Factor polynomial $$ 20x^{2}-72x+7 $$

$$ 20x^{2}-72x+7 = ( 2x-7 )( 10x-1 ) $$

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:

$$ a = 20 , b = -72 ~ \text{ and } ~ c = 7 $$

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

$$ a \cdot c = 140 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 140 $ and add to $ b = -72 $.

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 $\color{blue}{ b = -72 }$

PRODUCT = 140 and SUM = -72
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 $ -72 x $ with $ -2x-70x $:

$$ 20x^{2}-72x+7 = 20x^{2}-2x-70x+7 $$

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

$$ 20x^{2}-2x-70x+7 = 2x\left(10x-1\right) -7\left(10x-1\right) = \left(2x-7\right) \left(10x-1\right) $$

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