First we need to factor trinomial $ \color{blue}{ x^2-8x+15 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-8x+15 = 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:
$$ \color{blue}{ b = -8 } ~ \text{ and } ~ \color{red}{ c = 15 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -8 } $ and multiply to $ \color{red}{ 15 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 15 }$.
PRODUCT = 15 | |
1 15 | -1 -15 |
3 5 | -3 -5 |
Step 3: Find out which pair sums up to $\color{blue}{ b = -8 }$
PRODUCT = 15 and SUM = -8 | |
1 15 | -1 -15 |
3 5 | -3 -5 |
Step 4: Put -3 and -5 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-8x+15 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-8x+15 & = (x -3)(x -5) \end{aligned} $$Step 5: Set each factor to zero and solve equations.
$$ \begin{array}{ccc} \begin{aligned} x-3 &= 0 \\ x &= 3 \end{aligned} & ~ & \begin{aligned} x-5 &= 0 \\ x &= 5 \end{aligned} \end{array} $$