Factor the expression by grouping. First, the expression needs to be rewritten as $6x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-1$$ $$ab=6\left(-15\right)=-90$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-90$.
Rewrite $6x^{2}-x-15$ as $\left(6x^{2}-10x\right)+\left(9x-15\right)$.
$$\left(6x^{2}-10x\right)+\left(9x-15\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(3x-5\right)+3\left(3x-5\right)$$
Factor out common term $3x-5$ by using distributive property.
$$\left(3x-5\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$6x^{2}-x-15=0$$
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 $x=\frac{1±19}{12}$ when $±$ is plus. Add $1$ to $19$.
$$x=\frac{20}{12}$$
Reduce the fraction $\frac{20}{12}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{5}{3}$$
Now solve the equation $x=\frac{1±19}{12}$ when $±$ is minus. Subtract $19$ from $1$.
$$x=-\frac{18}{12}$$
Reduce the fraction $\frac{-18}{12}$ to lowest terms by extracting and canceling out $6$.
$$x=-\frac{3}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{3}$ for $x_{1}$ and $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{3x-5}{3}$ times $\frac{2x+3}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $6$
$$x ^ 2 -\frac{1}{6}x -\frac{5}{2} = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = \frac{1}{6} $$ $$ rs = -\frac{5}{2}$$
Two numbers $r$ and $s$ sum up to $\frac{1}{6}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{1}{6} = \frac{1}{12}$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = \frac{1}{12} - u$$ $$s = \frac{1}{12} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{5}{2}$
