Factor the expression by grouping. First, the expression needs to be rewritten as $3x^{2}+ax+bx-4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=3\left(-4\right)=-12$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-12$.
$$-1,12$$ $$-2,6$$ $$-3,4$$
Calculate the sum for each pair.
$$-1+12=11$$ $$-2+6=4$$ $$-3+4=1$$
The solution is the pair that gives sum $1$.
$$a=-3$$ $$b=4$$
Rewrite $3x^{2}+x-4$ as $\left(3x^{2}-3x\right)+\left(4x-4\right)$.
$$\left(3x^{2}-3x\right)+\left(4x-4\right)$$
Factor out $3x$ in the first and $4$ in the second group.
$$3x\left(x-1\right)+4\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(3x+4\right)$$
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 $ax^{2}+bx+c=0$.
$$3x^{2}+x-4=0$$
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 $x=\frac{-1±7}{6}$ when $±$ is plus. Add $-1$ to $7$.
$$x=\frac{6}{6}$$
Divide $6$ by $6$.
$$x=1$$
Now solve the equation $x=\frac{-1±7}{6}$ when $±$ is minus. Subtract $7$ from $-1$.
$$x=-\frac{8}{6}$$
Reduce the fraction $\frac{-8}{6}$ to lowest terms by extracting and canceling out $2$.
$$x=-\frac{4}{3}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $1$ for $x_{1}$ and $-\frac{4}{3}$ for $x_{2}$.
