Factor polynomial $$ 7n^{2}-39n+20 $$

$$ 7n^{2}-39n+20 = ( n-5 )( 7n-4 ) $$

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:

$$ a = 7 , b = -39 ~ \text{ and } ~ c = 20 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 7 }$ by the constant term $\color{blue}{c = 20} $.

$$ a \cdot c = 140 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 140 $ and add to $ b = -39 $.

Step 4: All pairs of numbers with a product of $ 140 $ are:

PRODUCT = 140
1     140	-1     -140
2     70	-2     -70
4     35	-4     -35
5     28	-5     -28
7     20	-7     -20
10     14	-10     -14

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -39 }$

PRODUCT = 140 and SUM = -39
1     140	-1     -140
2     70	-2     -70
4     35	-4     -35
5     28	-5     -28
7     20	-7     -20
10     14	-10     -14

Step 6: Replace middle term $ -39 x $ with $ -4x-35x $:

$$ 7x^{2}-39x+20 = 7x^{2}-4x-35x+20 $$

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

$$ 7x^{2}-4x-35x+20 = x\left(7x-4\right) -5\left(7x-4\right) = \left(x-5\right) \left(7x-4\right) $$

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