Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx-20$. To find $a$ and $b$, set up a system to be solved.
$$a+b=19$$ $$ab=6\left(-20\right)=-120$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-120$.
Rewrite $6x^{2}+19x-20$ as $\left(6x^{2}-5x\right)+\left(24x-20\right)$.
$$\left(6x^{2}-5x\right)+\left(24x-20\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(6x-5\right)+4\left(6x-5\right)$$
Factor out common term $6x-5$ by using distributive property.
$$\left(6x-5\right)\left(x+4\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$6x^{2}+19x-20=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Now solve the equation $x=\frac{-19±29}{12}$ when $±$ is plus. Add $-19$ to $29$.
$$x=\frac{10}{12}$$
Reduce the fraction $\frac{10}{12}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{5}{6}$$
Now solve the equation $x=\frac{-19±29}{12}$ when $±$ is minus. Subtract $29$ from $-19$.
$$x=-\frac{48}{12}$$
Divide $-48$ by $12$.
$$x=-4$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{6}$ for $x_{1}$ and $-4$ for $x_{2}$.
