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Here is my question:

Find the coefficient of $x^3$ in the expansion of $(1 + 2x - 3x^2)^5.$

I know that I am suppose to use the multinomial theorem, but I don't know how to. The proof that's in my textbook approaches it using partial derivatives, as opposed to combinations and I was expecting an example of how to find the coefficient of some expression, but there wasn't any. I've been reading a couple other threads about how to find coefficients of terms and I came across this link and this link as well. After reading through everything, here is my approach:

$$(1)^j(2x)^k(-3x^2)^p,$$ where $(j + k + p )= 5$. Since I want to look for the coefficient of the $x^3$ term, do I have to find combinations such that $(j + k + p) = 5?$ So for example, let $j = 1, k = 1$ and $p = 1$. But this doesn't add up to $5$... and this is where I got stuck Perhaps there is a different approach. By the way, wolfram gave me a coefficient of $-40$.

I don't know why this problem in particular is hard for me. I can do a similar problem of the same type.

EDIT So as the answer below suggests, this problem can be solved using the binomial theorem, but how would it look like if we used the multinomial theorem?

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    Well, first, you think about HOW you can form a x^3. – 2017-02-25
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    You can achieve this with an x and an x^2, or 3 x's. – 2017-02-25
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    Rewrite your expression as (p+q+r)^5 and compute two cases, then multiply by appropriate poewrs (the coefficients of x and x^2 aren't 1 – 2017-02-25
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    What two cases? – 2017-02-25
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    `Since I want to look for the coefficient of the x^3 term, do I have to find combinations such that (j+k+p)=5?` You are looking for the $3^{rd}$ power term. Why `=5`? – 2017-02-26
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    Ah so $j + k + p = 3$. But that still doesn't clear up my question. Am I suppose to figure out how many ways to select power such that they add up to $3$? – 2017-02-26
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    @i8Σπ_821 Indeed. In fact, that's been pointed out in previous comments already. – 2017-02-26
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    Ah I see. So there should be 10 ways of selecting variables such that they add up to 3. – 2017-02-26

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Ok but you could just use the Binomial Theorem instead

$$(1+x(2-3x))^5=1+5x(2-3x)+10x^2(2-3x)^2+10x^3(2-3x)^3+...$$

So you don't have to look very hard to find the coefficient of $x^3$ which is $$-120+80$$