Factor polynomial $$ 15x^{2}+8x-16 $$

$$ 15x^{2}+8x-16 = ( 5x-4 )( 3x+4 ) $$

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:

$$ a = 15 , b = 8 ~ \text{ and } ~ c = -16 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 15 }$ by the constant term $\color{blue}{c = -16} $.

$$ a \cdot c = -240 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -240 $ and add to $ b = 8 $.

Step 4: All pairs of numbers with a product of $ -240 $ are:

PRODUCT = -240
-1     240	1     -240
-2     120	2     -120
-3     80	3     -80
-4     60	4     -60
-5     48	5     -48
-6     40	6     -40
-8     30	8     -30
-10     24	10     -24
-12     20	12     -20
-15     16	15     -16

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 8 }$

PRODUCT = -240 and SUM = 8
-1     240	1     -240
-2     120	2     -120
-3     80	3     -80
-4     60	4     -60
-5     48	5     -48
-6     40	6     -40
-8     30	8     -30
-10     24	10     -24
-12     20	12     -20
-15     16	15     -16

Step 6: Replace middle term $ 8 x $ with $ 20x-12x $:

$$ 15x^{2}+8x-16 = 15x^{2}+20x-12x-16 $$

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

$$ 15x^{2}+20x-12x-16 = 5x\left(3x+4\right) -4\left(3x+4\right) = \left(5x-4\right) \left(3x+4\right) $$

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