Factor polynomial $$ 81b^{3}-24 $$

$$ 81b^{3}-24 = 3( 3b-2 )( 9b^{2}+6b+4 ) $$

Step 1 :

After factoring out $ 3 $ we have:

$$ 81b^{3}-24 = 3 ( 27b^{3}-8 ) $$

Step 2 :

To factor $ 27b^{3}-8 $ we can use difference of cubes formula:

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

After putting $ I = 3b $ and $ II = 2 $ , we have:

$$ 27b^{3}-8 = ( 3b-2 ) ( 9b^{2}+6b+4 ) $$

Step 3 :

Step 3: 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:

$$ a = 9 , b = 6 ~ \text{ and } ~ c = 4 $$

Step 4: Multiply the leading coefficient $\color{blue}{ a = 9 }$ by the constant term $\color{blue}{c = 4} $.

$$ a \cdot c = 36 $$

Step 5: Find out two numbers that multiply to $ a \cdot c = 36 $ and add to $ b = 6 $.

Step 6: All pairs of numbers with a product of $ 36 $ are:

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

Step 7: Find out which factor pair sums up to $\color{blue}{ b = 6 }$

Step 8: 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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