Factor polynomial $$ b^{2}-15b+56 $$

$$ b^{2}-15b+56 = ( b-7 )( b-8 ) $$

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:

$$ \color{blue}{ b = -15 } ~ \text{ and } ~ \color{red}{ c = 56 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -15 } $ and multiply to $ \color{red}{ 56 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 56 }$.

PRODUCT = 56
1     56	-1     -56
2     28	-2     -28
4     14	-4     -14
7     8	-7     -8

Step 3: Find out which pair sums up to $\color{blue}{ b = -15 }$

PRODUCT = 56 and SUM = -15
1     56	-1     -56
2     28	-2     -28
4     14	-4     -14
7     8	-7     -8

Step 4: Put -7 and -8 into placeholders to get factored form.

$$ \begin{aligned} b^{2}-15b+56 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ b^{2}-15b+56 & = (x -7)(x -8) \end{aligned} $$

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