Factor polynomial $$ a^{2}-11a+24 $$

$$ a^{2}-11a+24 = ( a-3 )( a-8 ) $$

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:

$$ \color{blue}{ b = -11 } ~ \text{ and } ~ \color{red}{ c = 24 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -11 } $ and multiply to $ \color{red}{ 24 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 24 }$.

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

Step 3: Find out which pair sums up to $\color{blue}{ b = -11 }$

PRODUCT = 24 and SUM = -11
1     24	-1     -24
2     12	-2     -12
3     8	-3     -8
4     6	-4     -6

Step 4: Put -3 and -8 into placeholders to get factored form.

$$ \begin{aligned} a^{2}-11a+24 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ a^{2}-11a+24 & = (x -3)(x -8) \end{aligned} $$

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