Step 1 :
To factor $ 64y^{3}+125 $ we can use sum of cubes formula:
$$ I^3 - II^3 = (I + II)(I^2 - I \cdot II + II^2) $$After putting $ I = 4y $ and $ II = 5 $ , we have:
$$ 64y^{3}+125 = ( 4y+5 ) ( 16y^{2}-20y+25 ) $$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 $\color{blue}{ a = 16 }$ by the constant term $\color{blue}{c = 25} $.
$$ a \cdot c = 400 $$Step 4: Find out two numbers that multiply to $ a \cdot c = 400 $ and add to $ b = -20 $.
Step 5: All pairs of numbers with a product of $ 400 $ are:
PRODUCT = 400 | |
1 400 | -1 -400 |
2 200 | -2 -200 |
4 100 | -4 -100 |
5 80 | -5 -80 |
8 50 | -8 -50 |
10 40 | -10 -40 |
16 25 | -16 -25 |
20 20 | -20 -20 |
Step 6: Find out which factor pair sums up to $\color{blue}{ b = -20 }$
Step 7: Because none of these pairs will give us a sum of $ \color{blue}{ -20 }$, we conclude the polynomial cannot be factored.