Factor polynomial $$ 5y^{2}+12y-32 $$

$$ 5y^{2}+12y-32 = ( 5y-8 )( y+4 ) $$

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:

$$ a = 5 , b = 12 ~ \text{ and } ~ c = -32 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = -32} $.

$$ a \cdot c = -160 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -160 $ and add to $ b = 12 $.

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 $\color{blue}{ b = 12 }$

PRODUCT = -160 and SUM = 12
-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 $ 12 x $ with $ 20x-8x $:

$$ 5x^{2}+12x-32 = 5x^{2}+20x-8x-32 $$

Step 7: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ -8 $ out of the last two terms.

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

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