Factor polynomial $$ -20y^{2}-96y-64 $$

$$ -20y^{2}-96y-64 = ( -4 )( 5y+4 )( y+4 ) $$

Step 1 :

After factoring out $ -4 $ we have:

$$ -20y^{2}-96y-64 = -4 ( 5y^{2}+24y+16 ) $$

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:

$$ a = 5 , b = 24 ~ \text{ and } ~ c = 16 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 5 }$ by the constant term $\color{blue}{c = 16} $.

$$ a \cdot c = 80 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 80 $ and add to $ b = 24 $.

Step 5: All pairs of numbers with a product of $ 80 $ are:

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

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 24 }$

PRODUCT = 80 and SUM = 24
1     80	-1     -80
2     40	-2     -40
4     20	-4     -20
5     16	-5     -16
8     10	-8     -10

Step 7: Replace middle term $ 24 x $ with $ 20x+4x $:

$$ 5x^{2}+24x+16 = 5x^{2}+20x+4x+16 $$

Step 8: Apply factoring by grouping. Factor $ 5x $ out of the first two terms and $ 4 $ out of the last two terms.

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

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