the polynomial is (x-12).
Therefore, the answer is d.
