Factor polynomial $$ p^{2}-13p+30 $$

$$ p^{2}-13p+30 = ( p-3 )( p-10 ) $$

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:

$$ \color{blue}{ b = -13 } ~ \text{ and } ~ \color{red}{ c = 30 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -13 } $ and multiply to $ \color{red}{ 30 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = 30 }$.

PRODUCT = 30
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 3: Find out which pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = 30 and SUM = -13
1     30	-1     -30
2     15	-2     -15
3     10	-3     -10
5     6	-5     -6

Step 4: Put -3 and -10 into placeholders to get factored form.

$$ \begin{aligned} p^{2}-13p+30 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ p^{2}-13p+30 & = (x -3)(x -10) \end{aligned} $$

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