SOLUTION: Will somepne please help me solve this problem? fector completely: y^3-27

Algebra ->  Real-numbers -> SOLUTION: Will somepne please help me solve this problem? fector completely: y^3-27      Log On


   



Question 110148: Will somepne please help me solve this problem? fector completely: y^3-27
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
You first need to realize that the problem is the difference of two cubes, i.e. the cube root of y%5E3 is exactly y, and the cube root of 27 is exactly 3.
:
The factorization of the difference of two cubes follows this pattern:
a%5E3-b%5E3=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29
:
So, substituting:
:
y%5E3-27=%28y-3%29%28y%5E2%2B3y%2B9%29
:
Which is the final answer because neither of the factors can be factored any further. How can you tell? Look at the second factor,%28y%5E2%2B3y%2B9%29, then calculate b%5E2-4ac where a, b, and c are the coefficients of the first, second, and third terms -- in this case, they are 1, 3, and 9. In this case b%5E2-4ac is a negative number. Anytime b%5E2-4ac is negative or not a perfect square, then the expression cannot be factored.
:
By the way, the sum of two cubes is factored as follows:
a%5E3%2Bb%5E3=%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29