Step 1 :
To factor $ 125x^{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 = 5x $ and $ II = 2 $ , we have:
$$ 125x^{3}+8 = ( 5x+2 ) ( 25x^{2}-10x+4 ) $$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:
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.