SOLUTION: what is the answer to 2y(cubed)-18y(squared)+28y=0

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Question 138394: what is the answer to 2y(cubed)-18y(squared)+28y=0
Found 2 solutions by checkley77, solver91311:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
2y^3-18y^2+28y
2y(y^2-9y+14)
2y(y-7)(y-2) answer.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Your question is more correctly put: "What are the answers to..." Since you have a cubic, or 3rd degree polynomial equation, the Fundamental Theorem of Algebra tells us there must be three roots.

In general terms, a cubic equation is a rather tiresome thing to have to solve. Fortunately, however, your equation has a pair of common factors in every term:

2y%5E3-18y%5E2%2B28y=0

You can factor out a 2 and a y from every term:

2y%28y%5E2-9y%2B14%29=0

You can divide both sides by 2 to get:

y%28y%5E2-9y%2B14%29=0

Now, the Zero Product Rule tells us that either y=0 or y%5E2-9y%2B14=0

But notice that y%5E2-9y%2B14=0 is just a quadratic equation, and an easily factorable one at that.

So, 0 is one of the elements of your solution set, and the roots of the quadratic are the other two.