Factor polynomial $$ t^{2}-18t-19 $$

$$ t^{2}-18t-19 = ( t+1 )( t-19 ) $$

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:

$$ \color{blue}{ b = -18 } ~ \text{ and } ~ \color{red}{ c = -19 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -18 } $ and multiply to $ \color{red}{ -19 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -19 }$.

PRODUCT = -19
-1     19	1     -19

Step 3: Find out which pair sums up to $\color{blue}{ b = -18 }$

PRODUCT = -19 and SUM = -18
-1     19	1     -19

Step 4: Put 1 and -19 into placeholders to get factored form.

$$ \begin{aligned} t^{2}-18t-19 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ t^{2}-18t-19 & = (x + 1)(x -19) \end{aligned} $$

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