Factor the expression by grouping. First, the expression needs to be rewritten as $3a^{2}+pa+qa+20$. To find $p$ and $q$, set up a system to be solved.
$$p+q=17$$ $$pq=3\times 20=60$$
Since $pq$ is positive, $p$ and $q$ have the same sign. Since $p+q$ is positive, $p$ and $q$ are both positive. List all such integer pairs that give product $60$.
Rewrite $3a^{2}+17a+20$ as $\left(3a^{2}+5a\right)+\left(12a+20\right)$.
$$\left(3a^{2}+5a\right)+\left(12a+20\right)$$
Factor out $a$ in the first and $4$ in the second group.
$$a\left(3a+5\right)+4\left(3a+5\right)$$
Factor out common term $3a+5$ by using distributive property.
$$\left(3a+5\right)\left(a+4\right)$$
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 $ax^{2}+bx+c=0$.
$$3a^{2}+17a+20=0$$
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 $a=\frac{-17±7}{6}$ when $±$ is plus. Add $-17$ to $7$.
$$a=-\frac{10}{6}$$
Reduce the fraction $\frac{-10}{6}$ to lowest terms by extracting and canceling out $2$.
$$a=-\frac{5}{3}$$
Now solve the equation $a=\frac{-17±7}{6}$ when $±$ is minus. Subtract $7$ from $-17$.
$$a=-\frac{24}{6}$$
Divide $-24$ by $6$.
$$a=-4$$
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 $-4$ 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 $3$
$$x ^ 2 +\frac{17}{3}x +\frac{20}{3} = 0$$
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 $rs = C$
$$r + s = -\frac{17}{3} $$ $$ rs = \frac{20}{3}$$
Two numbers $r$ and $s$ sum up to $-\frac{17}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{17}{3} = -\frac{17}{6}$. 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 $u$.
