Factor polynomial $$ x^{3}-8 $$

$$ x^{3}-8 = ( x-2 )( x^{2}+2x+4 ) $$

Step 1 :

To factor $ x^{3}-8 $ 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 = 2 $ , we have:

$$ x^{3}-8 = ( x-2 ) ( x^{2}+2x+4 ) $$

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:

$$ \color{blue}{ b = 2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 2 } $ and multiply to $ \color{red}{ 4 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.

PRODUCT = 4
1     4	-1     -4
2     2	-2     -2

Step 4: Because none of these pairs will give us a sum of $ \color{blue}{ 2 }$, we conclude the polynomial cannot be factored.

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