Solve $\color{blue}{x^2-8x-20 = 0}$ using factoring.
First we need to factor trinomial $ \color{blue}{ x^2-8x-20 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-8x-20 = 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:
$$ \color{blue}{ b = -8 } ~ \text{ and } ~ \color{red}{ c = -20 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -8 } $ and multiply to $ \color{red}{ -20 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -20 }$.
PRODUCT = -20 | |
-1 20 | 1 -20 |
-2 10 | 2 -10 |
-4 5 | 4 -5 |
Step 3: Find out which pair sums up to $\color{blue}{ b = -8 }$
PRODUCT = -20 and SUM = -8 | |
-1 20 | 1 -20 |
-2 10 | 2 -10 |
-4 5 | 4 -5 |
Step 4: Put 2 and -10 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-8x-20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-8x-20 & = (x + 2)(x -10) \end{aligned} $$Step 5: Set each factor to zero and solve equations.
$$ \begin{array}{ccc} \begin{aligned} x+2 &= 0 \\ x &= -2 \end{aligned} & ~ & \begin{aligned} x-10 &= 0 \\ x &= 10 \end{aligned} \end{array} $$