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Math question about reducing + my attempt but something is wrong

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data

    reduce this one

    f´(x)=(12x^3+12x)*(1/2 x^2+3)+(3x^4+6x^2+1)*(x)

    i have to make all the () disappear and make it shorter

    2. Relevant equations

    the thing i have to use are the normal rules about ^n

    3. The attempt at a solution
    ive tried to reduce it this way but i dont know why i just feel its wrong so if someone could show me how to do step by step

    f´(x)=6x^5+36x^3+6x^2+36+4x^4+7x^2+x
     
  2. jcsd
  3. Jan 29, 2012 #2
    Your answer is not correct. It'll be easier to help you if you show us exactly what you've tried. It's a straightforward problem, just expand the brackets with FOIL and combine like terms.

    [tex]
    a*(b+c+d)+b*(a+g+h) = ab+ac+ad+ab+ag+ah = 2ab + ac + ad + ag + ah
    [/tex]

    That's almost all you have to do.
     
  4. Jan 29, 2012 #3
    yes i know the rules you have posted but my problem is that i cant only say a*(b+c) since i have a*(b^n + c^n)
     
  5. Jan 29, 2012 #4
    I can't really post a solution, it's against forum rules. Show us what you've done, and we can see where you went wrong. Here's a better example:

    [tex]
    x^{2}(3x^{4}+6x^{3}-\frac{1}{2}x^{2}) = 3x^{6}+6x^{5}-\frac{1}{2}x^{4}

    [/tex]
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Of course you can: just set B = b^n and C = c^n; now you have a*(B+C). You need to realize that in rules such as a*(b+c) = a*b + a*c, the a, b and c can be ANY algebraic or numerical quantities.

    RGV
     
  7. Jan 29, 2012 #6
    so if we just use numbers as an example and lets say
    a*(b^n + c^n)

    2*(3^4+5^6) then that mean you are telling me that usally you have to solve the ^ first so it ends up being 2*(81+15625) ? and then 162+31250 ? but it seems so strange how i can use it since i have like x^n ? i dont really know what x is anyway

    f´(x)=(12x^3+12x)*(1/2 x^2+3)+(3x^4+6x^2+1)*(x)

    when i look at the (3x^4+6x^2+1)*(x)
    i dont know what (3x^4)*x is should i say 4x^4 or 3x^5 ? i dont know if my x goes to the 3x or to the ^4
     
  8. Jan 29, 2012 #7
    The rule is:

    [tex]
    ax^{n}\cdot x^{m} = ax^{n+m}
    [/tex]

    Now you can use this to expand your brackets, making sure to preserve the order of operations (Sometimes remembered using BEDMAS : brackets, exponents, division/multiplication, addition/subtraction).
     
  9. Jan 29, 2012 #8
    then it means my (3x^4+6x^2+1)*(x)
    becomes (3x^4+6x^2+1) since there isnt any x^n but only x on the right side
     
  10. Jan 29, 2012 #9

    Ray Vickson

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    Science Advisor
    Homework Helper

    No, I never said what you claim I am saying. I say that 2*(3^4+5^6) equals 2*(3^4) + 2*(5^6). Whether or not you choose to put 3^4 = 81 and/or 5^6 = 15625 before, or after the expansion, or not at all, is purely a matter of personal choice.

    I don't see you problem with having x^n; x^n is just another algebraic quantity, no better or worse than plain x itself. In your example y = x*(3x^4 + 6x^2 + 2), you just expand each term separately, so y = x*(3x^4) + x*(6x^2) + x*(2) = 3x^5 + 6x^3 + 2x. That's all there is to it.

    Finally, your question "i dont know what (3x^4)*x is" puzzles me. You just keep all the x's together and all the numbers separate, so x*(3x^4) = 3 x x^4 = 3x^5. You most certainly cannot get (3x^4)*x = 4x^4 as you suggest; just plug in a value such as x = 1 or x = 2 and see if it works (t doesn't).

    RGV
     
    Last edited: Jan 29, 2012
  11. Jan 29, 2012 #10

    Ray Vickson

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    Science Advisor
    Homework Helper

    No, I never said what you claim I am saying. I say that 2*(3^4+5^6) equals 2*(3^4) + 2*(5^6). Whether or not you choose to put 3^4 = 81 and/or 5^6 = 15625 before, or after the expansion, or not at all, is purely a matter of personal choice.

    I don't see you problem with having x^n; x^n is just another algebraic quantity, no better or worse than plain x itself. In your example y = x*(3x^4 + 6x^2 + 2), you just expand each term separately, so y = x*(3x^4) + x*(6x^2) + x*(2) = 3x^5 + 6x^3 + 2x. That's all there is to it.

    Finally, your question "i dont know what (3x^4)*x is" puzzles me. You just keep all the x's together and all the numbers separate, so x*(3x^4) = 3 x x^4 = 3x^5. You most certainly cannot get (3x^4)*x = 4x^4 as you suggest; just plug in a value such as x = 1 or x = 2 and see if it works (it doesn't).

    RGV
     
  12. Jan 29, 2012 #11
    ok sorry i misunderstood it

    if i have x*2x^6 = 2x^7 which means i just add it to the 6 right and if i have 5x^3*4x^2 itl make 20x^6 ? i really liked your last explanation thank you alot
     
  13. Jan 29, 2012 #12
    Almost there.
    [tex]
    5x^{3}\cdot4x^{2} = (4\cdot5)x^{3+2} = 20x^{5}
    [/tex]

    Here's the difference between adding polynomials and multiplying them:

    Adding:

    [tex]

    3x + 2x + 2x^{2} - x^{2} = (3x+2x)+(2x^{2} - x^{2}) = 5x + x^{2}

    [/tex]

    All the ones with equal powers (Ex: 2x and 3x have equal powers, 1 and 1: 2x=2x^1) are grouped together.
    You can add and subtract monomials/polynomials with the same base (i.e. x^{2} and 2x have the same base, x) AND the same powers. So:

    You can't add/subtract these two any more than this:

    [tex] 2x - 3x^{2} [/tex]

    But you can add/subtract these:

    [tex] 3x+4x [/tex]

    Because they have the same base and the same power.

    Now for multiplying:

    [tex] 3x^{2} \cdot 3x = (3)\cdot x^{2+1} = 3x^{3} [/tex]

    You can multiply two monomials as long as they have the same base ("x" in this example). You cannot further simplify/multiply these two:

    [tex] a^{2} and x^{3} [/tex] because they aren't the same monomials.

    Here's a final example to summarize all of this:

    [tex]


    \begin{align*}
    & x(1+3x+2x^{2}-3x^{4}) + 2x^{2}(1-\frac{1}{2}x^{2}) \\
    & = (x\cdot1)+(x\cdot3x)+(x\cdot2x^{2})+(x\cdot -3x^{4})+(2x^{2}\cdot1)+ (2x^{2}\cdot \frac{-1}{2}x^{2}) \\
    & = x+3x^{1+1}+2x^{2+1}-3x^{1+4} + 2x^{2} - (2\cdot \frac{1}{2})(x^{2+2}) \\
    & = x+ 3x^{2} + 2x^{3} - 3x^{5} + 2x^{2} - x^{4} \\
    & = x + 5x^{2} + 2x^{3} -x^{4} -3x^{5} \\
    \end{align*}


    [/tex]

    Does this help?
     
    Last edited: Jan 29, 2012
  14. Jan 29, 2012 #13
    Does it help ? No it doesnt help

    it does help me in the most perfect way ever !....i am so much sure that your explanation is even better than my book !! i really like the exampels you gave couse you can relate more instead uf just heavy rules ...its easier to understand and remember this way so i thank you very much for your help and assure you that im going to use your explanation as my personal notes to remember and train them so i can learn it all :=) thank you a thousand times again
     
  15. Jan 29, 2012 #14
    You're quite welcome! :smile:
    People here are always happy to help you as long as you've shown some effort. Good luck!
     
  16. Jan 29, 2012 #15
    thank you alot :=)
     
  17. Jan 30, 2012 #16

    NascentOxygen

    User Avatar

    Staff: Mentor


    I make this 9x3
     
  18. Jan 31, 2012 #17
    Oops typo. Thanks for spotting it! However, for some reason I can't go back and edit it.
     
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