Factor polynomial $$ 125a^{3}+8 $$

$$ 125a^{3}+8 = ( 5a+2 )( 25a^{2}-10a+4 ) $$

Step 1 :

To factor $ 125a^{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 = 5a $ and $ II = 2 $ , we have:

$$ 125a^{3}+8 = ( 5a+2 ) ( 25a^{2}-10a+4 ) $$

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:

$$ a = 25 , b = -10 ~ \text{ and } ~ c = 4 $$

Step 3: Multiply the leading coefficient $\color{blue}{ a = 25 }$ by the constant term $\color{blue}{c = 4} $.

$$ a \cdot c = 100 $$

Step 4: Find out two numbers that multiply to $ a \cdot c = 100 $ and add to $ b = -10 $.

Step 5: All pairs of numbers with a product of $ 100 $ are:

PRODUCT = 100
1     100	-1     -100
2     50	-2     -50
4     25	-4     -25
5     20	-5     -20
10     10	-10     -10

Step 6: Find out which factor pair sums up to $\color{blue}{ b = -10 }$

Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -10 }$, we conclude the polynomial cannot be factored.

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