Factor polynomial $$ 8x^{2}+4x-40 $$

$$ 8x^{2}+4x-40 = 4( x-2 )( 2x+5 ) $$

Step 1 :

After factoring out $ 4 $ we have:

$$ 8x^{2}+4x-40 = 4 ( 2x^{2}+x-10 ) $$

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:

$$ a = 2 , b = 1 ~ \text{ and } ~ c = -10 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -10} $.

$$ a \cdot c = -20 $$

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 $ 5x-4x $:

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

Step 8: Apply factoring by grouping. Factor $ x $ out of the first two terms and $ -2 $ out of the last two terms.

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

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