SOLUTION: y^3 =125 solve the equation for y. seeing if i am doing this right. my answer was y=5

Question 139381: y^3 =125 solve the equation for y. seeing if i am doing this right. my answer was y=5
Found 2 solutions by checkley77, solver91311:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
y^3=125
y=cuberoot125
y=cuberoot(5*5*5)
y=5 answer.

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Your answer is correct -- as far as it goes. If your assignment is to find just the real number roots, then you are done. But if you were asked to completely solve the equation, including finding any complex number roots, you are a long way from done.

The Fundamental Theorem of Algebra tells us a couple of important things in this context. First, any polynomial equation of degree n has exactly n roots. You have a 3rd degree equation, so there must be 3 roots. Furthermore, if is a root of the polynomial equation, then is a factor of the polynomial. That is to say that, given a polynomial function , if , then there exists a polynomial function such that .

Knowing that 5 is one of the roots of your equation, and we know this because and , makes the task of finding the other two roots much simpler. Solving the general cubic is a horror that I wouldn't wish on anyone, even if I particularly disliked them.

We can use the fact that is a factor of and the process of polynomial long division to find the other factor of which will be a quadratic polynomial. But first we need to put into standard form by putting in placeholders for the zero coefficient terms, thus:

goes into times. is the first term of the quotient.

times is . Subtract this from the first two terms of to get (remember: change the sign and add).

Bring down the .

goes into times. is the second term of the quotient.

times is . Subtract this from to get .

Bring down the .

goes into times. is the third term of the quotient.

times is . Subtract this from to get -- meaning no remainder.

The quotient polynomial is then , and the zeros of this polynomial are the other two zeros of your original polynomial.

Using the quadratic formula:

or

Where is the imaginary number defined by

In summary, your three roots are:

, ,
RELATED QUESTIONS
I would love it if you would tell me I'm doing this right, however, if I am not please... (answered by ankor@dixie-net.com)
What is the equation for the line represented by this graph? (I don't know how to put a... (answered by checkley71)
Hi! I am not sure if I got my answer right. The problem is Transforming Equations:... (answered by vleith)
Solve the equation: 4-5(y+2)=2(y-1)-7y-4 I keep attempting to solve for y and I keep... (answered by solver91311)
How do you work out x^2 + y^2 = 25; (3, -4) I am doing practice for my math 3 class.... (answered by Alan3354,MathTherapy)
I am having trouble using the substitute method to solve the simultaneous equation:... (answered by ReadingBoosters)
This is a question from my review sheet for my final also.. not sure if I am doing this... (answered by lyra)
I am stuck on this problem. I did the equation and my answer was y=0,y=9. The problem is... (answered by Fombitz,stanbon,josmiceli)
solve the equation for y (5/y-3)+ (10/y^2-y-6) = (y/y+2)] (five over y-3) plus (10... (answered by solver91311,mananth)