Factor polynomial $$ 20x^{2}-95x-5 $$

$$ 20x^{2}-95x-5 = 5( 4x^{2}-19x-1 ) $$

Step 1 :

After factoring out $ 5 $ we have:

$$ 20x^{2}-95x-5 = 5 ( 4x^{2}-19x-1 ) $$

Step 2 :

Step 2: 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:

$$ a = 4 , b = -19 ~ \text{ and } ~ c = -1 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -1} $.

$$ a \cdot c = -4 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -4 $ and add to $ b = -19 $.

Step 5: All pairs of numbers with a product of $ -4 $ are:

PRODUCT = -4
-1     4	1     -4
-2     2	2     -2

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -19 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -19 }$, we conclude the polynomial cannot be factored.

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