Factor polynomial $$ 20x^{2}+27x-8 $$

$$ 20x^{2}+27x-8 = ( 4x-1 )( 5x+8 ) $$

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:

$$ a = 20 , b = 27 ~ \text{ and } ~ c = -8 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 20 }$ by the constant term $\color{blue}{c = -8} $.

$$ a \cdot c = -160 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -160 $ and add to $ b = 27 $.

Step 4: All pairs of numbers with a product of $ -160 $ are:

PRODUCT = -160
-1     160	1     -160
-2     80	2     -80
-4     40	4     -40
-5     32	5     -32
-8     20	8     -20
-10     16	10     -16

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 27 }$

PRODUCT = -160 and SUM = 27
-1     160	1     -160
-2     80	2     -80
-4     40	4     -40
-5     32	5     -32
-8     20	8     -20
-10     16	10     -16

Step 6: Replace middle term $ 27 x $ with $ 32x-5x $:

$$ 20x^{2}+27x-8 = 20x^{2}+32x-5x-8 $$

Step 7: Apply factoring by grouping. Factor $ 4x $ out of the first two terms and $ -1 $ out of the last two terms.

$$ 20x^{2}+32x-5x-8 = 4x\left(5x+8\right) -1\left(5x+8\right) = \left(4x-1\right) \left(5x+8\right) $$

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