Factor polynomial $$ 2d^{3}-432 $$

$$ 2d^{3}-432 = 2( d-6 )( d^{2}+6d+36 ) $$

Step 1 :

After factoring out $ 2 $ we have:

$$ 2d^{3}-432 = 2 ( d^{3}-216 ) $$

Step 2 :

To factor $ d^{3}-216 $ we can use difference of cubes formula:

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

After putting $ I = d $ and $ II = 6 $ , we have:

$$ d^{3}-216 = ( d-6 ) ( d^{2}+6d+36 ) $$

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:

$$ \color{blue}{ b = 6 } ~ \text{ and } ~ \color{red}{ c = 36 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ 6 } $ and multiply to $ \color{red}{ 36 } $.

Step 4: Find out pairs of numbers with a product of $\color{red}{ c = 36 }$.

PRODUCT = 36
1     36	-1     -36
2     18	-2     -18
3     12	-3     -12
4     9	-4     -9
6     6	-6     -6

Step 5: Because none of these pairs will give us a sum of $ \color{blue}{ 6 }$, we conclude the polynomial cannot be factored.

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