Factor polynomial $$ 32y^{5}+108y^{2} $$

$$ 32y^{5}+108y^{2} = 4y^{2}( 2y+3 )( 4y^{2}-6y+9 ) $$

Step 1 :

After factoring out $ 4y^{2} $ we have:

$$ 32y^{5}+108y^{2} = 4y^{2} ( 8y^{3}+27 ) $$

Step 2 :

To factor $ 8y^{3}+27 $ we can use sum of cubes formula:

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

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

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

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:

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

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

$$ a \cdot c = 36 $$

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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