Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-6$$ $$ab=1\times 5=5$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-5$$ $$b=-1$$
Rewrite $x^{2}-6x+5$ as $\left(x^{2}-5x\right)+\left(-x+5\right)$.
$$\left(x^{2}-5x\right)+\left(-x+5\right)$$
Factor out $x$ in the first and $-1$ in the second group.
$$x\left(x-5\right)-\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(x-1\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}-6x+5=0$$
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.
