Factor polynomial $$ 5x^{2}-21x-20 $$

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

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:

$$ a = 5 , b = -21 ~ \text{ and } ~ c = -20 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = -20} $.

$$ a \cdot c = -100 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -100 $ and add to $ b = -21 $.

Step 4: All pairs of numbers with a product of $ -100 $ are:

PRODUCT = -100
-1     100	1     -100
-2     50	2     -50
-4     25	4     -25
-5     20	5     -20
-10     10	10     -10

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

PRODUCT = -100 and SUM = -21
-1     100	1     -100
-2     50	2     -50
-4     25	4     -25
-5     20	5     -20
-10     10	10     -10

Step 6: Replace middle term $ -21 x $ with $ 4x-25x $:

$$ 5x^{2}-21x-20 = 5x^{2}+4x-25x-20 $$

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

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

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