First we need to factor trinomial $ \color{blue}{ x^2-26x+153 } $ and than we use factored form to solve an equation $ \color{blue}{ x^2-26x+153 = 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:
$$ \color{blue}{ b = -26 } ~ \text{ and } ~ \color{red}{ c = 153 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -26 } $ and multiply to $ \color{red}{ 153 } $.
Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 153 }$.
PRODUCT = 153 | |
1 153 | -1 -153 |
3 51 | -3 -51 |
9 17 | -9 -17 |
Step 3: Find out which pair sums up to $\color{blue}{ b = -26 }$
PRODUCT = 153 and SUM = -26 | |
1 153 | -1 -153 |
3 51 | -3 -51 |
9 17 | -9 -17 |
Step 4: Put -9 and -17 into placeholders to get factored form.
$$ \begin{aligned} x^{2}-26x+153 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-26x+153 & = (x -9)(x -17) \end{aligned} $$Step 5: Set each factor to zero and solve equations.
$$ \begin{array}{ccc} \begin{aligned} x-9 &= 0 \\ x &= 9 \end{aligned} & ~ & \begin{aligned} x-17 &= 0 \\ x &= 17 \end{aligned} \end{array} $$