Factor polynomial $$ n^{2}-13n+40 $$

$$ n^{2}-13n+40 = ( n-5 )( n-8 ) $$

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:

$$ \color{blue}{ b = -13 } ~ \text{ and } ~ \color{red}{ c = 40 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -13 } $ and multiply to $ \color{red}{ 40 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 40 }$.

PRODUCT = 40
1     40	-1     -40
2     20	-2     -20
4     10	-4     -10
5     8	-5     -8

Step 3: Find out which pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = 40 and SUM = -13
1     40	-1     -40
2     20	-2     -20
4     10	-4     -10
5     8	-5     -8

Step 4: Put -5 and -8 into placeholders to get factored form.

$$ \begin{aligned} n^{2}-13n+40 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ n^{2}-13n+40 & = (x -5)(x -8) \end{aligned} $$

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