Melody

Yes Ninja, you left out a zero.  

But you can have my thumbs up anyway.

Ninja, I am not sure how well you thought this through.   It is correct but maybe there was a bit of luck involved here.  

"Since a 3/4 yard piece can't be cut out of a 1/3 yard-long piece, let's forget about this 1/3 yard piece"

This quote concerns me a little. There might be some left over from the 12 yard peice and together with the 1/3 yard maybe it would make another one.  Of course if you looked at the left over bit afterwards this would be alright.  That is probably what you were intending to do.

Otherwise it would be just

$$12\frac{1}{3}\div \frac{3}{4}$$

$${\frac{\left({\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}} = {\frac{{\mathtt{148}}}{{\mathtt{9}}}} = {\mathtt{16.444\: \!444\: \!444\: \!444\: \!444\: \!4}}$$

Obviously you only want the whole bits so the answer is 16pieces

Here is another young woman that Rosala probably knows all about.

Rosala, our Web2 ghost is right.  How can we do justice to something so powerful?

What you have shared with us is truly moving.  I am so glad that you chose to share this .  

You are already are a diplomat.  On this forum you are dealing with people from all over the world and it would have been impossible for you to have made a better impression.  You represent your country, in particular you represent the girls of your country.  There could be no better representative.  Thank you so much for being amoung us.

Melody.

Tuesday 24/6/14

This one is special.  Thank you Rosala.

if (x+2) is a factor of f(x) =x^3+3x^2-kx+4

then

(x+2)(something)=x^3+3x^2-kx+4

consider if x=-2  then

(-2+2)(something)=x^3+3x^2-kx+4

But -2+2=0 and if you multiply anything by zero then the answer is zero, that is;

(-2+2)(something)=x^3+3x^2-kx+4=0

So what this is saying is that  (x+2) is a factor IF x=-2 is a root, that is, f(-2) =0

so 

$$f(x)=x^3+3x^2-kx+4\\\\
f(-2)=(-2)^3+3(-2)^2-k(-2)+4\\\\
f(-2)=-8+12+2k+4\\\\
f(-2)=8+2k\\\\$$

Now if (x+2) is a factor then f(-2)=0 so

$$8+2k=0\\

2k=\:-8\\

k=\:-4\\$$

BY THE WAY this is called remainder theorum

If (x+a) is a factor of f(x) then f(-a)=0

I've just explained why.  If you have a good maths brain this will help a lot.

I am sure Heureka ( or someone else) will come along with a beautifully laid out explanation but in the mean time 

$$(x+1)(x-3)=x^2-2x-3$$

so you need to divide your polynomial by that to get the product of the 2 brackets containing a,b,and c.

You can so this with a polynomial long division.  There are lots of you tube videos explaining thiase techniques.  I can find one if you need me to.   (I've included one below)

after that you just factorise what is left by ordinary techniques to get the value of a,b and c.

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Long division

I am sure that Andre will love them Rosala.

Hi Ninja,

When I wrote my last letter to Andre I specifically passed on your thank you to him.

and yes, there are a lot of members, of course, most of them are not active.

Melody.

Maybe I did.  That is  a pretty Cool Green Lantern Ninja!