Factor polynomial $$ 5x^{2}+41x+42 $$

$$ 5x^{2}+41x+42 = ( 5x+6 )( x+7 ) $$

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:

$$ a = 5 , b = 41 ~ \text{ and } ~ c = 42 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 42} $.

$$ a \cdot c = 210 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 210 $ and add to $ b = 41 $.

Step 4: All pairs of numbers with a product of $ 210 $ are:

PRODUCT = 210
1     210	-1     -210
2     105	-2     -105
3     70	-3     -70
5     42	-5     -42
6     35	-6     -35
7     30	-7     -30
10     21	-10     -21
14     15	-14     -15

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 41 }$

PRODUCT = 210 and SUM = 41
1     210	-1     -210
2     105	-2     -105
3     70	-3     -70
5     42	-5     -42
6     35	-6     -35
7     30	-7     -30
10     21	-10     -21
14     15	-14     -15

Step 6: Replace middle term $ 41 x $ with $ 35x+6x $:

$$ 5x^{2}+41x+42 = 5x^{2}+35x+6x+42 $$

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

$$ 5x^{2}+35x+6x+42 = 5x\left(x+7\right) + 6\left(x+7\right) = \left(5x+6\right) \left(x+7\right) $$

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