Factor polynomial $$ 20k^{2}-96k-20 $$

$$ 20k^{2}-96k-20 = 4( k-5 )( 5k+1 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 20k^{2}-96k-20 = 4 ( 5k^{2}-24k-5 ) $$

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:

$$ a = 5 , b = -24 ~ \text{ and } ~ c = -5 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -25 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -25 $ and add to $ b = -24 $.

Step 5: All pairs of numbers with a product of $ -25 $ are:

PRODUCT = -25
-1     25	1     -25
-5     5	5     -5

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -24 }$

PRODUCT = -25 and SUM = -24
-1     25	1     -25
-5     5	5     -5

Step 7: Replace middle term $ -24 x $ with $ x-25x $:

$$ 5x^{2}-24x-5 = 5x^{2}+x-25x-5 $$

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 5x^{2}+x-25x-5 = x\left(5x+1\right) -5\left(5x+1\right) = \left(x-5\right) \left(5x+1\right) $$

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