Factor polynomial $$ 2x^{3}-686 $$

$$ 2x^{3}-686 = 2( x-7 )( x^{2}+7x+49 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2x^{3}-686 = 2 ( x^{3}-343 ) $$

Step 2 :

To factor $ x^{3}-343 $ we can use difference of cubes formula:

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

After putting $ I = x $ and $ II = 7 $ , we have:

$$ x^{3}-343 = ( x-7 ) ( x^{2}+7x+49 ) $$

Step 3 :

Step 3: 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:

$$ \color{blue}{ b = 7 } ~ \text{ and } ~ \color{red}{ c = 49 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 7 } $ and multiply to $ \color{red}{ 49 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 49 }$.

PRODUCT = 49
1     49	-1     -49
7     7	-7     -7

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 7 }$, we conclude the polynomial cannot be factored.

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