Step 1 :

After factoring out $ -1 $ we have:

$$ -18z^{2}-39z-7 = - ~ ( 18z^{2}+39z+7 ) $$

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:

$$ a = 18 , b = 39 ~ \text{ and } ~ c = 7 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 18 }$ by the constant term $\color{blue}{c = 7} $.

$$ a \cdot c = 126 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 126 $ and add to $ b = 39 $.

Step 5: All pairs of numbers with a product of $ 126 $ are:

PRODUCT = 126
1     126	-1     -126
2     63	-2     -63
3     42	-3     -42
6     21	-6     -21
7     18	-7     -18
9     14	-9     -14

Step 6: Find out which factor pair sums up to $\color{blue}{ b = 39 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ 39 }$, we conclude the polynomial cannot be factored.

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