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    I, as you know, am stupid...
    I really have no idea how to go about this question? Do I expand the brackets?

    Factorise fully: (w+4)^3 - (w+4)^2(w+1)

    Zacken
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    (Original post by homeland.lsw)
    I, as you know, am stupid...
    I really have no idea how to go about this question? Do I expand the brackets?

    Factorise fully: (w+4)^3 - (w+4)^2(w+1)

    Zacken
    Just factor out (w + 4)^2

    Cba with latex I'm too tired.

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    (Original post by ServantOfMorgoth)
    Just factor out (w + 4)^2

    Cba with latex I'm too tired.

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    :borat: What do you mean factor out?
    Ohhhhhhhhhh...one sec
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    (Original post by homeland.lsw)
    I, as you know, am stupid...
    I really have no idea how to go about this question? Do I expand the brackets?

    Factorise fully: (w+4)^3 - (w+4)^2(w+1)

    Zacken

    If it helps you to see, it - why don't you set u = (w+4)^2? Then since (w +4)^3 = (w+4)^2(w+4) you have:

    \displaystyle 

\begin{equation*}u(w+4) - u(w+1) = u(w+4 - w - 1) = u(3) \end{equation*}

    and then back-substitute what you initially set u to be.
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    (Original post by Zacken)
    If it helps you to see, it - why don't you set u = (w+4)^2? Then since (w +4)^3 = (w+4)^2(w+4) you have:

    \displaystyle 

\begin{equation*}u(w+4) - u(w+1) = u(w+4 - w - 1) = u(3) \end{equation*}

    and then back-substitute what you initially set u to be.
    Erm...ok? I'll try that...
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    (Original post by homeland.lsw)
    Erm...ok? I'll try that...
    You know how you have x^2 + 2x you can factor that out as x(x + 2)? This is basically the same thing, except instead of one thing in common (x) you have a bracket in common (w+4).
    • Thread Starter
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    (Original post by Zacken)
    You know how you have x^2 + 2x you can factor that out as x(x + 2)? This is basically the same thing, except instead of one thing in common (x) you have a bracket in common (w+4).
    :yy: makes sense...!
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    (Original post by homeland.lsw)
    :yy: makes sense...!
    You're welcome

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    (Original post by ServantOfMorgoth)
    You're welcome

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    LOL
 
 
 
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