Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=7$$ $$ab=2\times 3=6$$
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 $6$.
$$1,6$$ $$2,3$$
Calculate the sum for each pair.
$$1+6=7$$ $$2+3=5$$
The solution is the pair that gives sum $7$.
$$a=1$$ $$b=6$$
Rewrite $2x^{2}+7x+3$ as $\left(2x^{2}+x\right)+\left(6x+3\right)$.
$$\left(2x^{2}+x\right)+\left(6x+3\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(2x+1\right)+3\left(2x+1\right)$$
Factor out common term $2x+1$ by using distributive property.
$$\left(2x+1\right)\left(x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}+7x+3=0$$
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 $x=\frac{-7±5}{4}$ when $±$ is plus. Add $-7$ to $5$.
$$x=-\frac{2}{4}$$
Reduce the fraction $\frac{-2}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{1}{2}$$
Now solve the equation $x=\frac{-7±5}{4}$ when $±$ is minus. Subtract $5$ from $-7$.
$$x=-\frac{12}{4}$$
Divide $-12$ by $4$.
$$x=-3$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-\frac{1}{2}$ for $x_{1}$ and $-3$ for $x_{2}$.
