Consider $6a^{2}+13a-15$. Factor the expression by grouping. First, the expression needs to be rewritten as $6a^{2}+pa+qa-15$. To find $p$ and $q$, set up a system to be solved.
$$p+q=13$$ $$pq=6\left(-15\right)=-90$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-90$.
Rewrite $6a^{2}+13a-15$ as $\left(6a^{2}-5a\right)+\left(18a-15\right)$.
$$\left(6a^{2}-5a\right)+\left(18a-15\right)$$
Factor out $a$ in the first and $3$ in the second group.
$$a\left(6a-5\right)+3\left(6a-5\right)$$
Factor out common term $6a-5$ by using distributive property.
$$\left(6a-5\right)\left(a+3\right)$$
Rewrite the complete factored expression.
$$2\left(6a-5\right)\left(a+3\right)$$
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 $ax^{2}+bx+c=0$.
$$12a^{2}+26a-30=0$$
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 $a=\frac{-26±46}{24}$ when $±$ is plus. Add $-26$ to $46$.
$$a=\frac{20}{24}$$
Reduce the fraction $\frac{20}{24}$ to lowest terms by extracting and canceling out $4$.
$$a=\frac{5}{6}$$
Now solve the equation $a=\frac{-26±46}{24}$ when $±$ is minus. Subtract $46$ from $-26$.
$$a=-\frac{72}{24}$$
Divide $-72$ by $24$.
$$a=-3$$
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 $-3$ for $x_{2}$.
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 $12$
$$x ^ 2 +\frac{13}{6}x -\frac{5}{2} = 0$$
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 $rs = C$
$$r + s = -\frac{13}{6} $$ $$ rs = -\frac{5}{2}$$
Two numbers $r$ and $s$ sum up to $-\frac{13}{6}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{13}{6} = -\frac{13}{12}$. 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 $u$.
