First we need to factor trinomial $ \color{blue}{ x^2-4x-5 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-4x-5 = 0} $.
Step 1: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:
$$ \color{blue}{ b = -4 } ~ \text{ and } ~ \color{red}{ c = -5 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -4 } $ and multiply to $ \color{red}{ -5 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -5 }$.
PRODUCT = -5 | |
-1 5 | 1 -5 |
Step 3: Find out which pair sums up to $\color{blue}{ b = -4 }$
PRODUCT = -5 and SUM = -4 | |
-1 5 | 1 -5 |
Step 4: Put 1 and -5 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-4x-5 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-4x-5 & = (x + 1)(x -5) \end{aligned} $$Step 5: Set each factor to zero and solve equations.
$$ \begin{array}{ccc} \begin{aligned} x+1 &= 0 \\ x &= -1 \end{aligned} & ~ & \begin{aligned} x-5 &= 0 \\ x &= 5 \end{aligned} \end{array} $$