Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx+14$. To find $a$ and $b$, set up a system to be solved.
$$a+b=23$$ $$ab=3\times 14=42$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $42$.
$$1,42$$ $$2,21$$ $$3,14$$ $$6,7$$
Calculate the sum for each pair.
$$1+42=43$$ $$2+21=23$$ $$3+14=17$$ $$6+7=13$$
The solution is the pair that gives sum $23$.
$$a=2$$ $$b=21$$
Rewrite $3x^{2}+23x+14$ as $\left(3x^{2}+2x\right)+\left(21x+14\right)$.
$$\left(3x^{2}+2x\right)+\left(21x+14\right)$$
Factor out $x$ in the first and $7$ in the second group.
$$x\left(3x+2\right)+7\left(3x+2\right)$$
Factor out common term $3x+2$ by using distributive property.
$$\left(3x+2\right)\left(x+7\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}+23x+14=0$$
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 $x=\frac{-23±19}{6}$ when $±$ is plus. Add $-23$ to $19$.
$$x=-\frac{4}{6}$$
Reduce the fraction $\frac{-4}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{2}{3}$$
Now solve the equation $x=\frac{-23±19}{6}$ when $±$ is minus. Subtract $19$ from $-23$.
$$x=-\frac{42}{6}$$
Divide $-42$ by $6$.
$$x=-7$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{2}{3}$ for $x_{1}$ and $-7$ 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 $3$
$$x ^ 2 +\frac{23}{3}x +\frac{14}{3} = 0$$
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 $rs = C$
$$r + s = -\frac{23}{3} $$ $$ rs = \frac{14}{3}$$
Two numbers $r$ and $s$ sum up to $-\frac{23}{3}$ exactly when the average of the two numbers is $\frac{1}{2}*-\frac{23}{3} = -\frac{23}{6}$. 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 $u$.
