Factor polynomial $$ b^{2}-4b-45 $$

$$ b^{2}-4b-45 = ( b+5 )( b-9 ) $$

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:

$$ \color{blue}{ b = -4 } ~ \text{ and } ~ \color{red}{ c = -45 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -4 } $ and multiply to $ \color{red}{ -45 } $.

Step 2: Find out pairs of numbers with a product of $\color{red}{ c = -45 }$.

PRODUCT = -45
-1     45	1     -45
-3     15	3     -15
-5     9	5     -9

Step 3: Find out which pair sums up to $\color{blue}{ b = -4 }$

PRODUCT = -45 and SUM = -4
-1     45	1     -45
-3     15	3     -15
-5     9	5     -9

Step 4: Put 5 and -9 into placeholders to get factored form.

$$ \begin{aligned} b^{2}-4b-45 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ b^{2}-4b-45 & = (x + 5)(x -9) \end{aligned} $$

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