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Homework Help: Equation Help! 2r^3 + 6r^2=20r

  1. Feb 5, 2006 #1
    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. jcsd
  3. Feb 5, 2006 #2


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    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!
  4. Feb 5, 2006 #3
    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?
  5. Feb 5, 2006 #4


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    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... :)
  6. Feb 5, 2006 #5
    All you have to do is factor out a r which gives you r=0.
    Then what you are left with is [tex]2r^2+6r-20=0[/tex]
    After that do quadratic equation.
    Last edited: Feb 5, 2006
  7. Feb 5, 2006 #6


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    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.
  8. Feb 5, 2006 #7
    That is not necessary since once 2 is divide from the equation it may be factored.
  9. Feb 5, 2006 #8
    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
  10. Feb 5, 2006 #9

    (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.
  11. Feb 6, 2006 #10


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    If you think 2r*r= 2r3 rather than 2r2 you need to review the meaning of powers.
  12. Feb 6, 2006 #11
    I see what I was doing wrong now. I forgot to multiply in the GCF and I had a ton of small errors. Thanks:approve:
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