# Equation Help! 2r^3 + 6r^2=20r

1. Feb 5, 2006

### ramstin

O.k. I am stuck on this equation 2r^3 + 6r^2=20r
The answer to the equation is -5,0,2(according to my book I have not got it yet).

so far I got to 2(1r^2+5r)(r^2-2r) From here I keep getting different answers that make no sense.I can not figure out how to make the greatist common factor work into the answers. Any help?

2. Feb 5, 2006

### HallsofIvy

Excuse me?
Do you not see how to factor r2+ 5r and r2- 2r?

Hint: they both have the same factor which is especially easy to see knowing that r= 0 is a solution to the equation!

3. Feb 5, 2006

### ramstin

I get r^2+5r=0 r^2-2r=0
5r^3=0 -2r=0
5r^3/r^3=0 -2r/r=0
5=0 -2=0
5,2 is my final answer. I dont see what to do with my GCF of 2 or how 5 and -2 becomes -5 and 2 or how a zero ends up with the answer?

4. Feb 5, 2006

### VietDao29

Nooo, I don't really know what you mean...
Having a conclusion like 5 = 0, or -2 = 0, is like to say a monkey is a dog, or a cat is an elephant!!!
Okay, as far as I can tell, please open your text book. Read it again (read the chapters that teach you how to factor, or solve a quadratic equation), and see if you can understand it. If you don't understand some parts, just post it here.
Be sure that you understand all concepts before solving a problem.
Now, it's time to do some reading... :)

5. Feb 5, 2006

### SolidFist

All you have to do is factor out a r which gives you r=0.
Then what you are left with is $$2r^2+6r-20=0$$

Last edited: Feb 5, 2006
6. Feb 5, 2006

### HallsofIvy

I have no idea what you are doing! How do you get from
r2+ 5r= 0 to 5r3= 0??

For that matter, although I didn't notice it before, how did you get from the original equation, 2r3 + 6r2=20r, which is the same as 2r3+ 6r2- 20r, to 2(r2+5r)(r2-2r)?? Surely you can see that, if you multiply them together, you will get r4 as leading term, not r3.

From 2r3+ 6r2- 20r you should immediately see that there is at least one "r" in each term and so you can factor as
2r(r2+ 3r- 10). Now you might note that 10= 2(5) and 5- 2= 3. That is, r2+ 3r- 10= (r+ 5)(r- 2).
That is, 2r3+ 5r2- 20r= 2r(r+5)(r-2)= 0.

If r2+ 5r= r(r+ 5)= 0 then either r= 0 or r+ 5= 0. In other words, r= 0 and r= -5 are solutions.

If r2- 2r= r(r- 2)= 0 then either r= 0 or r- 2= 0. In other words, r= 0 and r= 2 are solutions.

7. Feb 5, 2006

### Plastic Photon

That is not necessary since once 2 is divide from the equation it may be factored.

8. Feb 5, 2006

### ramstin

O.k I went back over the problem. I then see how I forgot to take into account the r as part of my GCF.

Just wondering how did you get from 2r in 2r(r+5)(r-2)= 0 to r2+ 5r= r(r+ 5)= 0? I thought 2r* r = 2r^3?

Last edited: Feb 5, 2006
9. Feb 5, 2006

### Plastic Photon

rxr=r^2

(r+5)(r-2)=r²-2r+5r-10. multipling by 2r yields 2r^3-4r²+10r²-20r=2r^3+6r²-20r
or 2r(r+5)=2r²+10r. (2r²+10r)(r-2)=2r^3-4r²+10r²-20r=2r^3+6r²-20r

You can factor both an r and 2 from the equation which leaves you with r²+6r-10 which can further be factored into (r+5)(r-2) yielding the result of r=-5 r=2.

10. Feb 6, 2006

### HallsofIvy

If you think 2r*r= 2r3 rather than 2r2 you need to review the meaning of powers.

11. Feb 6, 2006

### ramstin

I see what I was doing wrong now. I forgot to multiply in the GCF and I had a ton of small errors. Thanks