Factor polynomial $$ -x^{3}+39304 $$

$$ -x^{3}+39304 = ( -x+34 )( x^{2}+34x+1156 ) $$

Step 1 :

To factor $ -x^{3}+39304 $ we can use difference of cubes formula:

$$ I^3 - II^3 = (I - II)(I^2 + I \cdot II + II^2) $$

After putting $ I = 34 $ and $ II = x $ , we have:

$$ -x^{3}+39304 = ( -x+34 ) ( x^{2}+34x+1156 ) $$

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:

$$ \color{blue}{ b = 34 } ~ \text{ and } ~ \color{red}{ c = 1156 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 34 } $ and multiply to $ \color{red}{ 1156 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 1156 }$.

PRODUCT = 1156
1     1156	-1     -1156
2     578	-2     -578
4     289	-4     -289
17     68	-17     -68
34     34	-34     -34

Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ 34 }$, we conclude the polynomial cannot be factored.

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