Factor polynomial $$ 6x^{2}+29x+20 $$

$$ 6x^{2}+29x+20 = ( 6x+5 )( x+4 ) $$

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:

$$ a = 6 , b = 29 ~ \text{ and } ~ c = 20 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 6 }$ by the constant term $\color{blue}{c = 20} $.

$$ a \cdot c = 120 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 120 $ and add to $ b = 29 $.

Step 4: All pairs of numbers with a product of $ 120 $ are:

PRODUCT = 120
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 29 }$

PRODUCT = 120 and SUM = 29
1     120	-1     -120
2     60	-2     -60
3     40	-3     -40
4     30	-4     -30
5     24	-5     -24
6     20	-6     -20
8     15	-8     -15
10     12	-10     -12

Step 6: Replace middle term $ 29 x $ with $ 24x+5x $:

$$ 6x^{2}+29x+20 = 6x^{2}+24x+5x+20 $$

Step 7: Apply factoring by grouping. Factor $ 6x $ out of the first two terms and $ 5 $ out of the last two terms.

$$ 6x^{2}+24x+5x+20 = 6x\left(x+4\right) + 5\left(x+4\right) = \left(6x+5\right) \left(x+4\right) $$

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