For the quadratic trinomial \(ax^2+bx+c\) the following formula
\(ax^2+bx+c=a(x-x_1)(x-x_2)\), where \(x_1, x_2\) are trinomial roots, holds.
1. For trinomial \(6x^2 - 13x - 5\) find the roots:
\(D=(-13)^2-4\cdot (-5)\cdot 6=169+120=289,\ \sqrt{D}=17,\\ \\
x_1=\dfrac{13-17}{2\cdot 6} = -\dfrac{4}{12} =-\dfrac{1}{3} ,\\ \\
x_2=\dfrac{13+17}{2\cdot 6} = \dfrac{30}{12} =\dfrac{5}{2}\).
2. The factoring form is
\(6x^2-13x-5=6\left(x+\dfrac{1}{3}\right)\left(x-\dfrac{5}{2}\right) =(3x+1)(2x-5)\)
Answer: correct option is B.
