Factor polynomial $$ 8a^{3}+1 $$

$$ 8a^{3}+1 = ( 2a+1 )( 4a^{2}-2a+1 ) $$

Step 1 :

To factor $ 8a^{3}+1 $ we can use sum of cubes formula:

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

After putting $ I = 2a $ and $ II = 1 $ , we have:

$$ 8a^{3}+1 = ( 2a+1 ) ( 4a^{2}-2a+1 ) $$

Step 2 :

Step 2: Identify constants $ a $ , $ b $ and $ c $.
$ a $ is a number in front of the $ x^2 $ term $ b $ is a number in front of the $ x $ term and $ c $ is a constant. In this case:

$$ a = 4 , b = -2 ~ \text{ and } ~ c = 1 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 4 }$ by the constant term $\color{blue}{c = 1} $.

$$ a \cdot c = 4 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 4 $ and add to $ b = -2 $.

Step 5: All pairs of numbers with a product of $ 4 $ are:

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

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -2 }$

Step 7: 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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