Step 1 :
After factoring out $ 2 $ we have:
$$ 2x^{3}+54 = 2 ( x^{3}+27 ) $$Step 2 :
To factor $ x^{3}+27 $ we can use sum of cubes formula:
$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$After putting $ I = x $ and $ II = 3 $ , we have:
$$ x^{3}+27 = ( x+3 ) ( x^{2}-3x+9 ) $$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:
$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = 9 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ 9 } $.
Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 9 }$.
PRODUCT = 9 | |
1 9 | -1 -9 |
3 3 | -3 -3 |
Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ -3 }$, we conclude the polynomial cannot be factored.