Factor polynomial $$ 4a^{2}+6a-40 $$

$$ 4a^{2}+6a-40 = 2( 2a-5 )( a+4 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 4a^{2}+6a-40 = 2 ( 2a^{2}+3a-20 ) $$

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:

$$ a = 2 , b = 3 ~ \text{ and } ~ c = -20 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 2 }$ by the constant term $\color{blue}{c = -20} $.

$$ a \cdot c = -40 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -40 $ and add to $ b = 3 $.

Step 5: 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 6: Find out which factor pair sums up to $\color{blue}{ b = 3 }$

PRODUCT = -40 and SUM = 3
-1     40	1     -40
-2     20	2     -20
-4     10	4     -10
-5     8	5     -8

Step 7: Replace middle term $ 3 x $ with $ 8x-5x $:

$$ 2x^{2}+3x-20 = 2x^{2}+8x-5x-20 $$

Step 8: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -5 $ out of the last two terms.

$$ 2x^{2}+8x-5x-20 = 2x\left(x+4\right) -5\left(x+4\right) = \left(2x-5\right) \left(x+4\right) $$

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