Factor polynomial $$ 20x^{2}-41x+20 $$

$$ 20x^{2}-41x+20 = ( 4x-5 )( 5x-4 ) $$

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:

$$ a = 20 , b = -41 ~ \text{ and } ~ c = 20 $$

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

$$ a \cdot c = 400 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 400 $ and add to $ b = -41 $.

Step 4: All pairs of numbers with a product of $ 400 $ are:

PRODUCT = 400
1     400	-1     -400
2     200	-2     -200
4     100	-4     -100
5     80	-5     -80
8     50	-8     -50
10     40	-10     -40
16     25	-16     -25
20     20	-20     -20

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

PRODUCT = 400 and SUM = -41
1     400	-1     -400
2     200	-2     -200
4     100	-4     -100
5     80	-5     -80
8     50	-8     -50
10     40	-10     -40
16     25	-16     -25
20     20	-20     -20

Step 6: Replace middle term $ -41 x $ with $ -16x-25x $:

$$ 20x^{2}-41x+20 = 20x^{2}-16x-25x+20 $$

Step 7: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 20x^{2}-16x-25x+20 = 4x\left(5x-4\right) -5\left(5x-4\right) = \left(4x-5\right) \left(5x-4\right) $$

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