Step 1 :

After factoring out $ 2 $ we have:

$$ 8x^{2}-2x-10 = 2 ( 4x^{2}-x-5 ) $$

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:

$$ a = 4 , b = -1 ~ \text{ and } ~ c = -5 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = -5} $.

$$ a \cdot c = -20 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -20 $ and add to $ b = -1 $.

Step 5: 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 6: Find out which factor pair sums up to $\color{blue}{ b = -1 }$

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

Step 7: Replace middle term $ -1 x $ with $ 4x-5x $:

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

Step 8: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -5 $ out of the last two terms.

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

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