Factor polynomial $$ 2b^{2}+18b+40 $$

$$ 2b^{2}+18b+40 = 2( b+4 )( b+5 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2b^{2}+18b+40 = 2 ( b^{2}+9b+20 ) $$

Step 2 :

Step 2: 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:

$$ \color{blue}{ b = 9 } ~ \text{ and } ~ \color{red}{ c = 20 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 9 } $ and multiply to $ \color{red}{ 20 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 20 }$.

PRODUCT = 20
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

Step 4: Find out which pair sums up to $\color{blue}{ b = 9 }$

PRODUCT = 20 and SUM = 9
1     20	-1     -20
2     10	-2     -10
4     5	-4     -5

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

$$ \begin{aligned} b^{2}+9b+20 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ b^{2}+9b+20 & = (x + 4)(x + 5) \end{aligned} $$

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