Factor polynomial $$ 3x^{2}-13x-56 $$

$$ 3x^{2}-13x-56 = ( x-7 )( 3x+8 ) $$

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:

$$ a = 3 , b = -13 ~ \text{ and } ~ c = -56 $$

Step 2: Multiply the leading coefficient $\color{blue}{ a = 3 }$ by the constant term $\color{blue}{c = -56} $.

$$ a \cdot c = -168 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -168 $ and add to $ b = -13 $.

Step 4: All pairs of numbers with a product of $ -168 $ are:

PRODUCT = -168
-1     168	1     -168
-2     84	2     -84
-3     56	3     -56
-4     42	4     -42
-6     28	6     -28
-7     24	7     -24
-8     21	8     -21
-12     14	12     -14

Step 5: Find out which factor pair sums up to $\color{blue}{ b = -13 }$

PRODUCT = -168 and SUM = -13
-1     168	1     -168
-2     84	2     -84
-3     56	3     -56
-4     42	4     -42
-6     28	6     -28
-7     24	7     -24
-8     21	8     -21
-12     14	12     -14

Step 6: Replace middle term $ -13 x $ with $ 8x-21x $:

$$ 3x^{2}-13x-56 = 3x^{2}+8x-21x-56 $$

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

$$ 3x^{2}+8x-21x-56 = x\left(3x+8\right) -7\left(3x+8\right) = \left(x-7\right) \left(3x+8\right) $$

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