Factor polynomial $$ 6x^{3}-10x^{2}-56x $$

$$ 6x^{3}-10x^{2}-56x = 2x( x-4 )( 3x+7 ) $$

Step 1 :

After factoring out $ 2x $ we have:

$$ 6x^{3}-10x^{2}-56x = 2x ( 3x^{2}-5x-28 ) $$

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:

$$ a = 3 , b = -5 ~ \text{ and } ~ c = -28 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -28} $.

$$ a \cdot c = -84 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = -84 $ and add to $ b = -5 $.

Step 5: All pairs of numbers with a product of $ -84 $ are:

PRODUCT = -84
-1     84	1     -84
-2     42	2     -42
-3     28	3     -28
-4     21	4     -21
-6     14	6     -14
-7     12	7     -12

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -5 }$

PRODUCT = -84 and SUM = -5
-1     84	1     -84
-2     42	2     -42
-3     28	3     -28
-4     21	4     -21
-6     14	6     -14
-7     12	7     -12

Step 7: Replace middle term $ -5 x $ with $ 7x-12x $:

$$ 3x^{2}-5x-28 = 3x^{2}+7x-12x-28 $$

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

$$ 3x^{2}+7x-12x-28 = x\left(3x+7\right) -4\left(3x+7\right) = \left(x-4\right) \left(3x+7\right) $$

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