Factor polynomial $$ 20x^{2}+21x+4 $$

$$ 20x^{2}+21x+4 = ( 4x+1 )( 5x+4 ) $$

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:

$$ a = 20 , b = 21 ~ \text{ and } ~ c = 4 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 20 }$ by the constant term $\color{blue}{c = 4} $.

$$ a \cdot c = 80 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = 80 $ and add to $ b = 21 $.

Step 4: All pairs of numbers with a product of $ 80 $ are:

PRODUCT = 80
1     80	-1     -80
2     40	-2     -40
4     20	-4     -20
5     16	-5     -16
8     10	-8     -10

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 21 }$

PRODUCT = 80 and SUM = 21
1     80	-1     -80
2     40	-2     -40
4     20	-4     -20
5     16	-5     -16
8     10	-8     -10

Step 6: Replace middle term $ 21 x $ with $ 16x+5x $:

$$ 20x^{2}+21x+4 = 20x^{2}+16x+5x+4 $$

Step 7: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ 1 $ out of the last two terms.

$$ 20x^{2}+16x+5x+4 = 4x\left(5x+4\right) + 1\left(5x+4\right) = \left(4x+1\right) \left(5x+4\right) $$

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