Factor polynomial $$ 4x^{2}-19x-5 $$

$$ 4x^{2}-19x-5 = ( x-5 )( 4x+1 ) $$

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:

$$ a = 4 , b = -19 ~ \text{ and } ~ c = -5 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -20 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -20 $ and add to $ b = -19 $.

Step 4: All pairs of numbers with a product of $ -20 $ are:

PRODUCT = -20
-1     20	1     -20
-2     10	2     -10
-4     5	4     -5

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -19 }$

PRODUCT = -20 and SUM = -19
-1     20	1     -20
-2     10	2     -10
-4     5	4     -5

Step 6: Replace middle term $ -19 x $ with $ x-20x $:

$$ 4x^{2}-19x-5 = 4x^{2}+x-20x-5 $$

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

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

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