Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-4$. To find $a$ and $b$, set up a system to be solved.
$$a+b=3$$ $$ab=1\left(-4\right)=-4$$
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 $-4$.
$$-1,4$$ $$-2,2$$
Calculate the sum for each pair.
$$-1+4=3$$ $$-2+2=0$$
The solution is the pair that gives sum $3$.
$$a=-1$$ $$b=4$$
Rewrite $x^{2}+3x-4$ as $\left(x^{2}-x\right)+\left(4x-4\right)$.
$$\left(x^{2}-x\right)+\left(4x-4\right)$$
Factor out $x$ in the first and $4$ in the second group.
$$x\left(x-1\right)+4\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(x+4\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+3x-4=0$$
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.
$$x=\frac{-3±\sqrt{3^{2}-4\left(-4\right)}}{2}$$
Square $3$.
$$x=\frac{-3±\sqrt{9-4\left(-4\right)}}{2}$$
Multiply $-4$ times $-4$.
$$x=\frac{-3±\sqrt{9+16}}{2}$$
Add $9$ to $16$.
$$x=\frac{-3±\sqrt{25}}{2}$$
Take the square root of $25$.
$$x=\frac{-3±5}{2}$$
Now solve the equation $x=\frac{-3±5}{2}$ when $±$ is plus. Add $-3$ to $5$.
$$x=\frac{2}{2}$$
Divide $2$ by $2$.
$$x=1$$
Now solve the equation $x=\frac{-3±5}{2}$ when $±$ is minus. Subtract $5$ from $-3$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
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 $-4$ for $x_{2}$.
