Consider $x^{2}+x-2$. Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-2$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=1\left(-2\right)=-2$$
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. The only such pair is the system solution.
$$a=-1$$ $$b=2$$
Rewrite $x^{2}+x-2$ as $\left(x^{2}-x\right)+\left(2x-2\right)$.
$$\left(x^{2}-x\right)+\left(2x-2\right)$$
Factor out $x$ in the first and $2$ in the second group.
$$x\left(x-1\right)+2\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
$$\left(x-1\right)\left(x+2\right)$$
Rewrite the complete factored expression.
$$5\left(x-1\right)\left(x+2\right)$$
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 $ax^{2}+bx+c=0$.
$$5x^{2}+5x-10=0$$
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.
