Factor the expression by grouping. First, the expression needs to be rewritten as $4x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=8$$ $$ab=4\times 3=12$$
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 $12$.
$$1,12$$ $$2,6$$ $$3,4$$
Calculate the sum for each pair.
$$1+12=13$$ $$2+6=8$$ $$3+4=7$$
The solution is the pair that gives sum $8$.
$$a=2$$ $$b=6$$
Rewrite $4x^{2}+8x+3$ as $\left(4x^{2}+2x\right)+\left(6x+3\right)$.
$$\left(4x^{2}+2x\right)+\left(6x+3\right)$$
Factor out $2x$ in the first and $3$ in the second group.
$$2x\left(2x+1\right)+3\left(2x+1\right)$$
Factor out common term $2x+1$ by using distributive property.
$$\left(2x+1\right)\left(2x+3\right)$$
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 $ax^{2}+bx+c=0$.
$$4x^{2}+8x+3=0$$
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 $x=\frac{-8±4}{8}$ when $±$ is plus. Add $-8$ to $4$.
$$x=-\frac{4}{8}$$
Reduce the fraction $\frac{-4}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{1}{2}$$
Now solve the equation $x=\frac{-8±4}{8}$ when $±$ is minus. Subtract $4$ from $-8$.
$$x=-\frac{12}{8}$$
Reduce the fraction $\frac{-12}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=-\frac{3}{2}$$
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 $-\frac{3}{2}$ for $x_{2}$.
Multiply $\frac{2x+1}{2}$ times $\frac{2x+3}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
