Equation Troubleshooting: 2r^3 + 6r^2 = 20r

[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?

 

[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!

 

[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 don't 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?

 

[VietDao29]

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 don't 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?

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 textbook. 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... :)

 

[SolidFist]

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

 

[HallsofIvy]

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 don't 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?

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.

 

[Plastic Photon]

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

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

 

[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.

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.

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?

 

[Plastic Photon]

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?

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.

 

[HallsofIvy]

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?

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

 

[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