Factor polynomial $$ 16p^{2}-2p-5 $$

$$ 16p^{2}-2p-5 = ( 8p-5 )( 2p+1 ) $$

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:

$$ a = 16 , b = -2 ~ \text{ and } ~ c = -5 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 16 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -80 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -80 $ and add to $ b = -2 $.

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 $\color{blue}{ b = -2 }$

PRODUCT = -80 and SUM = -2
-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 $ -2 x $ with $ 8x-10x $:

$$ 16x^{2}-2x-5 = 16x^{2}+8x-10x-5 $$

Step 7: Apply factoring by grouping. Factor $ 8x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 16x^{2}+8x-10x-5 = 8x\left(2x+1\right) -5\left(2x+1\right) = \left(8x-5\right) \left(2x+1\right) $$

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