bulgylau
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#1
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#1
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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ServantOfMorgoth
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#2
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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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bulgylau
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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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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

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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bulgylau
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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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Zacken
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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).
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bulgylau
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#7
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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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ServantOfMorgoth
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(Original post by homeland.lsw)
:yy: makes sense...!
You're welcome

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bulgylau
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(Original post by ServantOfMorgoth)
You're welcome

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LOL
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