Factor polynomial $$ 2s^{2}+13s+6 $$

$$ 2s^{2}+13s+6 = ( 2s+1 )( s+6 ) $$

Step 1: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 2 , b = 13 ~ \text{ and } ~ c = 6 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = 6} $.

$$ a \cdot c = 12 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 12 $ and add to $ b = 13 $.

Step 4: All pairs of numbers with a product of $ 12 $ are:

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

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 13 }$

PRODUCT = 12 and SUM = 13
1     12	-1     -12
2     6	-2     -6
3     4	-3     -4

Step 6: Replace middle term $ 13 x $ with $ 12x+x $:

$$ 2x^{2}+13x+6 = 2x^{2}+12x+x+6 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ 1 $ out of the last two terms.

$$ 2x^{2}+12x+x+6 = 2x\left(x+6\right) + 1\left(x+6\right) = \left(2x+1\right) \left(x+6\right) $$

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