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Thread: Rearranging Algebra

  1. #1
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    Rearranging Algebra

    Can someone show me how $\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3$ can be factorise or rearrange into $\displaystyle \frac{1}{4}(k+1)^2(k^2+4(k+1))$

    thanks
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  2. #2
    Super Member Anonymous1's Avatar
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    Sure.

    $\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)^3 = \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$ $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$
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  3. #3
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    Quote Originally Posted by Anonymous1 View Post
    Sure.

    $\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)^3 = \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$ $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$
    how did you go from

    Code:
    $\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$
    to

    Code:
    $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1))$
    and what did you do this bit


    $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$

    im quite un sure what's going on.

    thanks
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  4. #4
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    Hi BabyMilo,

    $\displaystyle (k+1)^2$ is common to both terms, hence

    $\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

    and for the last piece

    $\displaystyle k+1=\frac{4}{4}(k+1)=4\left(\frac{k+1}{4}\right)$
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Hi BabyMilo,

    $\displaystyle (k+1)^2$ is common to both terms, hence

    $\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

    and for the last piece

    $\displaystyle k+1=\frac{4}{4}(k+1)=4\left(\frac{k+1}{4}\right)$

    i get this bit now.

    $\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

    but still quite unsure how you get the last part.

    thanks as always.
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  6. #6
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    If you examine with numbers first, then it will be clearer with "k" introduced.

    $\displaystyle 3=\frac{4}{4}(3)=\frac{4(3)}{4}=\frac{12}{4}$

    If we divide any value by 4 and then multiply by 4, that will bring us back to the original value.

    Hence

    $\displaystyle \frac{4x}{4}=x$

    In terms of the example you have

    $\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^ 2+(k+1)(k+1)^2$

    $\displaystyle (k+1)^2$ is a common factor,

    and to make $\displaystyle \frac{1}{4}$ a common factor also, we write $\displaystyle (k+1)$ as $\displaystyle \frac{1}{4}\left(4[k+1]\right)$

    Hence we get $\displaystyle \frac{1}{4}(k+1)^2$ as a common factor

    $\displaystyle \color{red}\frac{1}{4}(k+1)^2\color{black}k^2+\col or{red}\frac{1}{4}(k+1)^2\color{black}(4[k+1])=\color{red}\frac{1}{4}(k+1)^2\color{black}\left( k^2+4[k+1]\right)$
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    If you examine with numbers first, then it will be clearer with "k" introduced.

    $\displaystyle 3=\frac{4}{4}(3)=\frac{4(3)}{4}=\frac{12}{4}$

    If we divide any value by 4 and then multiply by 4, that will bring us back to the original value.

    Hence

    $\displaystyle \frac{4x}{4}=x$

    In terms of the example you have

    $\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^ 2+(k+1)(k+1)^2$

    $\displaystyle (k+1)^2$ is a common factor,

    and to make $\displaystyle \frac{1}{4}$ a common factor also, we write $\displaystyle (k+1)$ as $\displaystyle \frac{1}{4}\left(4[k+1]\right)$

    Hence we get $\displaystyle \frac{1}{4}(k+1)^2$ as a common factor

    $\displaystyle \color{red}\frac{1}{4}(k+1)^2\color{black}k^2+\col or{red}\frac{1}{4}(k+1)^2\color{black}(4[k+1])=\color{red}\frac{1}{4}(k+1)^2\color{black}\left( k^2+4[k+1]\right)$
    thanks made it much easier to understand.
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