Factor polynomial $$ n^{2}-9n+14 $$

$$ n^{2}-9n+14 = ( n-2 )( n-7 ) $$

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:

$$ \color{blue}{ b = -9 } ~ \text{ and } ~ \color{red}{ c = 14 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -9 } $ and multiply to $ \color{red}{ 14 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 14 }$.

PRODUCT = 14
1     14	-1     -14
2     7	-2     -7

Step 3: Find out which pair sums up to $\color{blue}{ b = -9 }$

PRODUCT = 14 and SUM = -9
1     14	-1     -14
2     7	-2     -7

Step 4: Put -2 and -7 into placeholders to get factored form.

$$ \begin{aligned} n^{2}-9n+14 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ n^{2}-9n+14 & = (x -2)(x -7) \end{aligned} $$

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