Factor polynomial $$ 8x^{2}-6x-5 $$

$$ 8x^{2}-6x-5 = ( 4x-5 )( 2x+1 ) $$

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:

$$ a = 8 , b = -6 ~ \text{ and } ~ c = -5 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 8 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -40 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -40 $ and add to $ b = -6 $.

Step 4: All pairs of numbers with a product of $ -40 $ are:

PRODUCT = -40
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -6 }$

PRODUCT = -40 and SUM = -6
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 6: Replace middle term $ -6 x $ with $ 4x-10x $:

$$ 8x^{2}-6x-5 = 8x^{2}+4x-10x-5 $$

Step 7: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 8x^{2}+4x-10x-5 = 4x\left(2x+1\right) -5\left(2x+1\right) = \left(4x-5\right) \left(2x+1\right) $$

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