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Thread: Polynomials & more....Need help ASAP

  1. #1
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    Unhappy Polynomials & more....Need help ASAP

    Hi, I'm new to this forum. Currently taking math since it's a required pre-requisite. I must pass in order to move onto my college core classes. Since it's been over 10 years since I've done math (in high school) I am at a complete loss and feel like a REAL idiot when I'm in class.

    Here at the problems I am stumped on:

    (5x3+15x2-25x) 5x

    (a9)-3 (a-4)7

    -4w3-16w2+20w

    b3-216

    u3+125v3

    2h2-h-3=0

    2w(4w+1)=1

    m3+2m2-3m=0

    3x(2x+1)=18

    Please help???
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  2. #2
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    Nov 2007
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    Exclamation Correction...

    I just realized in my original post - the exponents aren't coming up. You can email me to get the correct equations??

    devonca@gmail.com
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  3. #3
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    Lexington, MA (USA)
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    Hello, devonca!

    Welcome aboard!
    I think I know what you meant by these problems.

    I'll take baby-steps in these problems.
    I hope this is an effective review for you.


    Simplify: .$\displaystyle \frac{5x^3+15x^2-25x}{5x}$
    Make three fractions: .$\displaystyle \frac{5x^3}{5x} + \frac{15x^2}{5x} - \frac{25x}{5x}$

    . . and reduce: .$\displaystyle x^2 + 3x - 5$



    Simplify: .$\displaystyle \frac{(a^9)^{-3}}{(a^{-4})^7}$
    We have: .$\displaystyle \frac{a^{-27}}{a^{-28}} \;=\;a^{(-27) - (-28)} \;=\;a^1 \;=\;a$


    Factor: .$\displaystyle -4w^3 - 16w^2 + 20w$
    We have: .$\displaystyle -4w(w^2 + 4w - 5) \;=\;-4w(w-1)(w+5)$


    Factor: .$\displaystyle b^3-216$ . . . . difference of cubes
    We have: .$\displaystyle (b)^3 - (6)^3$

    Therefore: .$\displaystyle (b - 6)(b^2 + 6b + 36)$



    Factor: .$\displaystyle u^3+125v^3$ . . . . sum of cubes
    We have: .$\displaystyle (u)^3 + (5v)^3$

    Therefore: .$\displaystyle (u + 5v)(u^2 - 5uv + 25v^2)$



    Solve: .$\displaystyle 2h^2 - h - 3\:=\:0$
    Factor: .$\displaystyle (h + 1)(2h - 3) \:=\:0$

    Then: .$\displaystyle \begin{array}{ccccccc}h + 1 & = & 0 & \Rightarrow & h & = &\text{-}1 \\
    2h - 3 & = & 0 & \Rightarrow & h & = & \frac{3}{2} \end{array}$



    Solve: .$\displaystyle 2w(4w+1)\:=\:1$
    We have: .$\displaystyle 8w^2 + 2w \:=\:1\quad\Rightarrow\quad 8w^2 + 2w - 1 \:=\:0$

    . . which factors: .$\displaystyle (2w-1(4w+1) \:=\:0$

    . . and has roots: .$\displaystyle \begin{array}{ccccccc}2w-1 & = & 0 & \Rightarrow & w & = & \frac{1}{2} \\
    4w + 1 & = & 0 & \Rightarrow & w & = & \text{-}\frac{1}{4}\end{array}$



    Solve: .$\displaystyle m^3 + 2m^2 - 3m\:=\:0$
    Factor: .$\displaystyle m(m^2 + 2m - 3)\:=\:0\quad\Rightarrow\quad m(m-1)(m+3) \:=\:0$

    And we have three roots: .$\displaystyle m \;=\;0,\:1,\:-3$



    Solve: .$\displaystyle 3x(2x+1)\:=\:18$

    We have: .$\displaystyle 6x^2 + 3x \:=\:18\quad\Rightarrow\quad 6x^2 + 3x - 18\:=\:0$

    . . which factors: .$\displaystyle 3(2x-3)(x+2) \:=\:0$

    . . and has roots: .$\displaystyle x \;=\;\frac{3}{2},\:-2$

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