Factor polynomial $$ n^{3}-9n^{2}-1000n $$

$$ n^{3}-9n^{2}-1000n = n( n^{2}-9n-1000 ) $$

Step 1 :

After factoring out $ n $ we have:

$$ n^{3}-9n^{2}-1000n = n ( n^{2}-9n-1000 ) $$

Step 2 :

Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -9 } ~ \text{ and } ~ \color{red}{ c = -1000 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -9 } $ and multiply to $ \color{red}{ -1000 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -1000 }$.

PRODUCT = -1000
-1     1000	1     -1000
-2     500	2     -500
-4     250	4     -250
-5     200	5     -200
-8     125	8     -125
-10     100	10     -100
-20     50	20     -50
-25     40	25     -40

Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ -9 }$, we conclude the polynomial cannot be factored.

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