Factor the expression by grouping. First, the expression needs to be rewritten as $3a^{2}+pa+qa-5$. To find $p$ and $q$, set up a system to be solved.
$$p+q=-2$$ $$pq=3\left(-5\right)=-15$$
Since $pq$ is negative, $p$ and $q$ have the opposite signs. Since $p+q$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-15$.
$$1,-15$$ $$3,-5$$
Calculate the sum for each pair.
$$1-15=-14$$ $$3-5=-2$$
The solution is the pair that gives sum $-2$.
$$p=-5$$ $$q=3$$
Rewrite $3a^{2}-2a-5$ as $\left(3a^{2}-5a\right)+\left(3a-5\right)$.
$$\left(3a^{2}-5a\right)+\left(3a-5\right)$$
Factor out $a$ in $3a^{2}-5a$.
$$a\left(3a-5\right)+3a-5$$
Factor out common term $3a-5$ by using distributive property.
$$\left(3a-5\right)\left(a+1\right)$$
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 $ax^{2}+bx+c=0$.
$$3a^{2}-2a-5=0$$
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 $a=\frac{2±8}{6}$ when $±$ is plus. Add $2$ to $8$.
$$a=\frac{10}{6}$$
Reduce the fraction $\frac{10}{6}$ to lowest terms by extracting and canceling out $2$.
$$a=\frac{5}{3}$$
Now solve the equation $a=\frac{2±8}{6}$ when $±$ is minus. Subtract $8$ from $2$.
$$a=-\frac{6}{6}$$
Divide $-6$ by $6$.
$$a=-1$$
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 $-1$ for $x_{2}$.
Cancel out $3$, the greatest common factor in $3$ and $3$.
$$3a^{2}-2a-5=\left(3a-5\right)\left(a+1\right)$$
Steps Using Direct Factoring Method
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 $3$
$$x ^ 2 -\frac{2}{3}x -\frac{5}{3} = 0$$
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 $rs = C$
$$r + s = \frac{2}{3} $$ $$ rs = -\frac{5}{3}$$
Two numbers $r$ and $s$ sum up to $\frac{2}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{2}{3} = \frac{1}{3}$. 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 $u$.
$$r = \frac{1}{3} - u$$ $$s = \frac{1}{3} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = -\frac{5}{3}$
