Factor polynomial $$ x^{2}+21x-72 $$

$$ x^{2}+21x-72 = ( x-3 )( x+24 ) $$

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:

$$ \color{blue}{ b = 21 } ~ \text{ and } ~ \color{red}{ c = -72 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 21 } $ and multiply to $ \color{red}{ -72 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -72 }$.

PRODUCT = -72
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

Step 3: Find out which pair sums up to $\color{blue}{ b = 21 }$

PRODUCT = -72 and SUM = 21
-1     72	1     -72
-2     36	2     -36
-3     24	3     -24
-4     18	4     -18
-6     12	6     -12
-8     9	8     -9

Step 4: Put -3 and 24 into placeholders to get factored form.

$$ \begin{aligned} x^{2}+21x-72 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}+21x-72 & = (x -3)(x + 24) \end{aligned} $$

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