Factor polynomial $$ 2z^{2}+9z-56 $$

$$ 2z^{2}+9z-56 = ( 2z-7 )( z+8 ) $$

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:

$$ a = 2 , b = 9 ~ \text{ and } ~ c = -56 $$

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

$$ a \cdot c = -112 $$

Step 3: Find out two numbers that multiply to $ a \cdot c = -112 $ and add to $ b = 9 $.

Step 4: All pairs of numbers with a product of $ -112 $ are:

PRODUCT = -112
-1     112	1     -112
-2     56	2     -56
-4     28	4     -28
-7     16	7     -16
-8     14	8     -14

Step 5: Find out which factor pair sums up to $\color{blue}{ b = 9 }$

PRODUCT = -112 and SUM = 9
-1     112	1     -112
-2     56	2     -56
-4     28	4     -28
-7     16	7     -16
-8     14	8     -14

Step 6: Replace middle term $ 9 x $ with $ 16x-7x $:

$$ 2x^{2}+9x-56 = 2x^{2}+16x-7x-56 $$

Step 7: Apply factoring by grouping. Factor $ 2x $ out of the first two terms and $ -7 $ out of the last two terms.

$$ 2x^{2}+16x-7x-56 = 2x\left(x+8\right) -7\left(x+8\right) = \left(2x-7\right) \left(x+8\right) $$

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