Step 1 :
To factor $ 27a^{3}+64 $ we can use sum of cubes formula:
$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$After putting $ I = 3a $ and $ II = 4 $ , we have:
$$ 27a^{3}+64 = ( 3a+4 ) ( 9a^{2}-12a+16 ) $$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:
Step 3: Multiply the leading coefficient $\color{blue}{ a = 9 }$ by the constant term $\color{blue}{c = 16} $.
$$ a \cdot c = 144 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 144 $ and add to $ b = -12 $.
Step 5: All pairs of numbers with a product of $ 144 $ are:
PRODUCT = 144 | |
1 144 | -1 -144 |
2 72 | -2 -72 |
3 48 | -3 -48 |
4 36 | -4 -36 |
6 24 | -6 -24 |
8 18 | -8 -18 |
9 16 | -9 -16 |
12 12 | -12 -12 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -12 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -12 }$, we conclude the polynomial cannot be factored.