Factorisation (1 Viewer)

Blitz_N7 · Oct 13, 2015

Is there a way to factorise:
m^3 -3m^2 +2m +3m^2n-3mn+3mn^2-3mn+n^3-3n^2+2n into:
(m+n)(m+n-1)(m+n-2)? Thx in advance.

braintic · Oct 13, 2015

Blitz_N7 said:
Is there a way to factorise:
m^3 -3m^2 +2m +3m^2n-3mn+3mn^2-3mn+n^3-3n^2+2n into:
(m-n)(m-n-1)(m-n-2)? Thx in advance.

I am going to guess that your first line is an intermediate stage in your working for a different question.
If so, could you post the actual question - there might be an easier way to do it.

Ekman · Oct 13, 2015

Blitz_N7 said:
Is there a way to factorise:
m^3 -3m^2 +2m +3m^2n-3mn+3mn^2-3mn+n^3-3n^2+2n into:
(m-n)(m-n-1)(m-n-2)? Thx in advance.

I got (m+n)(m+n-1)(m+n-2), is there something wrong with the question or am I simply wrong?

Edit: After doing some expansion, I realised that multiplying -n three times will not give positive n^3, so there must be something wrong with the answer

Blitz_N7 · Oct 13, 2015

The actual question is proving: (m+n)C3=mC3+mC2nC1+mC1nC2+nC3. Combinations :/

Blitz_N7 · Oct 13, 2015

It is actually (m+n)(m+n-1)(m+n-2). TY

InteGrand · Oct 13, 2015

Blitz_N7 said:
The actual question is proving: (m+n)C3=mC3+mC2nC1+mC1nC2+nC3. Combinations :/

Much more easily done combinatorially than algebraically.

rand_althor · Oct 13, 2015

It is easier if you consider the coefficient of $x^3$ in $(1+x)^{m+n}$ and $(1+x)^m(1+x)^n$ . You know that the coefficients of $x^3$ in each of these expansions will be equal as $(1+x)^m(1+x)^n=(1+x)^{m+n}$ .

InteGrand · Oct 13, 2015

Ekman · Oct 13, 2015

If you still want to know how to do it algebraically:

$m^3+n^3 - 3(m^2+2mn+n^2) + 2(m+n) +3mn(m+n) = (m+n)(m^2-mn+n^2 -3m -3n +2 +3mn)$

$(m+n)((m+n)^2 -3(m+n) + 2 )= (m+n)(m+n-1)(m+n-2)$

Blitz_N7 · Oct 13, 2015

InteGrand said:
$(This question only makes sense for $m,n\geq 3$.) The combinatorial method is to consider picking 3 objects from a total of $m+n$ distinct objects. The number of ways to do this is, on the one hand, $\binom{m+n}{3}$. On the other hand, we can count the number of ways as follows: split the group up in to the first $m$ objects and the last $n$ objects. We can choose 3 by picking $3$ from the first $m$ (done in $\binom{m}{3}$ ways), OR picking 2 from the first $m$ and 1 from the last $n$ (done in $\binom{m}{2}\times \binom{n}{1}$ ways), OR picking 1 from the first $m$ and 2 from the last $n$ (done in $\binom{m}{1}\times \binom{n}{2}$ ways) OR picking $3$ from the last $n$ (done in $\binom{n}{3}$ ways), so the total no. of ways to pick is also $\binom{m}{3} + \binom{m}{2}\times \binom{n}{1}+\binom{m}{1}\times \binom{n}{2} + \binom{n}{3}$. Hence $\binom{m+n}{3}=\binom{m}{3} + \binom{m}{2}\times \binom{n}{1}+\binom{m}{1}\times \binom{n}{2} + \binom{n}{3}$.$

Thank you!

Blitz_N7 · Oct 13, 2015

Ekman said:
If you still want to know how to do it algebraically:

$m^3+n^3 - 3(m^2+2mn+n^2) + 2(m+n) +3mn(m+n) = (m+n)(m^2-mn+n^2 -3m -3n +2 +3mn)$

$(m+n)((m+n)^2 -3(m+n) + 2 )= (m+n)(m+n-1)(m+n-2)$

Thank you!

YouCantMakeMe · Oct 13, 2015

InteGrand said:
$(This question only makes sense for $m,n\geq 3$.) The combinatorial method is to consider picking 3 objects from a total of $m+n$ distinct objects. The number of ways to do this is, on the one hand, $\binom{m+n}{3}$. On the other hand, we can count the number of ways as follows: split the group up in to the first $m$ objects and the last $n$ objects. We can choose 3 by picking $3$ from the first $m$ (done in $\binom{m}{3}$ ways), OR picking 2 from the first $m$ and 1 from the last $n$ (done in $\binom{m}{2}\times \binom{n}{1}$ ways), OR picking 1 from the first $m$ and 2 from the last $n$ (done in $\binom{m}{1}\times \binom{n}{2}$ ways) OR picking $3$ from the last $n$ (done in $\binom{n}{3}$ ways), so the total no. of ways to pick is also $\binom{m}{3} + \binom{m}{2}\times \binom{n}{1}+\binom{m}{1}\times \binom{n}{2} + \binom{n}{3}$. Hence $\binom{m+n}{3}=\binom{m}{3} + \binom{m}{2}\times \binom{n}{1}+\binom{m}{1}\times \binom{n}{2} + \binom{n}{3}$.$

How would you articulate this in an exam? Nice math skills

InteGrand · Oct 13, 2015

YouCantMakeMe said:
How would you articulate this in an exam? Nice math skills

Probably just write it similarly to that (though you probably wouldn't need full sentences), and maybe draw a diagram if it helps, like:

$\underbrace{***\cdots *}_{\text{first }m\text{ objects}}|\underbrace{**\cdots *}_{\text{last }n\text{ objects}}$

where each * represents a distinct object.

YouCantMakeMe · Oct 13, 2015

InteGrand said:
Probably just write it similarly to that (though you probably wouldn't need full sentences), and maybe draw a diagram if it helps, like:

$\underbrace{***\cdots *}_{\text{first }m\text{ objects}}|\underbrace{**\cdots *}_{\text{last }n\text{ objects}}$

where each * represents a distinct object.

And nice LaTex skills haha.

Factorisation (1 Viewer)

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