Factor polynomial $$ 15c^{3}+10c^{2}-40c $$

$$ 15c^{3}+10c^{2}-40c = 5c( 3c-4 )( c+2 ) $$

Step 1 :

After factoring out $ 5c $ we have:

$$ 15c^{3}+10c^{2}-40c = 5c ( 3c^{2}+2c-8 ) $$

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:

$$ a = 3 , b = 2 ~ \text{ and } ~ c = -8 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -8} $.

$$ a \cdot c = -24 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -24 $ and add to $ b = 2 $.

Step 5: All pairs of numbers with a product of $ -24 $ are:

PRODUCT = -24
-1     24	1     -24
-2     12	2     -12
-3     8	3     -8
-4     6	4     -6

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 2 }$

PRODUCT = -24 and SUM = 2
-1     24	1     -24
-2     12	2     -12
-3     8	3     -8
-4     6	4     -6

Step 7: Replace middle term $ 2 x $ with $ 6x-4x $:

$$ 3x^{2}+2x-8 = 3x^{2}+6x-4x-8 $$

Step 8: Apply factoring by grouping. Factor $ 3x $ out of the first two terms and $ -4 $ out of the last two terms.

$$ 3x^{2}+6x-4x-8 = 3x\left(x+2\right) -4\left(x+2\right) = \left(3x-4\right) \left(x+2\right) $$

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