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Math Help - need help for some factoring problems?

  1. #1
    Newbie ibirsen's Avatar
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    need help for some factoring problems?

    the first question is : (21.22.23.24+1)^(1/2) =? (without using calculator)

    A few more questions has been added to the end. thnx for all answers.
    Last edited by ibirsen; November 29th 2010 at 02:54 PM. Reason: explanation
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  2. #2
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    Quote Originally Posted by ibirsen View Post
    the first question is : (21.22.23.24+1)^(1/2) =? (without using calculator)
    is this the expression you posted? please clarify ...

    \left(21 \cdot 22 \cdot 23 \cdot 24 + 1\right)^{\frac{1}{2}}
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    Quote Originally Posted by skeeter View Post
    is this the expression you posted? please clarify ...

    \left(21 \cdot 22 \cdot 23 \cdot 24 + 1\right)^{\frac{1}{2}}
    If it is, then use \left(a+b\right)^n = a^n + 2ab + b^n so substitute \left(21 \cdot 22 \cdot 23 \cdot 24\right) for a, 1 for b and 1/2 for n.
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    Quote Originally Posted by skeeter View Post
    is this the expression you posted? please clarify ...

    \left(21 \cdot 22 \cdot 23 \cdot 24 + 1\right)^{\frac{1}{2}}
    If it is, then use \left(a+b\right)^n = a^n + 2ab + b^n so substitute \left(21 \cdot 22 \cdot 23 \cdot 24\right) for a, 1 for b and 1/2 for n.
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    Quote Originally Posted by Manfriedthetalkingpinata View Post
    If it is, then use \left(a+b\right)^n = a^n + 2ab + b^n so substitute \left(21 \cdot 22 \cdot 23 \cdot 24\right) for a, 1 for b and 1/2 for n.
    This is incorrect. (a+ b)^n= a^n+ 2ab+ b^n only if n= 2, which is not the case here.

    If we call that number "x", then 21(22)(23)(24)+ 1= x^2 so that x^2- 1= (x- 1)(x+ 1)= 21(22)(23)(24). Now notice that 21(24)= 504 and 22*23= 506.
    Last edited by HallsofIvy; November 30th 2010 at 05:00 AM.
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  6. #6
    Newbie ibirsen's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. (a+ b)^n= a^n+ 2ab+ b^n only if n= 2, which is not the case here.

    If we call that number "x", then 21(22)(23)(24)+ 1= x^2 so that x^2- 1= (x- 1)(x+ 1)= 21(22)(23(24). Now notice that 21(24)= 504 and 22*23= 506.
    Thank you for this very good solution
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  7. #7
    Newbie ibirsen's Avatar
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    I'm trying to solve the questions of factorization so i have got afew questions. I also wanted to continue here in order to avoid a new topic.
       y^3+1=3y^2
      3y-2=x^3 find y-x?
    A) -2 B) -1 C) 0 D) 1 E) 2
    ******************************************
    find the one of the factor of function 2x^2+y^2+3xy+y-2

    A) x+y+1 B) x+y-1 C) 2x+y+1 D) 2x+y-2 E) x+y+2

    *******************************************
    a , b ,c ∈ R and they r different eachother ;
    find the a+b+c if
    2a+b=3c
     2a^3 + b^3 =3c^3
    A) -2 B) -1 C) 0 D) 1 E)-2
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