Factor polynomial $$ x^{2}-14x+48 $$

$$ x^{2}-14x+48 = ( x-6 )( x-8 ) $$

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:

$$ \color{blue}{ b = -14 } ~ \text{ and } ~ \color{red}{ c = 48 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -14 } $ and multiply to $ \color{red}{ 48 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 48 }$.

PRODUCT = 48
1     48	-1     -48
2     24	-2     -24
3     16	-3     -16
4     12	-4     -12
6     8	-6     -8

Step 3: Find out which pair sums up to $\color{blue}{ b = -14 }$

PRODUCT = 48 and SUM = -14
1     48	-1     -48
2     24	-2     -24
3     16	-3     -16
4     12	-4     -12
6     8	-6     -8

Step 4: Put -6 and -8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-14x+48 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-14x+48 & = (x -6)(x -8) \end{aligned} $$

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