Step 1 :
After factoring out $ 5 $ we have:
$$ 5x^{3}+40 = 5 ( x^{3}+8 ) $$Step 2 :
To factor $ x^{3}+8 $ 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 = 2 $ , we have:
$$ x^{3}+8 = ( x+2 ) ( x^{2}-2x+4 ) $$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:
$$ \color{blue}{ b = -2 } ~ \text{ and } ~ \color{red}{ c = 4 }$$Now we must discover two numbers that sum up to $ \color{blue}{ -2 } $ and multiply to $ \color{red}{ 4 } $.
Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 4 }$.
PRODUCT = 4 | |
1 4 | -1 -4 |
2 2 | -2 -2 |
Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ -2 }$, we conclude the polynomial cannot be factored.